Question

A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as a box car passes him, the fugitive starts from rest and accelerates at a=1.4 m/s^2 to his maximum speed of 6.0 m/s, which he then maintains. (a) How long does he take to catch the box car? (b) how far does he take to reach the box car?

Answer 1

Answer:Given:

Initial speed of fugitive, v0 = 0 m/s

Final speed, vf = 6.1 m/s

acceleration, a = 1.4 m/s^2

Speed of train, v = 5.0 m/s

Solution:

t = (vf-v0)/a

t = (6.1-0)/1.4

t =4.36 s

Distance traveled by train, x_T =v*t

x_T =5*4.36 = 21.8 m

Distance travelled by fugitive, x_f = v0*t+1/2at^2

x_f = 0*4.36+1/2*1.4*4.36^2

x_f =13.31 m

5*t = v(t-4.36)+x_f

5*t=6.1*(t-4.36)+13.31

solve for t, we get

t = 12.08 s

The fugitive takes 12.08 s to catch up to the empty box car.

Distance traveled to reach the box car is

X_T = v*t

X_T = 5*12.08 s

X_T = 60.4 m

Explanation: