Question

A fugitive tries to hop on a freight train traveling at a constant speed of 5.5 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 4.2 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?
s
(b) What is the distance traveled to reach the box car?
m

Answer 1

Let the Fugitive catch the empty box after "t" s of time

So the distance moved by train in same time will be

$d = 5.5*t$

Now in the same way distance moved by fugitive who is accelerating with a = 4.2 m/s^2 to his maximum speed v = 8 m/s

SO the time taken by him to reach maximum speed

$vf = vi + at$

$8 = 0 + 4.2*t$

$t = 1.9 s$

distance moved by fugitive in t = 1.9 s

$d = vi* t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}*4.2* 1.9^2$

$d = 7.62 m$

So in 1.9 s the train will move more than 7.62 m so it will definitely catch the train after this

distance moved by fugitive in time "t" = distance moved by train in same time

Let say it will catch the train after t = 1.9 s so after that time it will move to his constant speed of 8 m/s

$7.62 + 8 * (t - 1.9) = 5.5 * t$

$8*t - 5.5*t = 8*1.9 - 7.62$

$2.5*t = 7.58$

$t = 3.03 s$

So it will take 3.03 s to catch the train

Part B)

Distance traveled by fugitive in same time

$d = 7.62 + 8*(t - 1.9)$

here it is because the fugitive will accelerate to reach at his maximum speed of 8 m/s till 1.9 s and cover a total distance of 7.62 m and after that it will move at his maximum speed for rest of the time.

$d = 7.62 + 8*(3.03 - 1.9)$

$d = 16.67 m$

so it will move a distance of 16.67 m