Question

An object has a mass of `5 kg`. The object's kinetic energy uniformly changes from `150 KJ` to `16KJ` over `t in [0,8s]`. What is the average speed of the object?

Answer 1

The average speed is `=176.4ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=5kg`

The initial velocity is `=u_1`

The initial kinetic energy is `1/2m u_1^2=150000J`

The final velocity is `=u_2`

The final kinetic energy is `1/2m u_2^2=16000J`

Therefore,

`u_1^2=2/5*150000=60000m^2s^-2`

and,

`u_2^2=2/5*16000=6400m^2s^-2`

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

The points are `(0,60000)` and `(8,6400)`

The equation of the line is

`v^2-60000=(6400-60000)/8t`

`v^2=-6700t+60000`

So,

`v=sqrt((-6700t+60000)`

We need to calculate the average value of `v` over `t in [0,8]`

`(8-0)bar v=int_0^8sqrt((-6700t+60000))dt`

`8 barv=[((-6700t+60000)^(3/2)/(-3/2*6700)]_0^8`

`=((-6700*8+60000)^(3/2)/(-10050))-((-6700*0+60000)^(3/2)/(-10050))`

`=60000^(3/2)/10050-6400^(3/2)/10050`

`=1411.4`

So,

`barv=1411.4/8=176.4ms^-1`

The average speed is `=176.4ms^-1`