An object has a mass of `4 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?

`E_k = 1/2mv^2`

150kJ = 150,000J

16kJ = 16,000J

`150,000 = 1/2*4v^2`
`v = sqrt((300,000)/4) = 273.861279ms^-1`

`16,000 = 1/2*4v^2`
`v = sqrt((32,000)/4) = 63.2455532 ms^-1`

Proof 1:
`(63.2455532+273.861279)/2 = 168.553416ms^-1`

Proof 2:

`s=(t(u+v))/2` s = distance, t = time, v = final velocity, u = initial velocity.

`(8(273.861279+63.2455532))/2=1348.42733m`

`"average speed" = ("total distance")/("total time") = 1348.42733/8 = 168.553416ms^-1`

The kinetic energy is

`KE=1/2mv^2`

The mass is `=4kg`

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=150000J`

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

Therefore,

`u_1^2=2/4*150000=75000m^2s^-2`

and,

`u_2^2=2/4*16000=8000m^2s^-2`

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

The points are `(0,75000)` and `(8,8000)`

The equation of the line is

`v^2-75000=(8000-75000)/8t`

`v^2=-8375t+75000`

So,

`v=sqrt((-8375t+75000)`

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

`(8-0)bar v=int_0^3sqrt(-8375t+75000))dt`

`8 barv=[((-8375t+75000)^(3/2)/(-3/2*8375)]_0^8`

`=((-8375*8+75000)^(3/2)/(-12562.5))-((-8375*0+75000)^(3/2)/(-12562.5))`

`=75000^(3/2)/12562.5-8000^(3/2)/12562.5`

`=1578.03`

So,

`barv=1578.03/8=197.3ms^-1`

The average speed is `=197.3ms^-1`