Question

A freight train leaving a yard must exert a force of 2530000 N in order to increase its speed from rest to 17 m/s. During this process, the train must do 1110000000 J of work. How far does the train travel? *

Answer 1

`\Delta s = 438.735\,m`

Step-by-step explanation:

Given that force exerted is constant and parallel to the railroad during acceleration, the equation of work is equal to:

`W = F\cdot \Delta s`

The travelled distance of the freight train is:

`\Delta s = (W)/(F)`

`\Delta s = (1110000000\,J)/(2530000\,N)`

`\Delta s = 438.735\,m`

Answer 2

Answer: The distance traveled in other to increase its speed from rest to 17m/s is 438.74m

Step-by-step explanation:

GIVEN the following :

Workdone = 1110000000 J

Force = 2530000 N

Workdone = force × distance

1110000000 = 2530000 × distance

Distance = 1110000000 ÷ 2530000

Distance = 438.735m

Distance = 438.74m

The distance traveled by the freight train in other to increase it's speed from rest to 17m/s is 438.74m