Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `0 KJ` to `96KJ` over `t in [0, 6 s]`. What is the average speed of the object?

Answer 1

The average speed is `=9.94ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=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=0J`

The final kinetic energy is `1/2m u_2^2=96000J`

Therefore,

`u_1^2=2/4*0=0m^2s^-2`

and,

`u_2^2=2/4*96000=48000m^2s^-2`

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

The points are `(0,0)` and `(6,48000)`

The equation of the line is

`v^2=(48000-0)/6t`

`v^2=8000t`

So,

`v=sqrt(8000t)`

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

`(6-0)bar v=int_0^6(sqrt(8000t))dt`

`6 barv= [(8000t)^(3/2)/(3/2*8000)] _( 0) ^ (6)`

`=((8000*6)^(3/2)/(12000))-((8000*0)^(3/2)/(12000))`

`=8000^(3/2)/12000`

`=59.63`

So,

`barv=58.63/6=9.94ms^-1`

The average speed is `=9.94ms^-1`