Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `120 KJ` to `64KJ` over `t in [0,6s]`. What is the average speed of the object?

Answer 1

The average speed is `=213.6ms^-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=120000J`

The final kinetic energy is `1/2m u_2^2=64000J`

Therefore,

`u_1^2=2/4*120000=60000m^2s^-2`

and,

`u_2^2=2/4*64000=32000m^2s^-2`

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

The points are `(0,60000)` and `(6,32000)`

The equation of the line is

`v^2-60000=(32000-60000)/6t`

`v^2=-4666.6t+60000`

So,

`v=sqrt((-4666.7t+60000)`

We need to calculate the average value of `v` over `t in [0,6]`

`(6-0)bar v=int_0^6(sqrt(-4666.7t+60000))dt`

`6 barv=[((-4666.7t+60000)^(3/2)/(-3/2*3500))]_0^6`

`=((-4666.7*6+60000)^(3/2)/(-7000))-((-4666.7*0+60000)^(3/2)/(-7000))`

`=60000^(3/2)/7000-32000^(3/2)/7000`

`=1281.8`

So,

`barv=1281.8/6=213.6ms^-1`

The average speed is `=213.6ms^-1`