A fugitive tries to hop a freight train traveling at a constant speed of 5.1 m/s

Question

A fugitive tries to hop a freight train traveling at a constant speed of 5.1 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a= 1.2 m/s2 to his maximum speed of 5.8 m/s.

[Part A] How long does it take him to catch up to the empty box car?

[Part B] What is the distance traveled to reach the box car?

A fugitive tries to hop a freight train traveling at a constant speed of 5.1 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.2 m/s2 to his maximum speed of 5.8 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?

Explanation / Answer

let t= total tim taken by fugitive t actch upto the train

distance travelled by car = Dc= v t= 5.1 t

maximum speed of fugitiv 5.8 m/s^2, initial speed=0 and acceleration= 1.2 m/s^2

let t1= time taken by fugitive to reach the maximum speed,

vf= vi+ at1

5.8= 0+ 1.2 (t1)

t1= 4.83 seconds

so time during with the fugitive is moving with constant maximum speed= t-4.83 seconds

total distance travelled by fugiitive= distance travelled with acceletaion+ distance travelled with constant maximum speed

Df= (vit1 +1/2 a t1^2) + Vmax (t-t1)

Df=0(4.83)+ 1/2 (1.2)(4.83)^2 + 5.8(t-4.83)

Df=14+5.8t-28

Df= 5.8t-14

now DF=Dc

5.8t-14= 5.1t

5.8t-5.1 t=14

0.7t= 14

t= 14/0.7

t= 20 seconds

so time taken to catch up to the empty car= 20seconds.

distance travlled to reach box car= 5.8t-24

distance= 5.8(20)-14 =102 m..

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