A fugitive tries to hop on a freight train traveling at a constant speed of 5.5 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 2.5 m/s2 to his maximum speed of 8 m/s. (a) How long does it take him to catch up to the empty box car? s (b) What is the distance traveled to reach the box car?

Question

Physics

Ian Brandt

28 May, 08:10

A fugitive tries to hop on a freight train traveling at a constant speed of 5.5 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 2.5 m/s2 to his maximum speed of 8 m/s. (a) How long does it take him to catch up to the empty box car? s (b) What is the distance traveled to reach the box car?

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Answers (1)

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Rhett Mckee · Answer 1

Th equations to be used here are the following:

a = (v - v₀) / t

x = v₀t + 0.5at²

The speed of the fugitive is the sum of his own speed plus the speed of the train. Thus,

v₀ = 0 + 5.5 m/s = 5.5 m/s

v = 8 m/s + 5.5 m/s = 13.5 m/s

a.) We use the first equation to determine time

2.5 m/s² = (13.5 m/s - 5.5 m/s) t

Solving for t,

t = 3.2 seconds

b.) We use the answer in a) and the 2nd equation:

x = (5.5 m/s) (3.2 s) + 0.5 (2.5 m/s²) (3.2 s) ²

x = 30.4 meters