Question

An object has a mass of `12 kg`. The object's kinetic energy uniformly changes from `64 KJ` to `24 KJ` over `t in [0, 5 s]`. What is the average speed of the object?

Answer 1

The average speed is `=84.9ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=12kg`

The initial velocity is `=u_1`

The initial kinetic energy is `1/2m u_1^2=64000J`

The final velocity is `=u_2`

The final kinetic energy is `1/2m u_2^2=24000J`

Therefore,

`u_1^2=2/12*64000=10666.7m^2s^-2`

and,

`u_2^2=2/12*24000=4000m^2s^-2`

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

The points are `(0,10666.7)` and `(5,4000)`

The equation of the line is

`v^2-10666.7=(4000-10666.7)/5t`

`v^2=-1333.3t+10666.7`

So,

`v=sqrt((-1333.3t+10666.7)`

We need to calculate the average value of `v` over `t in [0,5]`

`(5-0)bar v=int_0^5sqrt((-1333.3t+10666.7))dt`

`5 barv=[((-1333.3t+10666.7)^(3/2)/(-3/2*1333.3)]_0^5`

`=((-1333.3*5+10666.7)^(3/2)/(2000))-((-1333.3*0+10666.7)^(3/2)/(2000))`

`=10666.7^(3/2)/2000-4000^(3/2)/2000`

`=424.3`

So,

`barv=424.3/5=84.9ms^-1`

The average speed is `=84.9ms^-1`