Question

An object has a mass of `3 kg`. The object's kinetic energy uniformly changes from `96 KJ` to `240 KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed is `=332.01ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=3kg`

The initial velocity is `=u_1`

`1/2m u_1^2=96000J`

The final velocity is `=u_2`

`1/2m u_2^2=240000J`

Therefore,

`u_1^2=2/3*96000=64000m^2s^-2`

and,

`u_2^2=2/3*240000=160000m^2s^-2`

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

The points are `(0,64000)` and `(9,160000)`

The equation of the line is

`v^2-64000=(160000-64000)/9t`

`v^2=10666.67t+64000`

So,

`v=sqrt((10666.67t+64000)`

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

`(9-0)bar v=int_0^9sqrt((10666.67t+64000))dt`

`9 barv=[((10666.67t+64000)^(3/2)/(3/2*10666.67)]_0^9`

`=((10666.67*9+64000)^(3/2)/(16000))-((10666.67*0+64000)^(3/2)/(16000))`

`=160000^(3/2)/16000-64000^(3/2)/16000`

`=2988.07`

So,

`barv=2988.07/9=332.01ms^-1`