Question

An object has a mass of `8 kg`. The object's kinetic energy uniformly changes from `640 KJ` to `320 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=344.8ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=8kg`

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=640000J`

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

Therefore,

`u_1^2=2/8*640000=160000m^2s^-2`

and,

`u_2^2=2/8*320000=80000m^2s^-2`

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

The points are `(0,160000)` and `(12,80000)`

The equation of the line is

`v^2-160000=(80000-160000)/12t`

`v^2=-6666.7t+160000`

So,

`v=sqrt((-6666.7t+160000)`

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

`(12-0)bar v=int_0^12(sqrt(-6666.7t+160000))dt`

`12 barv=[((-6666.7t+160000)^(3/2)/(-3/2*6666.7))]_0^12`

`=((-6666.7*12+160000)^(3/2)/(-10000))-((-1666.7*0+160000)^(3/2)/(-10000))`

`=160000^(3/2)/10000-80000^(3/2)/10000`

`=4137.3`

So,

`barv=4137.3/12=344.8ms^-1`

The average speed is `=344.8ms^-1`