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a 600kg vehicle was propelled 1080m down a track by 22,700N of rocket thrust and...
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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` .
Posted by valentin68 on September 20, 2013 at 11:57 AM (Answer #2)
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