Fill in the Blanks

• If a body changes its position with respect to its surroundings, the body is said to be in …………..

If a body changes its position with respect to its surroundings, the body is said to be in motion.

Motion is a relative concept.

• The distance can never be ………., while displacement can be.

The distance can never be negative, while displacement can be.

Distance is always positive while displacement can be positive or negative or zero.

• If a body moves such that it covers equal distances in equal intervals of time, whatsoever be small, then the object is said to have …………. motion.

If a body moves such that it covers equal distances in equal intervals of time, whatsoever be small, then the object is said to have uniform motion.

• When a particle moves in a straight line from point A to point B the distance covered is ………. the magnitude of the displacement.

When a particle moves in a straight line from point A to point B the distance covered is equal to the magnitude of the displacement.

• Distance travelled divided by elapsed time gives ………….

Distance travelled divided by elapsed time gives speed.

• The ratio of the total displacement of a body to the total time taken is …………

The ratio of the total displacement of a body to the total time taken is velocity.

The average velocity of an object moving on a circle of radius R in one complete rotation if it takes t second in completing one rotation is …………..

The average velocity of an object moving on a circle of radius R in one complete rotation if it takes t second in completing one rotation is zero.

In one complete rotation displacement of the object is zero

Velocity = Displacement/time = 0/t = 0

• An object moving uniformly on a circular track. Its velocity ………. with time.

An object moving uniformly on a circular track. Its velocity changes with time.

In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity changes continuously. Hence body in circular motion moves with acceleration.

• If particle is moving along circular path such that in equal interval of time it describes the equal angle, then velocity vector …………

If particle is moving along circular path such that in equal interval of time it describes the equal angle, then velocity vector changes its direction continuously.

The particle describes the equal angle in equal interval of time. Hence particle is in uniform circular motion. In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity changes continuously.

• The magnitude of average velocity ………… equal to the average speed.

The magnitude of average velocity may or may not be equal to the average speed.

• When a body has unequal displacement in equal intervals of time, it is said to be moving with …………..

When a body has unequal displacement in equal intervals of time, it is said to be moving with non-uniform velocity.

• The acceleration of freely falling object is approximately equal to ……..

The acceleration of freely falling object is approximately equal to 9.8 ms^-2.

• If a body starts from rest and moves with uniform acceleration, then its displacement is proportional to ………….

If a body starts from rest and moves with uniform acceleration, then its displacement is proportional to square of time.

s = ut + 1/2 at²

now u = 0

s = 1/2 at²

• A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to …….

A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to ………

A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to time.

v = u + at

now u = 0

v = at

• All objects in free fall at a given place have the same

All objects in free fall at a given place have the same acceleration.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

• …………… acceleration is called retardation or deceleration.

Negative acceleration is called retardation or deceleration.

• A car is moving northward. The driver of the car applies brakes, the direction of acceleration is …………..

A car is moving northward. The driver of the car applies brakes, the direction of acceleration is southward.

When brakes are applied, the acceleration is negative and acts in the opposite direction to tht of motion at the instant of braking.

• A body moving with constant speed has zero acceleration only when the particle is in …… dimensional motion.

A body moving with constant speed has zero acceleration only when the particle is in one dimensional motion.

• In velocity-time graph, velocity is taken on ……… axis.

In velocity-time graph, velocity is taken on x-axis

• Area under velocity-time graph gives …………..

Area under velocity-time graph gives the displacement.

• The velocity-time graph of a moving object is a straight line parallel to the time axis. It means the velocity of the object is ………..

The velocity-time graph of a moving object is a straight line parallel to the time axis. It means the velocity of the object is constant.

• The velocity-time graph of a body moving with uniform velocity is a straight line parallel to …………..

The velocity-time graph of a body moving with uniform velocity is a straight line parallel to time axis.

The post Motion in a Straight Line Fill in the Blanks Questions appeared first on The Fact Factor.

In last few articles, we have seen the concept of a motion in a straight line. From this article we shall apply the concept. In this article we shall study true and false type questions based on a motion in a straight line.

State whether the following statements are true or false. If false correct the statement.

• The travel of a train from one station to another is an example of translatory motion.

True

In translational motion every particle of the body has the same displacement.

• Motion of an ant along one of the edge of table is translatory motion.

True

In translational motion every particle of the body has the same displacement.

• The magnitude of displacement can be equal to or lesser than the distance travelled.

True

The shortest distance from the initial position to the final position of the body is called the magnitude of the displacement. Thus in case of body moving in a straight line in the same direction, the maximum displacement can be equal to the distance travelled. In all other case, it will be less than the distance travelled.

• Ddistance covered by a moving body is always greater than zero.

True.

• Displacement of a particle can be less than or greater than or equal to zero.

True

• Uniform speed is a vector quantity

False

Correction: Speed is a scalar quantity

Speed = distance / time, in this formula, both the dstance travelled and time are scalar quantities, hence speed is a scalar quantity.

• A particle moving with a uniform velocity must be along a straight line.

True.

Since velocity is a vector quantity it has both the magnitude and direction and if direction changes it is not uniform velocity. In case of circular motion there is a continuous change in direction leading to accelarated motion which results in non uniform velocity.

• A body can have a constant speed and still have varying velocity.

True. 

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. A body can not have its velocity constant, while its speed varies.

• The magnitude of average velocity is always equal to the average speed.

False.

Correction: The magnitude of average velocity ned not be equal to the average speed.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. The magnitude of the average velocity of an object is equal to its average speed, only in one condition when an object is moving in a straight line.

• Average velocity can be calculated by taking the average of initial and final velocities for a given time interval irrespective of the type of acceleration.

True

• For a body moving along a circular path, the average velocity and average speed can never be equal.

True.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. Thus in case of uniform ciercular motion, average speed is constant and equal to the magnitude of the instantaneous velocity of the body, but average velocity is zero.

• Average velocity can be zero, but average speed of a moving body can not be zero in any finite time interval.

True

uniform ciercular motion, average speed is constant and equal to the magnitude of the instantaneous velocity of the body, but average velocity is zero.

• If a body moves with constant velocity, its displacement depends on depends on square of time

False

Correction: If a body moves with constant velocity, its displacement depends on depends on time

v = ds/dt = k = constant

Integrating both sides with time t

s = kt

Thus displacement varies directly with time (t)

• A particle speed is constant, acceleration of the particle must be zero.

False

Correction: A particle speed is constant, acceleration of the particle need not be zero.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. Thus particle possesses acceleration.

• When a particle moves with a constant speed in the same direction, neither the magnitude nor the direction of velocity changes and so acceleration is zero.

True

Since velocity is a vector quantity it has both the magnitude and direction. In this case both the speed and direction are the same. Hence the particle is moving with a constant velocity and has zero acceleration.

• A particle is known to be at rest at time t = 0. If its acceleration at t = 0 is zero.

False

Correction: A particle is known to be at rest at time t = 0. If its velocity at t = 0 is zero.

A body is said to be at rest if it does not change its position with respect to its immediate surroundings. Thus the velocity of the body decides its state of motion.

• An object covers distances in direct proportion to the square of the time elapsed. Its acceleration is increasing.

False

An object covers distances in direct proportion to the square of the time elapsed. Its acceleration is constant

s = kt² (given)

Differentiating both sides w.r.t. time t

velocity = v = ds/dt = 2kt

Differentiating both sides again w.r.t. time t

acceleration = a = dv/dt = 2k = constant

Thus in this case acceleration is constant. i.e. the object is moving with constant acceleration.

• A particle in one-dimensional motion with a positive value of acceleration must be speeding up.

False

A particle in one-dimensional motion with a positive value of acceleration may or may not be speeding up.

If the initial velocity of a body is negative then even in case of positive acceleration, the body speeds down. 

• There can be a motion in which speed is constant but velocity is variable.

True

In uniform circular motion, speed is constant but velocity is variable.

• A body moves with retardation when it is projected vertically upward.

True

Every body on the earth surface is acted upon by gravitational force acting in downward direction. When a body is projected vertically upward, due to the action of the gravitational force, its velocity goes on decreasing. Thus the body moves with retardation. at the highest point of its journey its velocity is zero.

• A body is projected vertically up. On reaching maximum height, its velocity becomes zero.

True

Every body on the earth surface is acted upon by gravitational force acting in downward direction. When a body is projected vertically upward, due to the action of the gravitational force, its velocity goes on decreasing. Thus the body moves with retardation. at the highest point of its journey its velocity is zero.

• A stone dropped from a height moves with constant velocity

False

Correction: A stone dropped from a height moves with constant acceleration.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

• When two balls of different masses are thrown vertically upwards with the same initial speed, the heavier body rises to greater height then the lighter body.

False

Correction: When two balls of different masses are thrown vertically upwards with the same initial speed, both the bodies will rise to the same maximum height.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. The height reached by the body epends on the initial speed by which they are thrown ipwards and the acceleration due to gravity at that place. Both the bodies will be acted upon by the same acceleration due to gravity and they are projected with same initial speed. hence both the bodies will rise to the same height.

• The distance travelled by a freely falling body in every successive second is the same.

False:

Correction: The distance travelled by a freely falling body in every successive second increases.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. Thus freely falling body moves with acceleration due to gravity in downward direction i.e. in the direction of motion, hence its speed increases continuously and thus it covers more distance in every successice second.

• The area under the velocity-time diagram shows the displacement of the body

True

• Velocity-time graph cannot be used to find the instantaneous velocity

False

Correction: Velocity-time graph can be used to find the instantaneous velocity

• Velocity-time graph can be used to find displacement of the body.

True

The area under the velocity-time diagram shows the displacement of the body

• Equations of motion are applicable only when a body moves with uniform velocity.

False

Correction: Equations of motion are applicable only when a body moves with uniform acceleration.

• Direction of motion is decided by the displacement of a body.

False

Direction of motion is decided by the velocity of a body. Positive value of velocity indicates body is moving in the direction of the displacement while neghative value of velocity indicates the body is moving in the opposite direction to that of displacement.

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The post Motion in a Straight Line True and False Questions appeared first on The Fact Factor.

In the last article we have studied the concept of speed and velocity. Motion is an important part of our life. Our daily activities involve motion of different kinds. When we study motion, we come across another important concept namely acceleration. In this article, we shall study the the concept of uniform acceleration.

Notes
Very Short Answer Type Questions
Short Answer Type Questions
Concept Application

Velocity:

The rate of change of displacement of a body with respect to time is called the velocity of the body.

Velocity = Displacement / Time

Velocity is a vector quantity, its S.I. unit is m/s and c.g.s. unit is cm/s. Its dimensions are [L¹M⁰T^-1].

Uniform Velocity: 

When the magnitude and direction of the velocity of a body remain the same at any instant, then the body is said to have uniform velocity. For uniform motion acceleration a = 0 and Displacement = velocity × time.

Example: The velocity of light in a particular medium is uniform velocity.  The velocity of sound in air at constant temperature is uniform velocity.

Non Uniform Velocity: 

When the magnitude of velocity or the direction of velocity or both changes at any instant the body is said to have the nonuniform velocity or variable velocity.

A body can have non-uniform velocity in the following three cases.

• When the direction of the velocity of a body remains the same but its magnitude changes continuously then the body has variable velocity. e.g. a ball is thrown vertically upward.
• When the magnitude of the velocity of a body remains the same but the direction changes continuously then the body has variable velocity. e.g. uniform circular motion of a body.
• When both the magnitude and direction of the velocity of body change continuously, then the body has variable velocity. e.g. ball thrown by making the acute angle with the horizontal (projectile motion)

When a body has variable velocity, then it has acceleration.

Acceleration:

The rate of change of velocity with respect to time is called acceleration.

Acceleration is vector quantity its S.I. unit is m/s². Its dimensions are [L¹M⁰T^-2].

Acceleration = (v – u)/t

Where, u = Initial velocity

v = Final velocity

t = Time in which the change takes place

Acceleration can be positive, negative or zero. If the velocity is increasing then acceleration is positive. If the velocity is decreasing acceleration is negative. If the velocity is the constant acceleration is zero. Negative acceleration is also called deceleration or retardation.

If the velocity is increasing then the direction of acceleration is same as that of the velocity of the body. If the velocity is decreasing then the direction of acceleration is opposite to that of the velocity of the body.

It is to be noted that the velocity and not the acceleration of the body determines the direction of motion.

Uniform Acceleration:

When equal changes take place in the velocity of a body in equal interval of time, then the acceleration is called uniform acceleration. e.g. the motion under gravity.

Variable Acceleration:

When The change in the velocity of a body in equal interval of time is not constant, then the acceleration is called non-uniform acceleration. Example: the motion of a vehicle on crowded road.

Acceleration Due To Gravity:

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

It is denoted by ‘g’. It varies from place to place. The average value of g at sea level is taken as 9.8 ms^-2 in S.I. system and 980 cms^-2 in c.g.s. system. When solving problems on the motion under gravity as per the convention the value of ‘g’ should be negative.

Relation Between Velocity and Acceleration:

• When both velocity and acceleration are positive, acceleration is in the direction of velocity and velocity increases.
• When velocity is positive and acceleration is negative, the acceleration is in the opposite direction of the velocity and velocity decreases.
• When velocity is negative and acceleration is positive, the acceleration is in opposite direction of velocity and velocity increases but body is moving in opposite direction.
• When both velocity and acceleration are negative, acceleration is in the direction of velocity and velocity decreases but body is moving in opposite direction.

Concepts:

Very Short Answer Type Questions

Q1. The average value of acceleration due to gravity at sea level is …… ms^-2

The average value of acceleration due to gravity at sea level is 9.8 ms^-2

Q2. Define acceleration.

The rate of change of velocity with respect to time is called acceleration

Q3. Give c.g.s., m.k.s. and S.I. units of acceleration.

Q4. Retardation is ……….. quantity

Retardation is a vector quantity

Q5. ……. acceleration is called retardation or deceleration.

Negative acceleration is called retardation or deceleration.

Q6. A freely falling body falls with …………..

A freely falling body falls with uniform acceleration, called acceleration due to gravity.

Q7. When is a body said to have zero acceleration?

If a body is at rest or moving with uniform velocity, then the body is said to have zero acceleration.

Q8. What is the acceleration of a body when its velocity remains constant?

In such case the acceleration of the body is zero.

Q9. What is acceleration due to gravity?

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

Q10. Rate of decrease in the magnitude of velocity is called ……..

Rate of decrease in the magnitude of velocity is called retadation or deceleration.

Q11. Give one example of each type of following motions

Uniform acceleration: free fall of a body under influence of gravity

Variable acceleration: A motion of vehicle on crowded road

Q12. What is retardation?

Negative acceleration is called retardation or deceleration.

Q13. Is acceleration due to gravity constant everywhere?

Acceleration due to gravity is not constant everywhere. It depends on altitude, depth, shape of earth, and latitude of the place. It is maximum at the poles and minimum at the equator.

Short Answer Type Questions

Q1. Explain the term ‘acceleration due to gravity’.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

It is denoted by ‘g’. It varies from place to place. The average value of g at sea level is taken as 9.8 ms^-2 in S.I. system and 980 cms^-2 in c.g.s. system. When solving problems on the motion under gravity as per the convention the value of ‘g’ should be negative.

Q2. Which of the following bodies does hit the ground first when realeased from the same height simultaneously: a body of mass 1 kg or a body of mass 10 kg.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. The value of g does not depend on mass of the body. If the two bodies of different masses are dropped from the same height simultaneusly, both will reach ground simultaneously, if the effect of air (friction and buoyancy) is neglected.

Q3. Distinguish between acceleration and retardation.

Q4. Distinguish between uniform acceleration and variable acceleration.

Concept Application:

Q1. A train moving with a speed 90 kmph is brought to rest in 10 s. Find its retardation.

u = Initial velocity = 90 kmph = 90 x (5/18) = 25 ms^-1

v = Final velocity = 0

Time in which change is brought = 10 s

We have acceleration = a = (v – u)/t

a = (0 – 25)/10 = – 2.5 ms^-2

Negative sign indicates retardation.

Q2. A car initially at rest attains velocity of 20 ms^-1 with uniform acceleration in 2.5 s. What is its acceleration?

u = Initial velocity = 0

v = Final velocity = 20 ms^-1

Time in which change is brought = 2.5 s

We have acceleration = a = (v – u)/t

a = (20 – 0)/2.5 = 8 ms^-2

Q3. The velocity of an object increases at a constant rate 20 m s^-1 to 50 m s^-1 in 10s. Find the acceleration.

Given: Initial velocity = u = 20 m s^-1, Final velocity = v = 50 m s^-1, Time elapsed = t = 10 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (50 – 20)/10 = 30/10 = 3 m s^-2

Acceleration of the object is 3 m s^-2

Q4. A stone is thrown vertically upwards with an initial velocity 50 m s^-1 comes to halt in 5 s. Find the acceleration.

Given: Initial velocity = u = 50 m s^-1, Final velocity = v = 0 m s^-1, Time elapsed = t = 5 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (0 – 50)/5 = – 50/5 = – 10 m s^-2

The negative sign indicates retardation

Retardation of the stone is 10 m s^-2

Q5. The velocity of an object increases at a constant rate 54 km h^-1 to 72 km h^-1 in 5 s. Find the acceleration.

Given: Initial velocity = u = 54 km h^-1= 54 x (5/18) = 15 m s^-1, Final velocity = v = 72 km h^-1= 72 x (5/18) = 20 m s^-1, Time elapsed = t = 5 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (20 – 15)/5 = 5/5 = 1 m s^-2

Acceleration of the object is 1 m s^-2

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In this article, we shall study to solve problems based on Newton’s equations of motion.

First Equation of Motion:

Let u =  initial velocity of a body, v =   final velocity of the body

t = time in which the change in velocity takes place.

By the definition of acceleration

Considering magnitudes only

v = u + at

This equation is known as Newton’s First equation of motion.

Second Equation of Motion:

Let u =  initial velocity of a body, v =   final velocity of the body

t = time in which the change in velocity takes place, a = acceleration of the body

This equation is known as Newton’s second equation of motion.

Third Equation of Motion:

Let u =  initial velocity of a body, v =   final velocity of the body

t = time in which the change in velocity takes place.

from equation (1) and (2)

Considering the magnitude only

v² = u² = 2 a s

This equation is known as Newton’s third equation of motion.

Expression for the Distance Travelled by Body in nth Second of its Motion:

By Newton’s Second equation of motion, s = ut + ½  at²

where s = displacement of body in ‘t’ seconds

u = initial velocity of the body, a = acceleration of the body, t = time

The distance travelled by body in ‘n’ seconds is given by

This distance by travelled by the body in (n-1) seconds is given by

∴ The distance travelled by the body in n th second

Motion Under Gravity:

A special case of uniform acceleration is the motion of a body under gravity. It is found that close to the surface of the earth, and in the absence of air resistance, all the bodies fall to the earth, at a given place, with the constant acceleration.  This constant acceleration is called acceleration due to gravity or gravitational acceleration.

It is denoted by “g”. It is always directed downward.  Its magnitude is approximately 9.8 m/s². For the motion of a body under gravity Newton’s equations of motion can be written as

Sign Convention (Cartesian):

• All vectors directed towards the right of the reference point are considered positive.
• All vectors directed towards the left of the reference point are considered negative.
• All vectors directed vertically upward the reference point are considered positive.
• All vectors directed vertically downward the reference point are considered negative.
• By this sign convention acceleration due to gravity “g” is always negative.

Numerical Problems:

Example – 01:

A car acquires a velocity of 72 kmph in 10 s starting from rest. Calculate its average velocity, acceleration and distance travelled during this period.

Given: Initial velocity = u = 0, Final velocity = v = 72 kmph = 72 x 5/18 = 20 m/s, Time taken = t = 10 s

To Find: average velocity = v_av =?, acceleration = a =?, distance travelled = s =?

Solution:

v_av = (u + v) /2 = (0 = 20)/2 = 10 m/s

Average velocity is 10 m/s

By first equation of motion we have

v = u + at

∴    20 = 0 + a x 10

∴    a = 20/10 = 2 m/s²

acceleration = 2 m/s²

By second equation of motion

s = ut + ½ at² = 0 x (10) + ½ x 2 x (10)² = 0 + 100 = 100 m

distance travelled = 100 m

Ans: Average velocity = 10 m/s, acceleration = 2 m/s² , distance travelled = 100 m

Example – 02:

A ball is moving with a velocity of 0.5 m/s. Its velocity is decreasing at the rate of 0.05 m/s². What is its velocity after 5 s? How much time will it take from start to stop? What is the distance travelled by it before stopping?

Given: Initial velocity = u = 0.5 m/s, acceleration = a = – 0.05 m/s²

To Find:  a) v = ? when t = 5 s b) t = ? when v = 0, c) distance travelled = s = ?

Solution:

By first equation of motion  (v = ? when t = 5)

v = u + at  = 0.5 + (- 0.05) x 5  = 0.5 – 0.25 = 0.25 m/s

Thus the velocity of ball after 5 s is 0.25 m/s

By First equation of motion (t = ? when v = 0)

v = u + at

∴    0 = 0.5 + (- 0.05) x t

∴    0.5  = – 0.05 x t

∴    t = 0.5/0.05 = 10 s

The ball will stop after 10 s from start

By the second equation of motion

s = ut + ½ at² = 0.5 x (10) + ½ x (-0.05) x (10)² = 5 – 2.5 = 2.5 m

The ball travels 2.5 m before coming to rest.

Ans: Velocity after 5 seconds = 0.25 m/s, time taken to come to rest = 10 s, distance travelled before coming to rest = 2.5 m

Example – 03:

A car initially at rest starts moving with acceleration 0.5 m/s² covers a distance of 25 m. Calculate the time required to cover this distance and the final velocity of the car.

Given: Initial velocity = u = 0 m/s, acceleration = a = 0.5 m/s², distance travelled = s = 25m.

To Find:  time taken = t =? , final velocity = v = ?

Solution:

By the second equation of motion

s = ut + ½ at²

∴     25 = 0 x (t) + ½ x (0.5) x t²

∴    25 = 0.25 t²

∴    t² = 25/ 0.25 = 100

∴    t = 10 s

Time required by car to cover distance of 25 m is 10 s

By first equation of motion

v = u + at  = 0 + ( 0.5) x 10  = 5 m/s

The final velocity of ball after is 5 m/s.

Ans: Time required to cover the distance = 10s, final velocity = 5 m/s

Example – 04:

A body starts from rest with a uniform acceleration of 2 m/s². Calculate the distance travelled by the body in 2 s.

Given: Initial velocity = u = 0 m/s, acceleration = a = 2 m/s², time taken = t = 2 s.

To Find:  Distance travelled = s =?

Solution:

By the second equation of motion

s = ut + ½ at² = 0 x (2) + ½ x (2) x 2²  = 4 m

Ans: The distance travelled by the body is 4 m

Example – 05:

A body is moving with 5m/s is accelerated at 5 m/s². Calculate the distance travelled by the body in 5 s.

Given: Initial velocity = u = 10 m/s, acceleration = a = 5 m/s², time taken = t = 5 s.

To Find:  Distance travelled = s =?

Solution:

By the second equation of motion

s = ut + ½ at² = 10 x (5) + ½ x (5) x 5²  = 50 + 62.5= 112.5 m

Ans: The distance travelled by the body is 4 m

Example – 06:

A vehicle is moving at a velocity of 30 kmph at some instant. After 2 s its velocity is found to be 33.6 kmph and after further 2 s the velocity is found to be 37.2 kmph. find the acceleration of the vehicle and comment on the result.

Solution:

Consider the change of velocity from 30 kmph to 33.6 kmph in 2 s.

u = 30 kmph = 30 x 5/18 = 150/18 m/s, v = 33.6 kmph = 33.6 x 5/18 = 168/18 m/s, t = 2 s

By first equation of motion

v = u + at

168/18 = 150/18 + a(2)

168/18 – 150/18 = 2a

2a = 18/18 = 1

a = 1/2 = 0.5 m/s²

Consider the change of velocity from 33.6 kmph to 37.2 kmph in 2 s.

u = 33.6 x 5/18 = 168/18 m/s, v = 37.2 kmph = 37.2 x 5/18 = 186/18 m/s, , t = 2 s

By first equation of motion

v = u + at

186/18 =  168/18 + a(2)

186/18 – 168/18 = 2a

2a = 18/18 = 1

a = 1/2 = 0.5 m/s²

Ans: The acceleration is same in both the cases thus the vehicle is moving with a constant acceleration of 0.5 m/s².

Example – 07:

A body initially at rest travels a distance of 100 m in 5 s with constant acceleration. calculate the acceleration and the final velocity at the end of 5 s.

Given: Initial velocity = u = 0 m/s, time taken = t = 5 s, distance travelled = s = 100 m

To Find:  Acceleration = a =?, final velocity = v =?

Solution:

By the second equation of motion

s = ut + ½ at²

100 = 0 x (5) + ½ x (a) x 5² 

200 = 25 a

a = 200/25 = 8 m/s²

Acceleration = 8 m/s²

By first equation of motion

v = u + at

v = 0 + 8 x 5 = 40 m/s

Ans: Acceleration = 8 m/s² and Final velocity = 40 m/s

Example – 08:

A body initially at moving with a speed of 18 kmph is accelerated uniformly at the rate of 9 cm/s² covers a distance of 200 m. Calculate the final velocity.

Given: Initial velocity = u = 18 kmph = 18 x 5/18 = 5 m/s, distance travelled = s = 100 m, acceleration = 9 cm/s² = 0.09 m/s².

To Find:  final velocity = v =?

Solution:

By the third equation of motion

v²  = u²  + 2as

v²  = 5²  + 2 x 0.09 x 200 = 25 + 36

v²  = 61

v = 7.81 m/s

Ans: The final velocity = 7.81 m/s

Example – 09:

A body initially at rest starts accelerating at the rate of 2 m/s². Find the final velocity and distance travelled by the body after 5s.

Given: Initial velocity = u = 0 m/s, acceleartion = a = 2 m/s², time taken = t = 5 s

To Find:  distance travelled = ?, final velocity = v =?

Solution:

By the second equation of motion

s = ut + ½ at² = 0 x 5 + ½  x 2 x 5² =  25 m

Distance travelled = 25 m

By first equation of motion

v = u + at

v = 0 + 2 x 5 = 10 m/s

Final velocity = 10 m/s

Ans: The final velocity = 10 m/s and distance travelled = 25 m

Example – 10:

A train moving with a velocity 72 kmph is brought to rest by applying brakes in 5 s. Calculate the retardation and distance travelled during this period.

Given: Initial velocity = u = 72 kmph = 72 x 5/18 = 20 m/s, final velocity = v = 0 m/s, time taken = t = 5 s

To Find:  distance travelled = ?, retardation = a =?

Solution:

By first equation of motion

v = u + at

0 = 20 + a x 5

5a = -20

a = -20/5 = -4 m/s²

Neagative sign indicates retardation

retardation = 4 m/s²

By the second equation of motion

s = ut + ½ at² = 20 x 5 + ½  x (- 4) x 5² =  100 – 50 = 50 m

Distance travelled = 50 m

Example – 11:

A body moves from rest with uniform acceleration and travels 270 m in 3 s. Find the velocity of the body after 5 s..

Given: Initial velocity = u = 0 m/s, distance travelled  = s = 270 m, time taken = t = 3 s

To Find:  velocity = ? when t = 5 s.

Solution:

By the second equation of motion

s = ut + ½ at²

270 = 0 x 5 + ½  x (a) x 3²

270 = 9/2 a

a = 540/9 = 60 m/s²

By first equation of motion

v = u + at  = 0 + 60 x 5 = 300 m/s

Ans: The velocity after 5s is 300 m/s

Example – 12:

A body moving with constant acceleration travels the distances 3 m and  8 m in 1 s and 2 s respectively. Calculate the initial velocity and acceleration of the body

Given:  Case – 1:  distance travelled  = s₁ = 3 m, time taken = t₁ = 1 s, Case – 2: distance travelled  = s₂ = 8 m, time taken = t₂ = 2 s,

To Find:  initial velocity = u = ? acceleration = a =?

Solution:

By the second equation of motion

s = ut + ½ at²

For first case

3 = u x 1 + ½  x (a) x 1²

3 = u + ½  x (a)

2u + a = 6 ……. (1)

For second case

8 = u x 2 + ½  x (a) x 2²

8  = 2u + 2 a

u + a = 4 ……. (2)

Solving equations (1) and (2) simultaneously

a = 2  and u = 2

Ans: Initial velocity = 2 m/s and acceleration = 2 m/s².

Example – 13:

A body moving with constant acceleration travels the distances 84 m and  264 m in 6 s and 11 s respectively. Calculate the initial velocity and acceleration of the body

Given:  Case – 1:  distance travelled  = s₁ = 84 m, time taken = t₁ = 6 s, Case – 2: distance travelled  = s₂ = 264 m, time taken = t₂ = 11 s,

To Find:  initial velocity = u = ? acceleration = a =?

Solution:

By the second equation of motion

s = ut + ½ at²

For first case

84 = u x 6 + ½  x (a) x 6²

84 = 6u + 18a

u + 3a = 14 ……. (1)

For second case

264 = u x 11 + ½  x (a) x 11²

264 = 11u + ½  x (a) x 121

22u + 121a = 528

2u + 11a = 48 ……. (2)

Solving equations (1) and (2) simultaneously

a = 4  and u = 2

Ans: Initial velocity = 2 m/s and acceleration = 4 m/s².

Example – 14:

A bus is moving with uniform velocity. The driver of the bus sees a pedestrian crossing the road at a distance of 60 m from the bus. He applied the brakes and reduce the speed with retardation of 25 cm/s2 and takes 20 s to stop the bus. Find the initial velocity of the bus and also decide the fate of the pedestrian.

Solution:

Given:  acceleration = – 25 cm/s² = – 0.25 m/s², time taken = t = 20 s, Final velocity = 0 m/s.

To Find:  initial velocity = u = ? Distance travelled = s = ?

By first equation of motion

v = u + at

0 = u + (-0.25)(20)

0 = u – 5

u = 5 m/s

Initial velocity = u = 5 m/s

By the second equation of motion

s = ut + ½ at²

s = 5 x 20 + ½  x (-0.25) x 20² = 100 – 50 = 50 m

As the distance travelled by the bus is less than the distance of pedestrian from the bus, there will be no accident

Example – 15:

A train is moving with a velocity of 90 kmph. When the brakes are applied the acceleration is found to be 0.5 m/s². Find the velocity after 10 s, the time taken to stop and the distance traveled before stopping from the application of brakes.

Given:  acceleration = – 0.5 m/s², time taken = t = 20 s, initial velocity = u = 90 kmph = 90 x 5/18 = 25 m/s.

To Find:  velocity = v = ? when t = 10 s, time taken = ? when v = 0,  Distance travelled = s = ?

Solution:

By first equation of motion

v = u + at  = 25 + (-0.5)(10) = 25 – 5 = 20 m/s

Velocity after 10 s is 20 m/s

By first equation of motion

v = u + at

0 = 25 + (-0.5)(t)

25 = – 0.5 t

t = 25/0.5 = 50 s

The train will take 50 s to stop

Velocity after 10 s is 20 m/s

By the second equation of motion

s = ut + ½ at²

s = 25 x 50 + ½  x (-0.5) x 50² = 1250 – 625 = 625 m

Ans: The train will civer 625 m before coming to rest

Example – 16:

A train is moving with a velocity of 54 kmph. It is accelerated at the rate 5 m/s². Find the distance travelled by the train in 5 seconds and in the  5th second of its journey.

Given:  acceleration = 5 m/s², time  = t = 5 s, initial velocity = u = 54 kmph = 54 x 5/18 = 15 m/s.

To Find:  Distance travelled in 5 s =? and distance travelled in 5th second = ?

Solution:

By the second equation of motion

s = ut + ½ at²

s = 15 x 5 + ½  x (5) x 5² = 75 + 62.5 = 137.5 m

Distance travelled in 5 s is 137.5 m

The distance travelled in nth second is given by

s_n = u +1/2a(2n – 1)

s₅ = 15 +1/2x 5 x (2 x 5 – 1)  = 15 +2.5 x 9

s₅ = 15 + 22.5 = 37.5 m

Ans: The distance travelled in 5 seconds = 137.5 m and the distance travelled in 5th second is 37.5 m

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Science > Physics > Motion in a Straight Line > Newton’s Equations of Motion

The post Newton’s Equations of Motion appeared first on The Fact Factor.

In this article, we shall study to solve problems to calculate displacement, Average Speed, and average Velocity.

Example – 01:

A train travels a distance of 100 m due east in 10 seconds. What is its speed and velocity?

Solution:

Speed = distance /time = 100/10 = 10 m/s

velocity = 10m/s  due east

Ans: The speed is 10 m/s and the velocity is 10m/s  due east

Example – 02:

A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate the speed of the train and the time taken to cover a distance of 240m.

Given: Distance travelled by train = s = 120 m, Time taken = 2 s

To Find: Speed of train = v =? and time t =? when s = 240 m.

Solution:

For uniform motion, speed = distance/time = 120/2 = 60 m/s

Time taken to cover 240 m, time = distance/speed = 240/60 = 4 s

Ans: The speed of the train is 60 m/s and it will take 4 s to cover a distance of 240 m

Example – 03:

A car takes 3 hours to travel from Delhi to Agra with a uniform speed of 65 kmph. Find the distance between the cities.

Given: Speed of car = v = 65 km/h, Time taken = t = 3 hours

To Find: Distance = s = ?.

Solution:

For uniform motion, Distance = speed x time = 65 x 3 = 195 km

Ans: The distance between delhi and Agra is 195 km

Example – 04:

A body travels at a uniform speed of 20 m/s. Find the distance travelled by the body in 10 s.

Given: Speed of body = v = 20 m/s, Time taken = t = 10 s

To Find: Distance = s = ?.

Solution:

For uniform motion, Distance = speed x time = 20 x 10 = 200 m

Ans: The distance travelled by the body in 10 s is 200 m

Example – 05:

A car is travelling with a uniform speed of 72 kmph. Find the distance travelled by it in 20 minutes

Given: Speed of car = v = 72 km/h  = 72x 5/18 = 20 m/s, Time = t = 20 min = 20 x 60 = 1200 s

To Find: Distance = s = ?.

Solution:

For uniform motion, Distance = speed x time = 20 x 1200 = 24000 m = 24 km

Ans: The distance travelled by car is 24 km

Example – 06:

A body rises vertically upward to a height of 100 m, in 5 seconds, then comes back at the same position after another 5 s. Find the distance travelled, displacement, average speed and average velocity of the body.

Given: Upward distance travelled = 100 m, time taken for upward journey = 5s

Solution:

Total distance travelled = distance travelled in upward journey + distance travelled in downward journey

Total distance travelled = 100 m + 100 m = 200 m

As the body is returning back to the same position.

Displacement = minimum distance between initial and final position = 0

Average speed = Total distance/ total time = 200/10 = 20 m/s

As displacement is zero, velocity is also zero

Example – 07:

A car travels at a uniform speed of 30 kmph for 30 minutes and then at a uniform speed of 40 kmph for the next 40 min. Calculate the total distance travelled by car and its average speed.

Given:  Speed for the first part of journey = v₁ = 30 kmph, time for first part of journey = t₁ = 30 min = 30/60 = 1/2 h, Speed for the second part of journey = v₂ = 40 kmph, time for first part of journey = t₂ = 40 min = 40/60 = 2/3 h,

To Find: Total distance travelled =? and average speed =?

Solution:

Total distance travelled = s = s₁ + s₂ = v₁t₁ + v₂t₂ = 30x (1/2) + 40 x (2/3)

Total distance travelled = 125/3 = 41.67 km

Total time taken = t = t₁ + t₂ = 1/2 + 2/3 = 7/6 h

Average speed = Total disrtance travelled / Total time taken

Average speed = (125/3)/(7/6) = 35.71 kmph

Ans: Total distance travelled = 41,67 km and average speed = 35.71 kmph

Example – 08:

A car travels at a uniform speed of 30 kmph for 30 minutes and then at a uniform speed of 60 kmph for next 30 min. Calculate the average speed of the car.

Given:  Speed for the first part of journey = v₁ = 30 kmph, time for first part of journey = t₁ = 30 min = 30/60 = 1/2 h, Speed for the second part of journey = v₂ = 60 kmph, time for first part of journey = t₂ = 30 min = 30/60 = 1/2 h,

To Find: Total distance travelled =? and average speed =?

Solution:

Total distance travelled = s = s₁ + s₂

Total distance travelled = v₁t₁ + v₂t₂

Total distance travelled = 30 x (1/2) + 60 x (1/2)

Total distance travelled = 15 + 30 = 45 km

Total time taken = t = t₁ + t₂ = 1/2 + 1/2 = 1 h

Average speed = Total disrtance travelled / Total time taken

Average speed = 45/1 = 45 kmph

Ans: The average speed = 45 kmph

Example – 09:

A car travels first 30 km at a uniform speed of 30 kmph and next 30 km at a uniform speed of 60 kmph. Calculate the average speed of the car.

Given:  Distance travelled in first part of journey = s₁ = 30km, Speed for the first part of journey = v₁ = 30 kmph, Distance travelled in second part of journey = s₂ = 30km, Speed for the second part of journey = v₂ = 60 kmph,

To Find:  average speed =?

Solution:

Total distance travelled = s = s₁ + s₂ = 30 km + 30 km = 60 km

Total time taken = t = t₁ + t₂ =  s₁/v₁ + s₂/v₂ = 30/30 + 30/60 = 1.5 h

Average speed = Total disrtance travelled / Total time taken

Average speed = 60/1.5 = 40 kmph

Ans: The average speed = 40 kmph

Example – 10:

A car travels first 50 km at a uniform speed of 25 kmph and next 60 km at a uniform speed of 20 kmph. Calculate the average speed of the car.

Given:  Distance travelled in first part of journey = s₁ = 50km, Speed for the first part of journey = v₁ = 25 kmph, Distance travelled in second part of journey = s₂ = 60km, Speed for the second part of journey = v₂ = 20 kmph,

To Find:  average speed =?

Solution:

Total distance travelled = s = s₁ + s₂ =50 km + 60 km = 110 km

Total time taken = t = t₁ + t₂ =  s₁/v₁ + s₂/v₂ = 50/25 + 60/20 = 5 h

Average speed = Total disrtance travelled / Total time taken

Average speed = 110/5 = 22 kmph

Ans: The average speed = 22 kmph

Example – 11:

A car is fitted with a speedometer which also gives reading of distance travelled by the car. At the start of the trio reading was found to be 1272 km and after 50 imutes at end of the trip  is 1352 km. Calculate the average speed of the car.

Given:  Initial reading = s₁ = 1272 km, Final reading = s₂ = 1352 km, time taken = 50 min = 50/60 = 5/6 h

To Find:  average speed =?

Solution:

Total distance travelled =  s₂ – s₁ = 1352 – 1272 = 80 km

Average speed = Total disrtance travelled / Total time taken

Average speed = 80/( 5/6)= 96 kmph

Ans: The average speed = 96 kmph

Example – 12:

A train takes 2 h to reach from station A to station B which is at a distance of 200 km from station A. It takes 3 h for the return journey. What is the average speed and average velocityof the train?

Given:  Distance travelled in first part of journey = s₁ = 200 km, time taken for the first part of journey = t₁ = 2 h, Distance travelled in second part of journey = s₂ = 200 km, time taken for the second part of journey = t₂ = 3 h.

To Find:  average speed =?

Solution:

Total distance travelled = s = s₁ + s₂ = 200 km + 200 km = 400 km

Total time taken = t = t₁ + t₂ =  2 h + 3 h = 5h

Average speed = Total disrtance travelled / Total time taken

Average speed = 400/5 = 80 kmph

As the train is coming back to starting point its displacement is zero.

Hence its average velocity = 0

Ans: The average speed = 80 kmph, average velocity = 0

Example – 13:

A driver of a car has a reaction time of 0.2 s. reaction time is a time between actually seeing the obstacle and applying the brake. He is moving with a uniform speed of 30 kmph. He spots a boy crossing the road. How much distance he travels before applying the brake.

Given: Reaction time = t = 0.2 s, speed of car = v = 36 kmph = 36 x 5/18 = 10 m/s

To Find: distance travelled = s =?

Solution:

Between seeing the boy and applying the brake car moves in uniform motion

Distance travelled before applying brake = s = v t = 10 x 0.2 = 2 m

Ans: Distance travelled before application of brakes is 2 m

Previous Topic: The Concept and Terminology of Motion

Next Topic: Newton’s Kinematical Equations of Motion

Science > Physics > Motion in a Straight Line > Numerical Problems on Displacement, Average Speed, Velocity

The post Numerical Problems on Displacement, Average Speed, Velocity appeared first on The Fact Factor.

Motion is an important part of our life. Our daily activities involve motion of different kinds. In this topic, we shall study the terminology of motion.

• Mechanics: The branch of physics which deals with the effects of forces on object is called mechanics. Mechanics is further classified into dynamics and statics.
• A Point Object: In the study of Mechanics we consider bodies or objects as particles or point objects. An object is said to be a point object if its dimensions are negligible as compared to the distance travelled by it. For example distance between stars is so large that for practical purpose those stars can be considered as particles or point objects.
• Body in Motion: A body is said to be in motion if it changes its position with respect to its immediate surroundings. It is to be noted that the motion of a body is a relative concept.
• Body at Rest: A body is said to be at rest if it does not change its position with respect to its immediate surroundings.
• Dynamics: The branch of physics (mechanics) which deals with the motion of the bodies and the forces causing it is called dynamics. It is further classified into kinematics and kinetics.
• Kinematics: The branch of physics (mechanics) which deals with the motion of the bodies without considering the forces causing it is called kinematics.
• Kinetics: The branch of physics which deals with the motion of bodies considering cause of their motion.
• Statics: It is the branch of physics which deals with objects at rest under the action of forces.

Motion is a Relative Concept:

The motion of a body is a relative concept. When we are specifying the motion it is with respect to some observer. Let us consider to persons say A and B in a lift moving upward. There is another person C standing outside the lift. Now though the lift is going up for A and B there is no change in positions with respect to each other. Thus for A and B, there is no motion with respect to each other but for both of them, C is moving downward. Now for C both A and B are moving upward. Hence motion is relative.

With respect to the earth surface, we may be at rest but we are moving about 100,000 km hr^-1 relative to sun.

Classification of Motion:

Motion can be classified into random motion, translational motion, rotational motion, and vibrational or oscillatory motion.

Random Motion:

In this motion particles move randomly in any possible direction and in any possible velocity. Thus, the path and the direction are not definite. Example: Motion of gas molecules. Such random motion of molecules of gas is called molecular chaos.

Translational Motion:

In this motion every particle of the body has the same displacement. The translational motion can be along straight line or along a curved path. Motion along a straight line is called a rectilinear motion  (e.g. motion of car in a straight line) and the motion of a body along a curved path is called curvilinear motion (e.g. the motion of a ball thrown in air, the motion of the earth around the Sun).

Rotational Motion:

In this motion the particles of body revolve in a circle about the same axis. Examples: Motion of a fan, the motion of a wheel of moving vehicle, the motion of merry go round, etc.

Oscillatory or Vibrational Motion:

In this motion the body moves to and fro about a fixed point along the same path. Examples: The motion of the bob of a pendulum, vibrating string of guitar, etc.

Rectilinear Motion or One Dimensional Motion:

When a body moves along a straight-line path, its motion is called the one-dimensional motion or motion in a straight line or rectilinear motion. Example: the motion of a car along a straight road.

The Position of a Body or Particle:

 Assuming the direction of the motion along the x-axis, the path of one-dimensional motion can be represented by a straight line parallel to the x-axis then each point on the straight line represents the position of the particle at a different instant of time. The position of the particle at any instant can be specified by its x-coordinate. The x-coordinate changes with time.

Distance and Displacement:

Distance:

The length of the path travelled by a body is called the distance travelled by it. The path of a body may not be straight. Distance is also referred as path length.

It is denoted by ‘s’ or ‘x’. Its S.I. unit is metre (m) and the c.g.s. unit is centimetre (cm). Its dimensions are [L¹M⁰T⁰]

Characteristics of Distance:

• It is the length of the path followed by the object in a certain time. The path followed may or may not be along a straight line.
• It is a scalar quantity.
• It depends on the path followed by the object.
• It is always positive.
• It can be more than or equal to displacement.
• It may not be zero even if the displacement is zero.

Displacement:

The shortest distance from the initial position to the final position of the body is called the magnitude of the displacement.

It is a vector quantity whose direction is from initial position to final position. Its S.I. unit is metre (m) and the c.g.s. unit is centimetre (cm). Its dimensions are [L¹M⁰T⁰]

Characteristics of Displacement:

• It is the shortest distance between the initial position to the final position of the body. It is always along a straight line.
• It is a vector quantity whose direction is from the initial position to final position.
• It is independent of the path followed by the object.
• It may be positive, negative or zero.
• It may be equal but cannot be more than the distance travelled.
• It is zero when the distance travelled is zero.

Displacement May be Positive or Negative or Zero.

Case – 1: When Distance travelled and displacement are equal.

If an object moves along the positive direction of the x-axis through 4m and further moves by 3 m in the same direction. In this case, the distance travelled by the object is 7m and displacement is also 7 m.

Case – 2: When Distance travelled and displacement are not equal and displacement is positive

If an object moves along the positive direction of the x-axis through 4m and further moves by 3 m in the opposite direction. In this case, the distance travelled by the object is 7m and displacement is also + 1 m (along the positive direction of the x-axis).

Case – 3: When Distance travelled and displacement are not equal and displacement is negative

If an object moves along the positive direction of the x-axis through 3m and further moves by 4 m in the opposite direction. In this case, the distance travelled by the object is 7m and displacement is also – 1 m (along the negative direction of the x-axis).

Case – 4: When Distance travelled and displacement are not equal and displacement is zero

If an object moves along the positive direction of the x-axis through 4m and further moves by 4 m in the opposite direction. In this case, the distance travelled by the object is 8 m and the displacement is also 0 m.

Understanding Concept of Displacement with Examples:

Example 01:

A cop gets information that a thief is 10 km away from the police station. Is it possible for cop to trace the thief with the given information?

This information is not sufficient as the direction in which the cop has to trace is not mentioned. The cop has to move along a circumference of a circle of radius 10 km through each and every point. At point T he can nab the thief. Thus, only specifying that a thief is 10 km away from the police station is not sufficient and it makes the task almost impossible. To trace the thief effectively with the distance direction also should be specified.

Thus for displacement both the magnitude and the direction is required. It is a vector quantity.

Example 02:

A horse is tied to a rope of length ‘r’ and the other end of the rope is tied to a pole. Find the displacement and the distance travelled by the horse in the following cases:

1. When the horse makes half revolution along a circular path.
2. When it makes one full revolution.
3. When it makes 3/4 th of the revolution.

When the horse makes half revolution along a circular path.

The starting point of journey is P and the end point is at Q

Distance travelled = Circumference/2 = 2πr/2 = πr units

Displacement = PQ = r + r = 2r units

When it makes one full revolution.

The starting point of journey is P and the end point is at Q

Distance travelled = Circumference = 2πr units

Displacement = PQ = 0 units

When it makes 3/4 th of the revolution.

The starting point of journey is P and the end point is at Q

Distance travelled = 3/4 x Circumference = 3/4(2πr) = 3πr/2 units

Applying Pythagoras theorem to Δ POQ

Displacement = PQ = √2 r units

Distinguishing Between Distance and Displacement:

Speed:

The rate of change of distance with time is called the speed of the body.

Mathematically, speed = Distance/Time

It is denoted by v. Its S.I. unit is m/s and c.g.s. unit is cm/s. Its dimensions are [L¹M⁰T^-1].

Uniform Speed:

A body is said to move with uniform speed if it covers equal distances in equal intervals of time throughout its motion.

Non-Uniform or Variable Speed:

A body is said to move at a non-uniform speed if it covers unequal distances in the same intervals of time.

Instantaneous Speed:

When a speed of a body changes continuously with time, its speed at any instant is known as instantaneous speed.

Average Speed:

The ratio of the total distance travelled by the body to the total time of the journey is called average speed.

When a body is moving with uniform speed, then the instantaneous speed and average speed are equal.

Characteristics of Speed:

• The rate of change of distance with time is called the speed of the body.
• It is a scalar quantity
• Speed is always positive.
• In a circular motion, after executing a complete circle, the average velocity of the body is zero but its average speed is not zero.

Velocity:

The rate of change of displacement of a body with respect to time is called the velocity of the body.

Velocity is a vector quantity, its S.I. unit is m/s and c.g.s. unit is cm/s.Its dimensions are [L¹M⁰T^-1]

Uniform Velocity: 

When the magnitude and direction of the velocity of a body remain the same at any instant, then the body is said to have uniform velocity. For uniform motion acceleration a = 0 and Displacement = velocity × time.

Example: The velocity of light in a particular medium is uniform velocity.  The velocity of sound in air at constant temperature is uniform velocity.

Non Uniform Velocity: 

When the magnitude of velocity or the direction of velocity or both changes at any instant the body is said to have the nonuniform velocity or variable velocity.

A body can have non-uniform velocity in the following three cases.

• When the direction of the velocity of a body remains the same but its magnitude changes continuously then the body has variable velocity. e.g. a ball is thrown vertically upward.
• When the magnitude of the velocity of a body remains the same but the direction changes continuously then the body has variable velocity. e.g. uniform circular motion of a body.
• When both the magnitude and direction of the velocity of body change continuously, then the body has variable velocity. e.g. ball thrown by making the acute angle with the horizontal (projectile motion)

When a body has variable velocity, then it has acceleration.

Instantaneous Velocity:

For a body moving with non-uniform velocity, the velocity of the body at an instant is called instantaneous velocity.

Average Velocity:

If the velocity of a body moving in particular direction changes with time, then the ratio of displacement to total time is called average velocity.

Characteristics of Velocity:

• The rate of change of displacement of a body with respect to time is called as the velocity of the body.
• It is a vector quantity.
• The velocity can be positive, negative or zero.
• In a circular motion, after executing a complete circle, the average velocity of the body is zero but its average speed is not zero.

Acceleration:

The rate of change of velocity with respect to time is called acceleration.

Acceleration is vector quantity its S.I. unit is m/s². Its dimensions are [L¹M⁰T^-2].

Acceleration can be positive, negative or zero. If the velocity is increasing then acceleration is positive. If the velocity is decreasing acceleration is negative. If the velocity is the constant acceleration is zero. Negative acceleration is also called deceleration or retardation.

If the velocity is increasing then the direction of acceleration is same as that of the velocity of the body. If the velocity is decreasing then the direction of acceleration is opposite to that of the velocity of the body.

It is to be noted that the velocity and not the acceleration of the body determines the direction of motion.

Uniform Acceleration:

When equal changes take place in the velocity of a body in equal interval of time, then the acceleration is called uniform acceleration. e.g. the motion under gravity.

Variable Acceleration:

When The change in the velocity of a body in equal interval of time is not constant, then the acceleration is called non-uniform acceleration.

Acceleration Due To Gravity:

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

It is denoted by ‘g’. Its value is g = 9.8 m/s2. When solving problems on the motion under gravity as per the convention the value of ‘g’ should be negative.

Next Topic: Numerical Problems on Displacement and Average Velocity

Science > Physics > Motion in a Straight Line > Terminology of Motion

The post Terminology of Motion appeared first on The Fact Factor.

The acceleration of gravity is constant at a particular place but it varies from place to place. In this article, we shall study this variation in acceleration due to gravity.

Variation in Acceleration Due to Gravity due to Shape of the Earth:

The acceleration due to gravity on the surface of the earth is given by

We can see that the acceleration due to gravity at a place is inversely proportional to the square of the distance of the point from the centre of the earth. Now, the earth is not perfectly spherical. It is flattened at the poles and elongated on the equatorial region. The radius of the equatorial region is approximately 21 km more than that at the poles. Hence acceleration due to gravity is maximum at the poles and minimum at the equator. As we move from the equator to the poles the distance of the point on the surface of the earth from the centre of the earth decreases. Hence the acceleration due to gravity increases.

Variation in acceleration due to gravity Due to Latitude of the Place:

The latitude of a point is the angle Φ between the equatorial plane and the line joining that point to the centre of the earth. Latitude of the equator is 0° and that of poles is 90°.

Let us consider a body of mass ‘m’ at a point P with latitude ‘Φ’ as shown on the surface of the earth. Let ‘g_Φ’ be the acceleration due to gravity at point P.

OP = Radius of earth = R

O’P = distance of point P from the axis of the earth

Due to rotational motion of the earth about its axis, the body at P experiences a centrifugal force which is given by mrω². Let us resolve this centrifugal force into two rectangular components. Its component along the radius of the earth is mrω² cosΦ.

Now the body is acted upon by two forces its weight mg acting towards the centre of the earth and the component mrω² cosΦ acting radially outward. The difference between the two forces gives the weight of that body at that point.

mg_Φ = mg – mrω²cosΦ ………….. (1)

Now cos Φ = O’P / OP = r/R

∴ r  = R cos Φ

Substituting in equation (1)

mg_Φ = mg – m(R cos Φ)ω²cos Φ

∴   g_Φ = g – R ω²cos² Φ

This is an expression for acceleration due to gravity at a point P on the surface of the earth having latitude Φ.

At the equator Φ = 0°. Hence ‘g’ is minimum on the equator. For the poles Φ = 90°. Hence ‘g’ is maximum on the poles.

Numerical Problems:

Example – 01:

Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg, R = 6400 km = 6.4 × 10⁶ m,

To find: W_E =? W_P = ? W_Φ =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²0

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (1)²

W_E = 980 – 3.386 = 976.6 N

At poles Φ =90°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²90

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0)²

W_E = 980 – 0 = 980 N

At latitude Φ = 30°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²30

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0.866)²

W_E = 980 – 2.539 = 977.5 N

Ans: Weight of the body on the equator, on the pole and on latitude 30° are 976.6 N, 980 N and 977.5 N respectively.

Example – 02:

Find the difference in weight of a body of mass 100 kg on equator and pole. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg,  R = 6400  km = 6.4 × 10⁶ m,

To find: W_P – W_E =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = mg – mRω²cos²0° = mg – mRω²(1)² = mg – mRω²    ………… (1)

At poles Φ = 90°

W_P = mg – mRω²cos²90° = mg – mRω²(0)² = mg    ………… (2)

Now, W_P – W_E = mg – (mg – mRω²) = mRω² 

W_P – W_E = mg – (mg – mRω²) = 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²   = 3.386 N

Ans: The required difference in Weight is 3.386 N

Variation in Acceleration Due to Gravity Due to Altitude:

Expanding binomially and neglecting terms of higher power of (h/R) we get

This is an expression for the acceleration due to gravity at small height ‘h’ from the surface of the earth. This expression shows acceleration due to gravity decreases as we move away from the surface of the earth.

Loss in Weight of a Body at Height h:

Numerical Problems:

Example – 03:

A mass of 5 kg is weighed on a balance at the top of a tower 20 m high. The mass is then suspended from the pan of the balance by a fine wire 20 m long and weighed. Find the change in the weight of a body in mgf assuming the radius of the earth as 6330 km.

Given: Mass of body = m = 5 kg, Radius of earth = R = 6330  km = 6.33 x 10⁶ m, height of tower =  h = 20 m.

To find: W – W_h =?

Solution:

Ans: The change in the weight of a body is 31.6 mgf

Example – 04:

At what height will a man’s weight become half his weight on the surface of the earth? Take the radius of the earth as R.

Given: W_h = 1/2W, Radius of earth = R

To find: h =?

Solution:

Now r = R + h

∴   h = r – R = 1.414 R – R = 0.414 R

Ans: At a height of 0.414R, the man’s weight become half his weight on the surface of the earth.

Example – 05:

A meteor is falling. How much gravitational acceleration it will experience when its height from the surface of the earth is equal to three times radius of the earth. Acceleration due to gravity on the surface of the earth is ‘g’.

Given: h = 3R.

To find: g_h =?

Solution:

Ans: The acceleration of the meteor is g/16.

Variation in Acceleration Due to Gravity Due to Depth:

The acceleration due to gravity on the surface of the earth is given by

Let ‘ρ’ be the density of the material of the earth.

Now, mass = volume x density

Substituting in the equation for g we get

Now, let the body be taken to the depth ‘d’ below the surface of the earth. Then acceleration due to gravity g_d at the depth ‘d’ below the surface of the earth is given by

Dividing equation (3) by (2) we get

This is an expression for the acceleration due to gravity at the depth ‘d’ below the surface of the earth. This expression shows acceleration due to gravity decreases as we move down into the earth. At the centre of the earth d = R, hence acceleration due to gravity at the centre of the earth is zero.

Relation between g_d and g_h:

We have

Thus the acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Variation of g with Altitude and Depth

Numerical Problems:

Example – 06:

Find percentage decrease in the weight of a body when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.

Given: depth = d = 16 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find: Percentage decrease in weight =?

Solution:

Ans: The percentage decrease in weight is 0.25.

Example – 07:

How much below the surface of the earth does acceleration due to gravity becomes 1 % of the value at the earth’s surface? Assume the radius of the earth as 6380 km.

Given: g_d = 1% g = 0.01 g, Radius of earth = R = 6380 km, g = 9.8 m/s².

To find: depth d =?

Solution:

Ans: At a depth of 6316 km below the surface of the earth the acceleration due to gravity becomes 1 % of the value at the earth’s surface.

Example – 08:

Compare the weight of 5 kg body 10 km above and 10 km below the surface of the earth. Given the radius of the earth = 6400 km.

Given: d = 10 km, d = 10 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Solution:

Ans: The required ratio of weights is 0.998 : 1

Example – 09:

Compare the weight of body 0.5 km above and 1 km below the surface of the earth. Given the radius of the earth = 6400 km.

Solution:

Given: d = 1 km, d = 0.5 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Ans: The required ratio of weights is 1:1

Example – 10:

The acceleration due to gravity at a height 1/20 th radius of the earth above the earth’s surface is 9 m/s². Find the value of acceleration due to gravity at an equal distance below the surface of the earth. Given the radius of the earth = 6400 km.

Given: h = 1/20 R, gh = 9 m/s², Radius of earth = R = 6400 km

To find:   g_d =? when d = 1/20 R

Solution:

Ans: The acceleration due to gravity at depth equal to R is 9.5 m/s²

Example – 11:

At what depth below the surface of the earth, a man’s weight becomes half his weight on the surface of the earth. Take the radius of the earth as R = 6400 km.

Given: W_d = 1/2 W, Radius of earth = R = 6400 km

To find:   d =?

Solution:

∴ R = 2R – 2d

∴ d = R i.e. d = R/2 = 6400/2 = 3200 km

Ans: At depth of 3200 km below the surface of the earth a man’s weight becomes half his weight on the surface of the earth.

Example – 12:

Find decrease in value of acceleration due to gravity at a point 1600 km below the earth’s surface. R = 6400 km, g = 9.8 m/s².

Given: d = 1600 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in value of acceleration due to gravity = g – g_d =?

Solution:

Ans: The decrease in acceleration due to gravity is 2.45 m/s²

Example – 13:

What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s².

Given: Mass of body = m = 500 kg, depth d = 1000 m = 1 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in weight = W – W_d =?

Solution:

Ans: The decrease in the weight of the body is 1 N.

Example – 14:

Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s².

Given: d = 2000 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   acceleration due to gravity =  g_d =?

Solution:

Ans: The acceleration at depth of 2000 km below the surface of the earth is 6,738 m/s²

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