An object has a mass of `2 kg`. The object's kinetic energy uniformly changes from `16 KJ` to `36KJ` over `t in [0, 6 s]`. What is the average speed of the object?

The kinetic energy is

`KE=1/2mv^2`

mass is `=2kg`

The initial velocity is `=u_1`

`1/2m u_1^2=16000J`

The final velocity is `=u_2`

`1/2m u_2^2=36000J`

Therefore,

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

and,

`u_2^2=2/2*36000=36000m^2s^-2`

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

The points are `(0,16000)` and `(6,36000)`

The equation of the line is

`v^2-16000=(36000-16000)/6t`

`v^2=3333.3t+16000`

So,

`v=sqrt((3333.3t+16000)`

We need to calculate the average value of `v` over `t in [0,6]`

`(6-0)bar v=int_0^8sqrt((3333.3t+16000))dt`

`6 barv=[((3333.3t+16000)^(3/2)/(3/2*3333.3)]_0^6`

`=((3333.3*6+16000)^(3/2)/(5000))-((3333.3*0+16000)^(3/2)/(5000))`

`=36000^(3/2)/5000-16000^(3/2)/5000`

`=961.3`

So,

`barv=961.3/6=160.2ms^-1`

During the period we can say that:

`(dT)/(dt) = (36,000-16,000)/(6) = (10,000)/3 " Joules / s"`

Integrating:

`T(t) = 1/2 m v(t) ^2 = int (10,000)/3 \ dt = (10,000)/3 t + alpha`

`T(0) = 16,000 implies alpha = 16,000`

And as `m = 2`, we have:

`v(t) = sqrt ((10,000)/3 t + 16,000)`

For average velocity `bar v`:

`bar v = (int_0^6 v(t) \ dt)/(6 - 0)` (simply put, this calculates that constant velocity that would cover the same distance (sense displacement!) in the same time)

`=([ ((10,000 t)/3 + 16,000)^(3/2) ]_0^6 )/(6 cdot 5,000) approx 160.2 " m/s"`

Reality check:

`v = sqrt ((2T)/m)`

`v(0) approx 126.5 " m/s"`

`v(6) approx 190 " m/s"`