Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `64 KJ` to `180 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=244.5ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=4kg`

The initial velocity is `=u_1`

The initial kinetic energy is `1/2m u_1^2=64000J`

The final velocity is `=u_2`

The final kinetic energy is `1/2m u_2^2=180000J`

Therefore,

`u_1^2=2/4*64000=32000m^2s^-2`

and,

`u_2^2=2/4*180000=90000m^2s^-2`

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

The points are `(0,32000)` and `(12,90000)`

The equation of the line is

`v^2-32000=(90000-32000)/12t`

`v^2=4833.3t+32000`

So,

`v=sqrt((4833.3t+32000)`

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

`(12-0)bar v=int_0^12sqrt((4833.3t+32000))dt`

`12 barv=[((4833.3t+32000)^(3/2)/(3/2*4833.3)]_0^12`

`=((4833.3*12+32000)^(3/2)/(7250))-((4833.3*0+32000)^(3/2)/(7250))`

`=90000^(3/2)/7250-32000^(3/2)/7250`

`=2934.6`

So,

`barv=2934.6/12=244.5ms^-1`

The average speed is `=244.5ms^-1`