Question

An object has a mass of `8 kg`. The object's kinetic energy uniformly changes from `24 KJ` to `42KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed is `90.5ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=8kg`

The initial velocity is `=u_1`

`1/2m u_1^2=24000J`

The final velocity is `=u_2`

`1/2m u_2^2=42000J`

Therefore,

`u_1^2=2/8*24000=6000m^2s^-2`

and,

`u_2^2=2/8*42000=10500m^2s^-2`

The graph of `v^2=f(t)` is a straight line

The points are `(0,6000)` and `(9,10500)`

The equation of the line is

`v^2-6000=(10500-6000)/9t`

`v^2=500t+6000`

So,

`v=sqrt((500t+6000)`

We need to calculate the average value of `v` over `t in [0,9]`

`(9-0)bar v=int_0^9sqrt((500t+6000))dt`

`9 barv=[((500t+6000)^(3/2)/(3/2*500)]_0^9`

`=((500*9+6000)^(3/2)/(750))-((500*0+6000)^(3/2)/(750))`

`=10500^(3/2)/750-6000^(3/2)/750`

`=814.9`

So,

`barv=814.9/9=90.5ms^-1`