Science > Physics > Motion in a Straight Line > Numerical Problems on Uniformly Accelerated Bodies Set – 01
In this article we shall study the use of newton’s equation of motion applied to uniformly accelerated bodies. The equations of motion are: the first equation of motion ( v = u + at ), the second equation of motion ( s = ut + ½ at2 ) and he second equation of motion ( v2 = u2 + 2as )
Example 01:
A body is starting from rest accelerates uniformly and attains a velocity of 20 ms-1 in 30 seconds. Find its acceleration and the distance covered in that time.
Given: Initial velocity = u = 0 ms-1, Final velocity = v = 20 ms-1, Time taken = t = 30 s.
To Find: Acceleration = a =?, Distance travelled = s =?
Solution:
By Newton’s first equation of motion
v = u + at
∴ 20 = 0 + (a) x 30
∴ 20 = 30 a
∴ a = 20/30 = 2/3 = 0.67 ms-2
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (0) (30) + ½ x (2/3)(30)2
∴ s = 0 + ½ x (2/3)(900)
∴ s = 300 m
Ans: Acceleration of the body = 0.67 ms-2 and distance covered in that time = 300 m
Example 02:
A body starting from rest accelerates uniformly and attains a velocity of 15 ms-1 in 15 s. Find its acceleration and the distance it covers in that time.
Given: Initial velocity = u = 0 ms-1, Final velocity = v = 15 ms-1, Time taken = t = 15 s.
To Find: Acceleration = a =?, Distance travelled = s =?
Solution:
By Newton’s first equation of motion
v = u + at
∴ 15 = 0 + (a) x 15
∴ 15 = 15 a
∴ a = 1 ms-2
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (0) (15) + ½ x (1)(15)2
∴ s = 0 + ½ x (1)(225)
∴ s = 112.5 m
Ans: Acceleration of the body = 1 ms-2 and distance covered in that time = 112.5 m
Example 03:
A body has an initial velocity of 3 ms-1 and has an acceleration of 2 ms-2. Find the distance travelled by it in 5 s and the velocity then.
Given: Initial velocity = u = 3 ms-1, acceleration = a = 2 ms-2, Time taken = t = 5 s.
To Find: Distance travelled = s =?, Final velocity = v =?
Solution:
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (3) (5) + ½ x (2)(5)2
∴ s = 15 + ½ x (2)(25)
∴ s = 15 + 25
∴ s = 40 m
By Newton’s first equation of motion
v = u + at
∴ v = 3 + (2) x 5
∴ v = 3 + 10
∴ v = 13 ms-1
Ans: Distance covered in that time = 40 m and velocity then is 13 ms-1
Example 04:
A body moving with an initial velocity of 1 ms-1 accelerates uniformly at 0.5 ms-2. Find its displacement and velocity at the end of 10 s. What is its velocity when it has a displacement of 48m?
Given: Initial velocity = u = 1 ms-1, Acceleration = a = 0.5 ms-2, Time taken = t = 10 s.
To Find: Displacement when t = 10s, s =?, velocity when t = 10s, v =?, Velocity when s = 48 m, v =?
Solution:
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (1) (10) + ½ x (0.5)(10)2
∴ s = 10 + ½ x (0.5)(100)
∴ s = 10 + 25
∴ s = 35 m
By Newton’s first equation of motion
v = u + at
∴ v = 1 + (0.5) x 10
∴ v = 1 + 5
∴ v = 6 ms-1
By Newton’s third equation of motion
v2 = u2 + 2as
∴ v2 = (1)2 + 2(0.5)(48)
∴ v2 = 1 + 48
∴ v2 = 49
∴ v = 7 ms-1
Ans: Displacement when t = 10 s is 35 m, velocity when t = 10s is 6 ms-1, Velocity when s = 48 m is 7 ms-1.
Example 05:
A body moving with an initial velocity of 2 ms-1 accelerates uniformly at 0.5 ms-2 for 20 s. What is the displacement then? What is its velocity when the displacement is 96 m?
Given: Initial velocity = u = 2 ms-1, Acceleration = a = 0.5 ms-2, Time taken = t = 20 s.
To Find: Displacement when t = 20s, s =?, velocity when t = 10s, v =?, Velocity when s = 96 m, v =?
Solution:
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (2) (20) + ½ x (0.5)(20)2
∴ s = 20 + ½ x (0.5)(400)
∴ s = 20 + 100
∴ s = 120 m
By Newton’s first equation of motion
v = u + at
∴ v = 2 + (0.5) x 20
∴ v = 2 + 10
∴ v = 12 ms-1
By Newton’s third equation of motion
v2 = u2 + 2as
∴ v2 = (2)2 + 2(0.5)(96)
∴ v2 = 4 + 96
∴ v2 = 100
∴ v = 10 ms-1
Ans: Displacement when t = 20 s is 120 m, velocity when t = 20s is 12 ms-1, Velocity when s = 96 m is 10 ms-1.
Example 06:
A body moving with a velocity of 10 m/s accelerates uniformly and covers 300 m when its velocity becomes 20 m/s. Find the time taken to achieve this Velocity.
Given: Initial velocity = u = 10 ms-1, Final velocity = v = 20 ms-1, Distance travelled = s = 300 m.
To Find: Time taken = t =?
Solution:
By Newton’s third equation of motion
v2 = u2 + 2as
∴ 202 = (10)2 + 2(a)(300)
∴ 400 = 100 + 600 a
∴ 300 = 600 a
∴ a = 300/600 = 0.5 ms-2
By Newton’s first equation of motion
v = u + at
∴ 20 = 10 + (0.5) x t
∴ 10 = 0.5 t
∴ t = 10/0.5 = 20 s
Ans: The time taken to achieve velocity of 20 ms-1 is 20 s
Example 07:
An airplane takes off the ground after covering a length of 800 m in 16 s. Find its acceleration and the velocity of takes off.
Given: Initial velocity = u = 0 ms-1, Distance travelled = s = 800 m, Time taken = t = 16 s.
To Find: Acceleration =?, Final velocity =?
Solution:
By Newton’s second equation of motion
s = ut + ½ at2
∴ 400 = (0) (16) + ½ x (a)(16)2
∴ 400 = 0 + ½ x a x 256
∴ 400 = 128 a
∴ a = 400/128 = 3.125 ms-2
By Newton’s first equation of motion
v = u + at
∴ v = 0 + (3.125) x 16
∴ v = 50 ms-1
Ans: Acceleration = 3.125 ms-2 and velocity of take-off = 50 ms-1.
Example 08:
An aeroplane touches the runway at 270 km h-1 and stops after a run of one kilometre. How long does it take to stop?
Given: Initial velocity = u = 270 km h-1 = 270 (5/18) = 75 ms-1, Distance travelled = s = 1 km = 1000 m, Final velocity = v = 0 ms-1.
To Find: Time taken to stop = t =?
Solution:
By Newton’s third equation of motion
v2 = u2 + 2as
∴ 02 = (75)2 + 2(a)(1000)
∴ 0 = 5625 + 2000 a
– 5625 = 2000 a
∴ a = – (5625/2000) = – 2.8125 ms-2
Negative sign indicate retardation.
By Newton’s first equation of motion
v = u + at
∴ 0 = 75 + (-2.8125) x t
∴ – 75 = – 2.8125 x t
∴ t = 75/2.8125 = 26.67 s
Ans: Plane will take 26.67 s to stop
Example 09:
A car travelling at 80 ms-1 is braked so as to produce a retardation of 4 ms-2. After what time its speed will be 40 ms-1 and what will be the distance travelled during this time?
Given: Initial velocity = u = 80 ms-1, Final velocity = v = 40 ms-1, Acceleration = a = – 4 ms-2.
To Find: Time taken = t =?, Distance travelled = s = ?
Solution:
By Newton’s first equation of motion
v = u + at
∴ 40 = 80 + (- 4) x t
∴ – 40 = – 4 t
∴ t = 40/4 = 10 s
By Newton’s third equation of motion
v2 = u2 + 2as
∴ 402 = (80)2 + 2(- 4)(s)
∴ 1600 = 6400 – 8 s
∴ – 4800 = – 8s
∴ s = 4800/8 = 600 m
Ans: Time taken = 10s and distance travelled during this time = 600 m
Example 10:
A car accelerates from rest at 2 ms-2. What will be its speed and how far will it be from its starting point after 10 s? After what distance will its speed be 40 ms-1?
Given: Initial velocity = u = 0 ms-1, Acceleration = a = 2 ms-2, Time =t = 10 s.
To Find: Final velocity = v =?, Distance travelled = s = ?, when v = 40 ms-1.
Solution:
By Newton’s first equation of motion
v = u + at
∴ v = 0 + (2) x (10)
∴ v = 20 ms-1
By Newton’s third equation of motion
v2 = u2 + 2as
∴ (40)2 = (0)2 + 2(2)(s)
∴ 1600 = 4 s
∴ s = 1600/4 = 400 m
Ans: velocity after 10 s is 20 ms-1 and distance travelled when velocity is 40 ms-1 is 400 m.
Example 11:
A train travelling at 126 km h-1 has its speed reduced to 54 km h-1 while it passes over 500 m. Assuming the track is straight and the retardation is uniform, find the retardation. How much farther will the train travel before coming to rest?
Given: Initial velocity = u = 126 km h-1 = 126 x (5/18) = 35 ms-1, Final velocity = v = 54 km h-1 = 54 x (5/18) = 15 ms-1, Distance travelled = s = 500 m.
To Find: Retardation = a =?, Distance travelled = s = ?, when v = 0 ms-1.
Solution:
By Newton’s third equation of motion
v2 = u2 + 2as
∴ (15)2 = (35)2 + 2(a)(500)
∴ 225 = 1225 + 1000 a
∴ – 1000 = 1000 a
∴ a = – 1000/1000 = – 1 ms-2
By Newton’s third equation of motion
v2 = u2 + 2as
∴ (0)2 = (35)2 + 2(- 1)(s)
∴ 0 = 1225 – 2 s
∴ 2s = 1225
∴ s = 1225/2 = 612.5 m
Ans: Retardation is – 1 ms-2 and distance travelled when train comes to rest is 612.5 m.
Example 12:
A car travelling at 72 km h-1 has its velocity reduced to 36 km h-1 in 5 s. If the retardation is uniform, find how much distance it has covered during this time. How much farther would it travel before coming to rest assuming the same uniform retardation?
Given: Initial velocity = u = 72 km h-1 = 72 x (5/18) = 20 ms-1, Final velocity = v = 36 km h-1 = 36 x (5/18) = 10 ms-1, Time taken = t = 5 s.
To Find: Distance travelled = s =? Distance travelled =?, when v = 0 ms-1.
Solution:
By Newton’s first equation of motion
v = u + at
∴ 10 = 20 + a (5)
∴ – 10 = 5a
∴ a = -10/5 = – 2 ms-2
By Newton’s second equation of motion
s = ut + ½ at2
∴ s = (20) (5) + ½ x (- 2)(5)2
∴ s = 100 + ½ x (- 2)(25)
∴ s = 100 – 25
∴ s = 75 m
By Newton’s third equation of motion
v2 = u2 + 2as
∴ (0)2 = (10)2 + 2(- 2)(s)
∴ 0 = 100 – 4 s
∴ 4s = 100
∴ s = 100/4 = 25 m
Ans: The distance it has covered in 5 s is 75 m and further travel before coming to rest is 25 m.
Example 13:
The traffic at the entrance of a bridge limits the speed to 10 km h-1. The driver of a car travelling at 60 km h-1 began applying the brakes 7s before reaching the entrance to the bridge. If the brakes produced an average deceleration 2 ms-2 did the driver break the rule?
Given: Permissible velocity = v = 10 km h-1 = 10 x (5/18) = (25/9) ms-1 = 2.78 ms-1, Initial velocity = u = 60 km h-1 = 60 x (5/18) = (50/3) ms-1 = 16.67 ms-1, Time taken = t = 7 s, deceleration = a = – 2 ms-2
To Find: Final velocity = v =?, Is it greater than permissible velocity?.
Solution:
By Newton’s first equation of motion
v = u + at
∴ v = 16.67 + (-2) (7)
∴ v = 16.67 – 14
∴ v = 2.67 s
Ans: The speed at entrance of the bridge is less than the permissible speed. Hence the driver didn’t break the rule.