An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 240 KJ to 64KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 23, 2017

The average speed is =192ms^-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=240000J

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

Therefore,

u_1^2=2/8*240000=60000m^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 (0,60000) and (3,16000)

The equation of the line is

v^2-60000=(16000-60000)/3t

v^2=-14666.7t+60000

So,

v=sqrt((-14666.7t+60000)

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

(3-0)bar v=int_0^3sqrt(-14666.7t+60000))dt

3 barv=[((-14666.7t+60000)^(3/2)/(-3/2*14666.7)]_0^3

=((-14666.7*3+60000)^(3/2)/(22000))-((14666.7*0+60000)^(3/2)/(22000))

=60000^(3/2)/22000-16000^(3/2)/22000

=576

So,

barv=576/3=192ms^-1

The average speed is =192ms^-1

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