Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `72 KJ` to `720 KJ` over `t in [0, 3 s]`. What is the average speed of the object?

Answer 1

The average speed is `=351.4ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=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=72000J`

The final kinetic energy is `1/2m u_2^2=720000J`

Therefore,

`u_1^2=2/6*72000=24000m^2s^-2`

and,

`u_2^2=2/6*720000=24000m^2s^-2`

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

The points are `(0,24000)` and `(3,240000)`

The equation of the line is

`v^2-24000=(240000-24000)/3t`

`v^2=72000t+24000`

So,

`v=sqrt((72000t+24000)`

We need to calculate the average value of `v` over `t in [0,3]`

`(3-0)bar v=int_0^3sqrt(72000t+24000))dt`

`3 barv=[((72000t+24000)^(3/2)/(3/2*72000)]_0^3`

`=((72000*3+24000)^(3/2)/(108000))-((72000*0+24000)^(3/2)/(108000))`

`=240000^(3/2)/108000-24000^(3/2)/108000`

`=1054.2`

So,

`barv=1054.2/3=351.4ms^-1`

The average speed is `=351.4ms^-1`