Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `640 KJ` to `360 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=406.9ms^-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=640000J`

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

Therefore,

`u_1^2=2/6*640000=213333.3m^2s^-2`

and,

`u_2^2=2/6*360000=120000m^2s^-2`

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

The points are `(0,213333.3)` and `(12,120000)`

The equation of the line is

`v^2-213333.3=(120000-213333.3)/12t`

`v^2=-7777.8t+213333.3`

So,

`v=sqrt(-7777.8t+213333.3)`

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

`(12-0)bar v=int_0^12(sqrt(-7777.8t+213333.3))dt`

`12 barv= (-7777.8t+213333.3)^(3/2)/(3/2*-7777.8)| _( 0) ^ (12)`

`=((-7777.8*12+213333.3)^(3/2)/(-11666.7))-((-7777.8*0+213333.3)^(3/2)/(-11666.7))`

`=213333.3^(3/2)/11666.7-120000^(3/2)/11666.7`

`=4882.7`

So,

`barv=4882.7/12=406.9ms^-1`

The average speed is `=406.9ms^-1`