# Homework Help: One Dimension Kinematics Question

1. Jun 10, 2010

### iurod

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

2. Jun 10, 2010

### tiny-tim

Welcome to PF!

Hi iurod! Welcome to PF!

(try using the X2 tag just above the Reply box )
You've found that, after 2s, he is 4m behind the train.

From then on, he is going 2m/s faster than the train, so the time to catch it is … ?

3. Jun 10, 2010

### iurod

so I came up with this but im not entirely sure if this is right

distance of the man accelerating + velocity(t2) = Velocity of the train (t1 +t2)

4m/s^2 + (8t) = 6(2 + t2)
4 + 8t = 12 +6t
2t = 8

t = 4 seconds????

if this is correct then the distance that the fugitive had to travel is:

d = vt
d = 6(4)
d = 24 meters

is this correct?

4. Jun 10, 2010

### tiny-tim

Sorry, I'm not following this …

you seem to be using t to mean two different things.

Write it out again, more carefully.

5. Jun 10, 2010

### iurod

yeah it doesn't make that much sense when I wrote it.

what I meant was:

the distance of the man accelerating + the velocity of the man(time2) = Velocity of the train (time 1 + time 2)

I need two times because it takes the fugitive 2 seconds to reach maximum velocity, after that I set the:

.5(acceleration)(2 seconds squared; that it took to get to maximum velocity) + the distance of the fugitive (vt) (i dont know this time so im calling it time 2) equal to the velocity(time 1 + time 2) of the train.

so:
.5(a)(time1)^2 + v(time2) = v(time1 + time2)

.5(4m/s^2)(2^2) + 8m/s (time 2) = 6m/s (2+time 2)
8 + 8time2 = 12 +6time2
2time2 =4
time2 = 2

time1 + time2
2+2 = 4 seconds

if this is correct then the distance that the fugitive had to travel is:
d = vt
d = 6(4)
d = 24 meters

hopefully this makes sense, its hard putting my thoughts into words. hopefully its correct also :)

6. Jun 11, 2010

### tiny-tim

hi iurod!

Yes, that's much better!

It's a bit long-winded, though …

you didn't need to define two "time"s, since you'd already worked out that the first one was 2 s (so it's no longer unknown, and there's no point giving names or labels to things that are known) …

so you could have immediately called the second one "t" (instead of "time2") …

alternatively, you could have simply said "after 2 s, the fugitive is 4 m behind, and is catching up at a constant relative speed of 2 m/s, so the further time he takes is 4/2 s = 2 s, making a total of 4 s"

7. Jun 11, 2010

### iurod

Re: One Dimension Kinematics Question SOLVED

Tiny Tim,
Your explanation makes a whole lot more sense than what I did :)

I really appreciate all your help on this, I think PF is going to be my new home for Physics Questions :)

Thanks again!