An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 150 KJ to 640KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 26, 2017

The average speed is =308.8ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =8kg

The initial velocity is =u_1

1/2m u_1^2=150000J

The final velocity is =u_2

1/2m u_2^2=640000J

Therefore,

u_1^2=2/8*150000=37500m^2s^-2

and,

u_2^2=2/8*640000=160000m^2s^-2

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

The points are (0,37500) and (3,160000)

The equation of the line is

v^2-37500=(160000-37500)/3t

v^2=40833.3t+37500

So,

v=sqrt((40833.3t+37500)

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

(3-0)bar v=int_0^3sqrt((40833.3t+37500)dt

3 barv=[((40833.3t+37500)^(3/2)/(3/2*40833.3)]_0^3

=((40833.3*3+37500)^(3/2)/(61250))-((40833.3*0+37500)^(3/2)/(61250))

=160000^(3/2)/61250-37500^(3/2)/61250

=926.3

So,

barv=926.3/3=308.8ms^-1

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