Question

An object has a mass of `2 kg`. The object's kinetic energy uniformly changes from `64 KJ` to `28 KJ` over `t in [0, 15 s]`. What is the average speed of the object?

Answer 1

The average speed is `=319.6ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=2kg`

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=64000J`

The final kinetic energy is `1/2m u_2^2=28000J`

Therefore,

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

and,

`u_2^2=2/2*28000=28000m^2s^-2`

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

The points are `(0,64000)` and `(15,28000)`

The equation of the line is

`v^2-64000=(28000-64000)/15t`

`v^2=-2400t+64000`

So,

`v=sqrt(-2400t+64000)`

We need to calculate the average value of `v` over `t in [0,15]`

`(15-0)bar v=int_0^15(sqrt(-2400t+64000))dt`

`15 barv= (-2400t+64000)^(3/2)/(3/2*-2400)| _( 0) ^ (15)`

`=((-2400*15+64000)^(3/2)/(-2400))-((-2400*0+64000)^(3/2)/(-2400))`

`=64000^(3/2)/2400-28000^(3/2)/2400`

`=4794`

So,

`barv=4794/15=319.6ms^-1`

The average speed is `=319.6ms^-1`