Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `480 KJ` to `12 KJ` over `t in [0, 8 s]`. What is the average speed of the object?

Answer 1

The average speed is `=272.4ms^-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=480000J`

The final kinetic energy is `1/2m u_2^2=12000J`

Therefore,

`u_1^2=2/6*480000=160000m^2s^-2`

and,

`u_2^2=2/6*12000=4000m^2s^-2`

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

The points are `(0,160000)` and `(8,4000)`

The equation of the line is

`v^2-160000=(4000-160000)/8t`

`v^2=-19500t+160000`

So,

`v=sqrt(-19500t+160000)`

We need to calculate the average value of `v` over `t in [0,8]`

`(8-0)bar v=int_0^8(sqrt(-19500t+160000))dt`

`8 barv= [(-19500t+160000)^(3/2)/(-3/2*19500)] _( 0) ^ (8)`

`=((-19500*8+160000)^(3/2)/(-29250))-((-19500*0+160000)^(3/2)/(-29250))`

`=160000^(3/2)/29250-4000^(3/2)/29250`

`=2179.4`

So,

`barv=2179.4/8=272.4ms^-1`

The average speed is `=272.4ms^-1`