Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `144 KJ` to `120KJ` over `t in [0, 3 s]`. What is the average speed of the object?

Answer 1

Average speed = `(v_0 + v ) /2 approx 256.64 m/s`

Average acceleration = `(v_0 - v ) /3 approx -7.794 m/(s^2)`

Explanation:

For this question we must cosnider the formula for `K.E`

`K.E = 1/2 * m * v^2`

Initial Kinetic energy`=>` `K.E_0= 144*1000 J`

`=> 1/2 * 4 * v_0^2 = 144000`

`=> v_0 approx 268.33 m/s`

Final Kinetic energy, `=> K.E = 120*1000J`

`=> 1/2 * 4 * v^2 = 120000`

`=> v approx 244.95 m/s`

Average speed = `(v_0 + v ) /2 approx 256.64 m/s`

Average acceleration = `(v_0 - v ) /3 approx -7.794 m/(s^2)`

Answer 2

The average speed is `=256.8ms^-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=144000J`

The final kinetic energy is `1/2m u_2^2=120000J`

Therefore,

`u_1^2=2/4*144000=72000m^2s^-2`

and,

`u_2^2=2/4*120000=60000m^2s^-2`

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

The points are `(0,72000)` and `(3,60000)`

The equation of the line is

`v^2-72000=(60000-72000)/3t`

`v^2=-4000t+72000`

So,

`v=sqrt(-4000t+72000)`

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

`(3-0)bar v=int_0^3(sqrt(-4000t+72000))dt`

`3 barv= (-4000t+72000)^(3/2)/(3/2*-4000)| _( 0) ^ (3)`

`=((-4000*3+72000)^(3/2)/(-6000))-((-4000*0+72000)^(3/2)/(-6000))`

`=72000^(3/2)/6000-60000^(3/2)/6000`

`=770.4`

So,

`barv=770.4/3=256.8ms^-1`

The average speed is `=256.8ms^-1`