One Dimension Kinematics Question

[iurod]

Homework Statement

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?

Homework Equations

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

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 I am 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 I am 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.

 

[tiny-tim]

Welcome to PF!

Hi iurod! Welcome to PF!

(try using the X² tag just above the Reply box )

So Now I am 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??

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 … ?

 

[iurod]

Thanks for your super fast reply Tiny Tim..


so I came up with this but I am 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?

 

[tiny-tim]

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

t = 4 seconds?

Sorry, I'm not following this …

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

Write it out again, more carefully.

 

[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 don't know this time so I am 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 :)

 

[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"

 

[iurod]

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!