A fugitive tries to hop a freight train traveling at a constant speed Of 4.5 m/s. Just as a empty box car passes him, the fugitive starts from rest and accelerates at a = 3.6 m/s squared to his maximum speed of 8.0 m/s. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car?

Question

Physics

Alisha Ray

13 June, 20:31

A fugitive tries to hop a freight train traveling at a constant speed Of 4.5 m/s. Just as a empty box car passes him, the fugitive starts from rest and accelerates at a = 3.6 m/s squared to his maximum speed of 8.0 m/s. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car?

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Answers (1)

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Caleb Casey · Answer 1

We use the kinematics equation:

Vf = Vi + a*t

8 = 0 + 3.6 * t

t=2.222s to reach 8.0 m/s

At that time the train has moved

4.5 m/s * 2.222s = 9.999 m

He travelled (another kinematics equation)

Vf^2 = Vi^2 + (2*a*d)

(8.0) ^2 = (0) ^2 + (2 * 3.6 * d)

d=8.888 m

The train is 9.999m, the fugitive is 8.888m,

He still needs to travel

9.999-8.888 = 1.111m

He needs to cover the rest of the distance in a smaller amount of time, however hes at his maximum velocity, so ...

8m/s (man) - 4.5m/s (train) = 3.5 m/s more

(1.111m) / (3.5m/s) =.317seconds more to reach the train

So if it takes 2.222 seconds to approach the train at 8.888m, it should take

2.222 +.317 = 2.529 seconds to reach the train completely

Last but not least is to figure out the total distance traveled in that time frame:

(Trains velocity) * (total time)

(4.5m/s) * (2.529s) = 11.3805m