# a 600kg vehicle was propelled 1080m down a track by 22,700N of rocket thrust and obtained its maximum speed in 5s before stopped 200m and 1.4s Calculate the kinetic energy of the sled, and final acceleration.

The vehicle is first accelerated by the rocket from rest to a speed `V_max` then it is decelerated with a final deceleration, until it stops.

The initial acceleration is (as the second law of physics states)

`a = F/m = 22700/600 = +37.83 m/s^2`

This acceleration is constant, hence the motion is uniform accelerated. The equation relating the initial and final speed to the space travelled is

`V^2 = V_0^2 +2*a*s`

Since the vehicle starts from rest, `V_0 = 0`, hence

`V_max = sqrt(2*a*s) = sqrt(2*37.83*1080) = 285.85 m/s`

The kinetic energy of the vehicle at this maximum speed is

`E_k = (m*V_max^2)/2 = 600*285.85^2/2 =24513840 J =24.51 MJ`

On the final part of the motion the deceleration is constant. We can write

`V_("final") = V_max +a*t`

and since `V_("final") =0` and `t =1.4 s` we have

`a = -V_(max)/t = -285.85/1.4 = -204.17 m/s^2 `

The kinetic energy at its maximum speed is 24.51 MJ and the final acceleration is -204.17 `m/s^2` .

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