Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `450 KJ` to `129 KJ` over `t in [0, 8 s]`. What is the average speed of the object?

Answer 1

The average speed is `=306.4ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=6kg`

The initial velocity is `=u_1`

`1/2m u_1^2=450000J`

The final velocity is `=u_2`

`1/2m u_2^2=129000J`

Therefore,

`u_1^2=2/6*450000=150000m^2s^-2`

and,

`u_2^2=2/6*129000=43000m^2s^-2`

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

The points are `(0,150000)` and `(8,43000)`

The equation of the line is

`v^2-150000=(43000-150000)/8t`

`v^2=-13375t+150000`

So,

`v=sqrt((-13375t+150000)`

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

`(8-0)bar v=int_0^8sqrt((-13375t+150000))dt`

`8 barv=[((-13375t+150000)^(3/2)/(-3/2*13375)]_0^8`

`=((-13375*8+150000)^(3/2)/(-20062.5))-((-13375*0+150000)^(3/2)/(-20062.5))`

`=43000^(3/2)/-20062.5-150000^(3/2)/-20062.5`

`=2451.2`

So,

`barv=22451.2/8=306.4ms^-1`