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?
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 $E=\frac 12 mv^2=500(3t^2-6t+6)^2$