Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `24 KJ` to `42KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

`4 m/s`

Explanation:

Let the rate of increase of kinetic energy with time be some `m (KJ)/s` where `m` is a constant as given in the question.

`therefore (dE(t))/(dt) = m` where `E` is Kinetic Energy and `t` is time.

`implies dE(t)=m*dt`
`implies` `intdE = int m*dt`

`implies` `E(t) = m*t + c` where `c` is a constant of integration.
We have two conditions:-
1. `E(0) = 24` &
2. `E(9) = 42`

from 1. & 2.
`24 = c` & `42 = 9*m + 24 implies m=2`
`therefore E(t) = 2*t + 24`

`implies (1/2)*m*v^2 = 2*t+24` where `m` is the mass & `v` is the velocity (speed) of body.

`implies v^2 = t+12`
`implies v = sqrt(t+12)`

Average speed = `(int_(t_1)^(t_2)v*dt)/(int_(t_1)^(t_2)dt)`
where
`t_1 = 0` & `t_2=9`

`therefore` Avg. speed `= (int_(0)^(9)sqrt(t+12)*dt)/(int_(0)^(9)dt)`
`= (2/3*(t+12)^(3/2))/(t)]_0^9 = 2/3*(21sqrt21 - 12sqrt12)/9 approx 4`

Answer 2

`128.05 ms^-1`, rounded to two decimal places.

Explanation:

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`
`24xx10^3 = Ccdot0+c`
`=>c=24xx10^3`
We have expression for kinetic energy as
`KE= Ccdott + 24xx10^3`

To evaluate `C`, use final condition at `t=9`
`42 xx10^3= Ccdot9+24xx10^3`
`=>C=2xx10^3`

Equation (1) becomes
`KE= (2cdott + 24)xx10^3` ......(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) = (2*t+24)xx10^3`
Given is `m=4kg`. Above expression becomes
`=> 1/2xx4v^2(t) = (2*t+24)xx10^3`
`=> v(t)^2 = (t+12)xx10^3`
`=> v= +-sqrt((t+12)xx10^3)`
`=> "speed "|v|= sqrt((t+12)xx10^3)`

Average speed `="Total Distance traveled"/"Time taken"=(int_0^9|v(t)|*dt)/(9-0)`

`:.` Average speed `= (int_(0)^(9)sqrt((t+12)xx10^3)*dt)/9`
`= sqrt(10^3)/9[(t+12)^(3/2)/(3/2)]_0^9`
`= (2sqrt(10^3))/27(21^(3/2) - 12^(3/2))`
`= 128.05 ms^-1`, rounded to two decimal places.