Question

An object has a mass of `5 kg`. The object's kinetic energy uniformly changes from `36 KJ` to `135 KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed is `=182.2ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=5kg`

The initial velocity is `=u_1`

`1/2m u_1^2=36000J`

The final velocity is `=u_2`

`1/2m u_2^2=135000J`

Therefore,

`u_1^2=2/5*36000=14400m^2s^-2`

and,

`u_2^2=2/5*135000=54000m^2s^-2`

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

The points are `(0,14400)` and `(9,54000)`

The equation of the line is

`v^2-14400=(54000-14400)/9t`

`v^2=4400t+14400`

So,

`v=sqrt((4400t+14400)`

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

`(9-0)bar v=int_0^9sqrt((4400t+14400))dt`

`9 barv=[((4400t+14400)^(3/2)/(3/2*4400)]_0^9`

`=((4400*9+14400)^(3/2)/(6600))-((4400*0+14400)^(3/2)/(6600))`

`=54000^(3/2)/6600-14400^(3/2)/6600`

`=1639.5`

So,

`barv=1639.5/9=182.2ms^-1`