Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `128 KJ` to `720 KJ` over `t in [0, 3 s]`. What is the average speed of the object?

Answer 1

The average speed is `=367.4ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

mass is `=6kg`

The initial velocity is `=u_1`

`1/2m u_1^2=128000J`

The final velocity is `=u_2`

`1/2m u_2^2=720000J`

Therefore,

`u_1^2=2/6*128000=42666.7m^2s^-2`

and,

`u_2^2=2/6*720000=240000m^2s^-2`

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

The points are `(0,42666.7)` and `(3,240000)`

The equation of the line is

`v^2-42666.7=(240000-42666.7)/3t`

`v^2=65777.8t+42666.7`

So,

`v=sqrt((65777.8t+42666.7)`

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

`(3-0)bar v=int_0^8sqrt((65777.8t+42666.7))dt`

`3 barv=[((65777.8t+42666.7)^(3/2)/(3/2*65666.7)]_0^3`

`=((65777.8*3+42666.7)^(3/2)/(98666.7))-((65777.8*0+42666.7)^(3/2)/(98666.7))`

`=240000^(3/2)/98666.7-42666.7^(3/2)/98666.7`

`=1102.3`

So,

`barv=1103.2/3=367.4ms^-1`