(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A fugitive tries to hop a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a=4.0 m/s2

to his maximum speed of 8.0 m/s. (a) How long does it take him to catch

up to the empty box car? (b) What is the distance traveled to reach the

box car?

2. Relevant equations

v = v(initial) +at

x = x(initial) + v(initial)t + .5at^2

v^2 = v^2(initial) + 2a(x - x(initial)

x=vt

3. The attempt at a solution

I first found out how long it took the fugitive to get to 8.0 m/s since this is his maximum speed. so i used the first equation above 8.0= 0 +4.0t... t=2s.

now i know it takes the fugitive 2 seconds to reach max velocity. now im trying to figure out how far the fugitive traveled and how far the train traveled. For the fugitive i used the second equation above x= 0 + 0 + .5(4)(2^2) x=8m

Now I want to find the distance the train traveled:

x =vt

x = 6(2) = 12

So Now Im left with the problem that the fugitive traveled 8m in 2 seconds and the train traveled 12m in 2 seconds. a difference of 4m. So i looked at the question and said to myself since the fugitive is traveling at a higher velocity than the train the will eventually meet each other. but how to find that time??

I broke the fugitives distance into two parts

1. getting up to speed

2. traveling a constant speed

but I have no clue on what equation to use or how to approach the problem...

p.s. sorry this is sooo long.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: One Dimension Kinematics Question

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