Question

train starting from rest moves along a straight track and attains a velocity of 72 kmph in 3
minutes and 20 seconds. Assuming that the acceleration is uniform, find
i. The acceleration
ii. The distance travelled by the train for attaining this velocity

Answer 1

Explanation:

As the object began to move from rest starting with the velocity 20ms-1 the initial speed is more than the final speed which is 0ms-1.

if the object moves and then comes to rest the initial speed will be less than the final speed which will then be 0ms-1 - 20ms-1 instead of 20ms-1 - 0ms-1. Then the acceleration can be found as well as time.

hope this helps

Answer 2

Answer:

Acceleration = 0.1 m/s²

distance = 2 kilometres

Explanation:

Given:

• Initial velocity of the train = u = 0 m/s
• Final velocity of the train = v = 72 km/hr = 72×5/18 = 20 m/s
• Time taken = t = 3 minutes 20 seconds = 200 seconds

To find:

• Acceleration of the train
• Distance travelled by train after attaining this velocity

Using the first equation of motion : v = u+at

20=0+a×200

20=200a

a = 0.1 m/s²

The acceleration of the train is equal to 0.1 m/s²

Using the second equation of motion: s=ut+1/2 at²

s=0×200+1/2×0.1×200²

s=0+1/2×40000×0.1

s=20000×0.1

s=2000 metres (or) 2 kilometres

The acceleration of the train is equal to 0.1 m/s² and distance travelled by it is equal to 2 kilometres