Question

An object has a mass of `8 kg`. The object's kinetic energy uniformly changes from `240 KJ` to `640KJ` over `t in [0, 3 s]`. What is the average speed of the object?

Answer 1

The average speed is `=328.7ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=8kg`

The initial velocity is `=u_1`

`1/2m u_1^2=240000J`

The final velocity is `=u_2`

`1/2m u_2^2=640000J`

Therefore,

`u_1^2=2/8*240000=60000m^2s^-2`

and,

`u_2^2=2/8*640000=160000m^2s^-2`

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

The points are `(0,60000)` and `(3,160000)`

The equation of the line is

`v^2-60000=(160000-60000)/3t`

`v^2=33333.3t+60000`

So,

`v=sqrt((33333.3t+60000)`

We need to calculate the average value of `v` over `t in [0,3]`

`(3-0)bar v=int_0^3sqrt((33333.3t+60000))dt`

`3 barv=[((33333.3t+60000)^(3/2)/(3/2*33333.3)]_0^3`

`=((33333.3*3+60000)^(3/2)/(50000))-((33333.3*0+60000)^(3/2)/(50000))`

`=160000^(3/2)/50000-60000^(3/2)/50000`

`=986.1`

So,

`barv=986.1/3=328.7ms^-1`