A fugitive tries to hop on a freight train traveling at a constant speed of 5.4m

Question

A fugitive tries to hop on a freight train traveling at a constant speed of 5.4m/s . Just as an empty box car passes him, the fugitive starts from rest and accelerates at a= 1.1m/s2 to his maximum speed of 6.1m/s , which he then maintains.

Part A

How long does it take him to catch up to the empty box car?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the distance traveled to reach the box car?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Velocity = 6.1, Rate of accelaration = 1.1

Therefore time to reach final velcity = 6.1/1.1 s

In this time the box car moves = 5.4 * 6.1 / 1.1 = 29.94 m

And the person travels ?V.dt = ?(u+ft).dt = ut + 1/2*ft2 = 0 + 1/2*1.1*(6.1/1.1)2 = 16.91 m

Difference = 29.94 - 16.91 = 13.03m Difference in speed = 6.1 - 5.4 = 0.7

Time taken = 13.03/0.7 = 18.61 s

In this time he travels 6.1 * 18.61 = 113.54 m
From start he travels 16.91 + 113.54 = 130.45 m

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