Question

Using the equation for kinetic energy: `k=1/2mv^2` Find the instantaneous rate of change of the kinetic energy of a `1500 kg` car which has a velocity of `80 m/s` and an acceleration of `10m/s^2`?

Answer 1

Kinetic energy `KE =1/2mv^2`
Instantaneous rate of change of KE is its differential with time
`dot (KE)=d/dt (1/2mv^2)`
`=>dot (KE)=1/2m*2v* (dv)/dt`
`=>(dot KE)=m*v*a`
Where `dot v=a`, acceleration.
Inserting given instantaneous values we get

`=>dot (KE)=1500*80*10`
`=>dot (KE)=1.2*10^6Js^-1`