Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `54 KJ` to `225 KJ` over `t in [0, 8 s]`. What is the average speed of the object?

Answer 1

The average speed is `=212.0ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=6kg`

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=54000J`

The final kinetic energy is `1/2m u_2^2=225000J`

Therefore,

`u_1^2=2/6*54000=18000m^2s^-2`

and,

`u_2^2=2/6*225000=75000m^2s^-2`

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

The points are `(0,18000)` and `(8,75000)`

The equation of the line is

`v^2-18000=(75000-18000)/8t`

`v^2=7125t+18000`

So,

`v=sqrt(7125t+18000)`

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

`(8-0)bar v=int_0^12(sqrt(7125t+18000))dt`

`8 barv= (7125t+18000)^(3/2)/(3/2*7125)| _( 0) ^ (8)`

`=((7125*8+18000)^(3/2)/(10687.5))-((7125*0+18000)^(3/2)/(10687.5))`

`=75000^(3/2)/10687.5-18000^(3/2)/10687.5`

`=1695.9`

So,

`barv=1695.9/8=212.0ms^-1`

The average speed is `=212.0ms^-1`