Question

An object has a mass of `3 kg`. The object's kinetic energy uniformly changes from `36 KJ` to `135 KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed is `=235.2ms^-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=36000J`

The final kinetic energy is `1/2m u_2^2=135000J`

Therefore,

`u_1^2=2/3*36000=24000m^2s^-2`

and,

`u_2^2=2/3*135000=90000m^2s^-2`

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

The points are `(0,24000)` and `(9, 90000)`

The equation of the line is

`v^2-24000=(90000-24000)/9t`

`v^2=7333.3t+24000`

So,

`v=sqrt(7333.3t+24000)`

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

`(9-0)bar v=int_0^9(sqrt(7333.3t+24000))dt`

`9 barv= [(7333.3t+24000)^(3/2)/(3/2*7333.3)] _( 0) ^ (9)`

`=((7333.3*9+24000)^(3/2)/(11000))-((7333.3*0+24000)^(3/2)/(11000))`

`=90000^(3/2)/11000-24000^(3/2)/11000`

`=2116.5`

So,

`barv=2116.5/9=235.2ms^-1`

The average speed is `=235.2ms^-1`