Question

A train moving with a velocity of 45 m/sec comes to rest in 9 sec. find i)acceleration produces by the train during its last 9 seconds and ii)the distance covered by train in these 9 sec
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Answer 1

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Answer 2

Answer:

a=-5 m/s²

Distance = 202.5 metres

Explanation:

Given :

• Initial velocity = u = 45 m/s

• Final velocity = v = 0 m/s

• Time taken = t = 9 seconds

To find :

• Acceleration produced by the train during 9 seconds

• Distance covered in 9 seconds

Using the first equation of motion :

V=u+at

0=45+a×9

-45=9a

a = - 45/9

a = - 5 m/s²

The acceleration of the train is equal to - 5 m/s²

Using the third equation of motion :

V²-u²=2as

0²-45²=2×-5×s

0-2025=-10s

-2025=-10s

s = - 2025/-10

s = 202.5 metres

The acceleration of the train is - 5 m/s² and distance covered is 202.5 metres