Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `18 KJ` to `4KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed of the object is `=72.8ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=4kg`

The initial velocity is `=u_1`

`1/2m u_1^2=18000J`

The final velocity is `=u_2`

`1/2m u_2^2=4000J`

Therefore,

`u_1^2=2/4*18000=9000m^2s^-2`

and,

`u_2^2=2/4*4000=2000m^2s^-2`

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

The points are `(0,9000)` and `(9,2000)`

The equation of the line is

`v^2-9000=(2000-9000)/9t`

`v^2=-777.8t+9000`

So,

`v=sqrt((-777.8t+9000)`

We need to calculate the average value of `v` over `t in [0,9]`

`(9-0)bar v=int_0^9sqrt((-777.8t+9000))dt`

`9 barv=[((-777.8t+9000)^(3/2)/(-3/2*777.8)]_0^9`

`=((-777.8*9+9000)^(3/2)/(-1166.7))-((-777.8*0+9000)^(3/2)/(-1166.7))`

`=-2000^(3/2)/1166.7+9000^(3/2)/1166.7`

`=655.2`

So,

`barv=655.2/9=72.8ms^-1`