Question

An object has a mass of `3 kg`. The object's kinetic energy uniformly changes from `48 KJ` to `38KJ` over `t in [0,12s]`. What is the average speed of the object?

Answer 1

The average speed is `=169.2ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=3kg`

The initial velocity is `=u_1ms^-1`

The final velocity is `=u_2 ms^-1`

The initial kinetic energy is `1/2m u_1^2=48000J`

The final kinetic energy is `1/2m u_2^2=38000J`

Therefore,

`u_1^2=2/3*48000=32000m^2s^-2`

and,

`u_2^2=2/3*38000=25333.3m^2s^-2`

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

The points are `(0,32000)` and `(12,25333.3)`

The equation of the line is

`v^2-32000=(25333.3-32000)/12t`

`v^2=-555.6t+32000`

So,

`v=sqrt(-555.6t+32000)`

We need to calculate the average value of `v` over `t in [0,12]`

`(12-0)bar v=int_0^12(sqrt(-555.6t+32000))dt`

`12 barv= [(-555.6t+32000)^(3/2)/(-3/2*555.6)] _( 0) ^ (12)`

`=((-555.6*8+32000)^(3/2)/(-833.3))-((-555.6*0+32000)^(3/2)/(-833.3))`

`=32000^(3/2)/833.3-25333.3^(3/2)/833.3`

`=2030.7`

So,

`barv=2030.7/12=169.2ms^-1`

The average speed is `=169.2ms^-1`