An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 24 KJ to 64KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Jun 21, 2017

The average speed is =103.9ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =8kg

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=24000J

The final kinetic energy is 1/2m u_2^2=64000J

Therefore,

u_1^2=2/8*24000=6000m^2s^-2

and,

u_2^2=2/8*64000=16000m^2s^-2

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

The points are (0,6000) and (3,16000)

The equation of the line is

v^2-6000=(16000-6000)/3t

v^2=3333.3t+6000

So,

v=sqrt((3333.3t+6000)

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

(3-0)bar v=int_0^3sqrt((3333.3t+6000))dt

3 barv=[((3333.3t+6000)^(3/2)/(3/2*3333.3)]_0^3

=((3333.3*3+6000)^(3/2)/(5000))-((3333.3*0+6000)^(3/2)/(5000))

=16000^(3/2)/5000-6000^(3/2)/5000

=311.8

So,

barv=311.8/3=103.9ms^-1

The average speed is =103.9ms^-1