Calculating Time and Distance in a Kinematics Train Problem

[joe215]

Homework Statement

A fugitive tries to hop on a freight train traveling at a constant speed of 6.0m/s. The fugitive starts from rest and accelerates at 4.0m/s^2 to his maximum speed of 8.0 m/s. How long does it take him to catch the train? What is the distance traveled to catch the train?

The Attempt at a Solution

Displacement of fugitive=Displacement of train
D=volt +1/2aT^2 So...

volt +1/2aT^2 = volt +1/2aT^2 Plug in some values...

0 + 1/2aT^2 = volt + 0 Now I will rearrange the equation to solve for T (I'm going to eliminate a double root here, but it really doesn't matter I think)

T=2Vo(of the train) / a (of the fugitive)

T=3s

Now to find the displacement of the fugitive when he catches the train, I plug in 3s into the equation d=1/2(Vo+V)

d=1/2(0+8)3
d=12m

It takes the fugitive 3s to catch up to the boxcar. The fugitive travels 12m to catch the car.

 

[alphysicist]

Hi joe215,

Homework Statement

A fugitive tries to hop on a freight train traveling at a constant speed of 6.0m/s. The fugitive starts from rest and accelerates at 4.0m/s^2 to his maximum speed of 8.0 m/s. How long does it take him to catch the train? What is the distance traveled to catch the train?

The Attempt at a Solution

Displacement of fugitive=Displacement of train
D=volt +1/2aT^2 So...

volt +1/2aT^2 = volt +1/2aT^2 Plug in some values...

0 + 1/2aT^2 = volt + 0 Now I will rearrange the equation to solve for T (I'm going to eliminate a double root here, but it really doesn't matter I think)

T=2Vo(of the train) / a (of the fugitive)

T=3s

I don't believe this is right. The factor of 1/2aT^2 only describes the fugitives motion while he is accelerating; however, he will reach maximum speed before he reaches the train, so you'll need to calculate his displacement in two steps (or two terms in the equation).

 

[joe215]

Ok is this right...

First I figure out how long it takes the fugitive to attain a velocity of 8m/s and the distance he travels in that interval. So...

t=(V-Vo)/a
8/4= 2s

In 2s he reaches 8m/s.

d=volt + 1/2at^2
d=1/2(4)(2^2)
d=8m

Now, when the distance of the train and fugitive are equal, their time is...

velocity of the fugitive * time of the fugitive=velocity of the train * time of the train +3s

8T=6(t+2)
2T=12
T=6s

The fugitives displacement= his velocity * time
D= 8m/s * 4s
D=32m

Now i add the the first 8m the fugitive traveled. 32m+40m

It takes him 6s to catch the train and he travels 40m.

 

[alphysicist]

Ok is this right...

First I figure out how long it takes the fugitive to attain a velocity of 8m/s and the distance he travels in that interval. So...

t=(V-Vo)/a
8/4= 2s

In 2s he reaches 8m/s.

d=volt + 1/2at^2
d=1/2(4)(2^2)
d=8m

Now, when the distance of the train and fugitive are equal, their time is...

velocity of the fugitive * time of the fugitive=velocity of the train * time of the train +3s

8T=6(t+2)

This equation is close, but not quite correct. The quantity on the right side is the distance the train has traveled from the time the man began running, but the quantity on the left is only the distance the man has run since the man reached full speed.

The left side needs to be the total distance the man has run since he began running, so you need to add the 8m term to the left side in this equation to solve for t (not later on in the problem).

2T=12
T=6s

The fugitives displacement= his velocity * time
D= 8m/s * 4s
D=32m

Now i add the the first 8m the fugitive traveled. 32m+40m

It takes him 6s to catch the train and he travels 40m.

By the way, you can check and see that this value does not work; in 6 seconds the train (moving 6m/s) has moved 36m, so this time and distance do not match for this problem.