Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `12 KJ` to `36 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

Answer 1

The average speed is `=108.3ms^-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=36000J`

Therefore,

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

and,

`u_2^2=2/4*36000=18000m^2s^-2`

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

The points are `(0,6000)` and `(12,18000)`

The equation of the line is

`v^2-6000=(18000-6000)/12t`

`v^2=1000t+6000`

So,

`v=sqrt(1000t+6000)`

We need to calculate the average value of `v` over `t in [0,12]`

`(12-0)bar v=int_0^12(sqrt(1000t+6000))dt`

`12 barv=[1000t+6000)^(3/2)/(3/2*1000))] _( 0) ^ (12)`

`=((1000*12+6000)^(3/2)/(1500))-((1000*0+6000)^(3/2)/(1500))`

`=18000^(3/2)/1500-6000^(3/2)/1500`

`=1300.1`

So,

`barv=1300.1/12=108.3ms^-1`

The average speed is `=108.3ms^-1`

Answer 2

Kinetic energy `KE` of an object of mass `m` moving with velocity `v` is given by the expression

`KE=1/2mv^2`

Given `m=4kg`
Let initial velocity be `=u" "ms^-1`
The final velocity be `=v" " ms^-1`

Initial kinetic energy

`1/2m u^2=12000J`

Final kinetic energy is

`1/2m v^2=36000J`

As kinetic energy changes uniformly, the graph of `t" vs "KE` is a straight line.

The points on the graph are `(0, 12000)` and `(12, 36000)`

The equation of the line is of the type `y=mx+c`

`1/2m(v(t))^2=(36000-12000)/12t+12000`
or `1/2xx4(v(t))^2=2000t+12000`
`=>(v(t))^2=1000t+6000`
`=>v(t)=sqrt(1000t+6000)`

We know that velocity can be written as `(ds)/dt` where `s` is distance moved

`=>(ds(t))/dt=sqrt(1000t+6000)`

We need to calculate total distance traveled in time `t` from `0` to`12s` by integrating over the interval

`:.s=int_0^12sqrt(1000t+6000)cdotdt`
`=>s=[(1000t+6000)^(3/2)/(3/2xx1000)] _( 0) ^ (12)`
`=>s=1/1500[(1000xx12+6000)^(3/2)-(1000xx0+6000)^(3/2)]`
`=>s=1/1500(18000^(3/2)-6000^(3/2))`
`=>s=1300.13m`

We know that
`"Average speed"="Total distance traveled"/"Time of travel"`
`=>barv=1300.13/12=108.3ms^-1`, rounded to one decimal place