Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `18 KJ` to `64KJ` over `t in [0,12s]`. What is the average speed of the object?

Answer 1

The average speed is `=115.3ms^-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=18000J`

The final kinetic energy is `1/2m u_2^2=64000J`

Therefore,

`u_1^2=2/6*18000=6000m^2s^-2`

and,

`u_2^2=2/6*64000=21333.3m^2s^-2`

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

The points are `(0,6000)` and `(12,21333.3)`

The equation of the line is

`v^2-6000=(21333.3-6000)/12t`

`v^2=1277.8t+6000`

So,

`v=sqrt(1277.8t+6000)`

We need to calculate the average value of `v` over `t in [0,12]`

`(12-0)bar v=int_0^12(sqrt(1277.8t+6000))dt`

`12 barv= [(1277.8t+6000)^(3/2)/(3/2*1277.8)] _( 0) ^ (12)`

`=((1277.8*12+6000)^(3/2)/(1916.7))-((1277.8*0+6000)^(3/2)/(1916.7))`

`=21333.3^(3/2)/1916.7-6000^(3/2)/1916.7`

`=1383.2`

So,

`barv=1383.2/12=115.3ms^-1`

The average speed is `=115.3ms^-1`