Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `48 KJ` to `15 KJ` over `t in [0, 8 s]`. What is the average speed of the object?

Answer 1

The average speed is `=101.23ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=6kg`

The initial velocity is `=u_1`

`1/2m u_1^2=48000J`

The final velocity is `=u_2`

`1/2m u_2^2=15000J`

Therefore,

`u_1^2=2/6*48000=16000m^2s^-2`

and,

`u_2^2=2/6*15000=5000m^2s^-2`

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

The points are `(0,16000)` and `(8,5000)`

The equation of the line is

`v^2-16000=(5000-16000)/8t`

`v^2=-1375t+16000`

So,

`v=sqrt((-1375t+16000)`

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

`(8-0)bar v=int_0^8sqrt((-1375t+16000))dt`

`8 barv=[((-1375t+16000)^(3/2)/(-3/2*1375)]_0^8`

`=((-1375*8+16000)^(3/2)/(-2062.5))-((-1375*0+16000)^(3/2)/(-2062.5))`

`=16000^(3/2)/2062.5-5000^(3/2)/2062.5`

`=809.84`

So,

`barv=809.84/8=101.23ms^-1`