Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `96 KJ` to `320 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=318.4ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=4kg`

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=96000J`

The final kinetic energy is `1/2m u_2^2=320000J`

Therefore,

`u_1^2=2/4*96000=48000m^2s^-2`

and,

`u_2^2=2/4*320000=160000m^2s^-2`

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

The points are `(0,48000)` and `(12,160000)`

The equation of the line is

`v^2-48000=(160000-48000)/12t`

`v^2=9333.3t+48000`

So,

`v=sqrt(9333.3t+48000)`

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

`(12-0)bar v=int_0^12(sqrt(9333.3t+48000))dt`

`12 barv= [(9333.3t+48000)^(3/2)/(3/2*9333.3)] _( 0) ^ (12)`

`=((9333.3*12+48000)^(3/2)/(14000))-((9333.3*0+48000)^(3/2)/(14000))`

`=16000^(3/2)/14000-48000^(3/2)/14000`

`=3820.3`

So,

`barv=3820.3/12=318.4ms^-1`

The average speed is `=318.4ms^-1`