An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `640 KJ` to `160 KJ` over `t in [0, 12 s]`. What is the average speed of the object?

The kinetic energy is

`KE=1/2mv^2`

mass is `=4kg`

The initial velocity is `=u_1`

`1/2m u_1^2=640000J`

The final velocity is `=u_2`

`1/2m u_2^2=160000J`

Therefore,

`u_1^2=2/4*640000=320000m^2s^-2`

and,

`u_2^2=2/4*160000=80000m^2s^-2`

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

The points are `(0,320000)` and `(12,80000)`

The equation of the line is

`v^2-320000=(80000-320000)/12t`

`v^2=-20000t+320000`

So,

`v=sqrt((-20000t+320000)`

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

`(12-0)bar v=int_0^12sqrt((-20000t+320000))dt`

`12 barv=[((-20000t+320000)^(3/2)/(-3/2*20000)]_0^12`

`=((-20000*12+320000)^(3/2)/(-30000))-((-20000*0+320000)^(3/2)/(-30000))`

`=320000^(3/2)/30000-80000^(3/2)/30000`

`=5279.7`

So,

`barv=5279.7/12=440ms^-1`

the kinetic energy of a body of mass m moving with speed v is defined as : `1/2mv^2`
thus you can calculate velocity of body at the starting and the end of interval of time :
lets say initially the velocity was `v_1` , `1/2*4*v_1^2 = 640000`
`v_1 = 565.68 ms^-1`
let at the end of interval of time the velocity was `v_2` , `1/2 * 4 * v_2^2 = 160000`
`v_2 = 282.84 ms^-1`
now , total distance travelled by the body over given time interval can be calculated by using `v_2^2 - v_1^2 = 2*a*s` and `v_2 = v_1 + a*t`
thus, `282.84 = 565.68 + 12a`
`a= -23.57ms^-2`
`282.84^2 - 565.68^2 = 2*-23.57*s`
`s= 5091.12m`
average speed is defined as ratio of total distance travelled and total time taken :
`v= s/t`
`v= 5091.12/12`
`v= 424.26ms^-1`

Let rate of change of kinetic energy be
`(dKE(t))/(dt) = C`
where `C` is constant with time `t`.

Integrating both sides with `t` we get
`int (dKE(t))/(dt) cdot dt = int C cdotdt`

`=>KE= Ccdott + c` ......(1)
where `c` is a constant of integration.

To evaluate `c`, use initial condition at `t=0`
`640xx10^3 = Ccdot0+c`
`=>c=640xx10^3`
We have expression for kinetic energy as
`KE= Ccdott + 640xx10^3`

To evaluate `C`, use final condition at `t=12`
`160 xx10^3= Ccdot12+640xx10^3`
`=>C=-40xx10^3`

Equation (1) becomes
`KE= (-t + 16)xx4xx10^4` ......(2)

If `m` is the mass and `v(t)` is the velocity of body, Kinetic energy is given as
`KE=1/2mv^2`
`=> 1/2mv^2(t) = (-t + 16)xx4xx10^4`
Given is `m=4kg`. Above expression becomes
`=> 1/2xx4v^2(t) = (-t + 16)xx4xx10^4`
`=> v(t)^2 = (-t+16)xx2xx10^4`
`=> v= +-sqrt((-t+16)xx2xx10^4)`
`=> "speed "|v|= sqrt((-t+16)xx2xx10^4)`

Average speed `="Total Distance traveled"/"Time taken"=(int_0^12|v(t)|*dt)/(12-0)`

`:.` Average speed `= (int_(0)^(12)sqrt((-t+16)xx2xx10^4)*dt)/12`
`= 100sqrt(2)/12[-(16-t)^(3/2)/(3/2)]_0^12`
`=- 50/9sqrt(2)(4^(3/2) - 16^(3/2))`
`= 439.98 ms^-1`, rounded to two decimal places.