An object has a mass of 1 kg. The object's kinetic energy uniformly changes from 48 KJ to 12 KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
May 13, 2017

The average speed is =240.99ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =1kg

The initial velocity is =u_1

1/2m u_1^2=48000J

The final velocity is =u_2

1/2m u_2^2=12000J

Therefore,

u_1^2=2/1*48000=96000m^2s^-2

and,

u_2^2=2/1*120000=24000m^2s^-2

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

The points are (0,96000) and (3,24000)

The equation of the line is

v^2-96000=(24000-96000)/3t

v^2=-24000t+96000

So,

v=sqrt((-24000t+96000)

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

(3-0)bar v=int_0^3sqrt((-24000t+96000))dt

3 barv=[((-24000t+96000)^(3/2)/(-3/2*24000)]_0^3

=((-24000*3+96000)^(3/2)/(-36000))-((-24000*0+96000)^(3/2)/(-36000))

=96000^(3/2)/36000-24000^(3/2)/36000

=722.96

So,

barv=722.96/3=240.99ms^-1