A rocket of mass m = 1000 kg is traveling in a straight line for a short time. The distance in meters covered by the rocket during this time is described by the function

$r(t)=t^3 −3t^2 +6t$

where $t > 0$ is the time in seconds.

The kinetic energy E of the rocket is given by $E = mv^2/2$ and $v=2$ is the rocket’s speed. Find a function that describes the kinetic energy of the rocket.

I thought you might use the second derivative? But have no idea what is the function im looking for?

  • $\begingroup$ The derivative of the position (here, distance covered) is the speed. $\endgroup$ – chubakueno Jun 1 '14 at 4:03
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    $\begingroup$ I thought the kinetic energy was $\frac{1}{2}mv^2$? $\endgroup$ – Cookie Jun 1 '14 at 4:09
  • $\begingroup$ It is, and OP reflects it properly now. $\endgroup$ – Alfred Yerger Jun 1 '14 at 7:17
  • $\begingroup$ the question i have has kinetic energy at e= mv^2/2 not 1/2mv^2 $\endgroup$ – user152431 Jun 1 '14 at 7:41

First note that $$v(t)=\frac{dr}{dt}=\frac{d}{dt}(t^3-3t^2+6t)=3t^2-6t+6$$ Then the kinetic energy $E$ is $E=\frac 12 mv^2=500(3t^2-6t+6)^2$

  • $\begingroup$ But the velocity should be squared...(and we are also supposing that the OP lives in a universe with doubled kinetic energy :) ) $\endgroup$ – chubakueno Jun 1 '14 at 4:12
  • $\begingroup$ I think the kinetic energy is $mv^2/2$! $\endgroup$ – Robert Lewis Jun 1 '14 at 4:18
  • $\begingroup$ @RobertLewis That's what I said as well (if you read my comment to OP's question) $\endgroup$ – Cookie Jun 1 '14 at 4:55
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    $\begingroup$ Yeah, it bears repeating! $\endgroup$ – Robert Lewis Jun 1 '14 at 5:01

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