Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `12 KJ` to `72KJ` over `t in [0,5s]`. What is the average speed of the object?

Answer 1

The average speed is `=141.5ms^-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=12000J`

The final kinetic energy is `1/2m u_2^2=72000J`

Therefore,

`u_1^2=2/4*12000=6000m^2s^-2`

and,

`u_2^2=2/4*72000=36000m^2s^-2`

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

The points are `(0,6000)` and `(5,36000)`

The equation of the line is

`v^2-6000=(36000-6000)/5t`

`v^2=6000t+6000`

So,

`v=sqrt(6000t+6000)`

We need to calculate the average value of `v` over `t in [0,5]`

`(5-0)bar v=int_0^5(sqrt(6000t+6000))dt`

`5 barv= (6000t+6000)^(3/2)/(3/2*6000)| _( 0) ^ (5)`

`=((6000*5+6000)^(3/2)/(9000))-((6000*0+6000)^(3/2)/(9000))`

`=36000^(3/2)/9000-6000^(3/2)/9000`

`=707.31`

So,

`barv=707.31/5=141.5ms^-1`

The average speed is `=141.5ms^-1`

Answer 2

133.63 m/s

Explanation:

Using equation of kinetic energy, i.e E =`(1/2)m(v^2)`
We get initial velocity=u was 77.46 m/sec and final velocity=v was 189.73 m/sec
Now,using v=u+at(all symbols are bearing their conventional meaning) we get a is 22.45 `m/(s^2)`
Hence putting it in `v^2 = u^2 +2as` (all symbols are bearing their conventional meaning)
We get total distance covered in 5 sec is 668.152 meter.
So, average speed = (total distance covered/total time) = `(668.152/5)` i.e 133.63 m/s