Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `48 KJ` to `360 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=310.5ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=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=48000J`

The final kinetic energy is `1/2m u_2^2=360000J`

Therefore,

`u_1^2=2/4*48000=24000m^2s^-2`

and,

`u_2^2=2/4*360000=180000m^2s^-2`

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

The points are `(0,24000)` and `(12,180000)`

The equation of the line is

`v^2-24000=(180000-24000)/12t`

`v^2=13000t+24000`

So,

`v=sqrt((13000t+24000)`

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

`(12-0)bar v=int_0^12sqrt((13000t+24000))dt`

`12 barv=[((13000t+24000)^(3/2)/(3/2*13000)]_0^12`

`=((13000*12+24000)^(3/2)/(19500))-((13000*0+24000)^(3/2)/(19500))`

`=180000^(3/2)/19500-24000^(3/2)/19500`

`=3725.6`

So,

`barv=3725.6/12=310.5ms^-1`

The average speed is `=310.5ms^-1`