An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 48 KJ over t in [0, 15 s]. What is the average speed of the object?

1 Answer
Narad T.
Feb 20, 2018

The average speed is =202.1ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=3kg

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=75000J

The final kinetic energy is 1/2m u_2^2=48000J

Therefore,

u_1^2=2/3*75000=50000m^2s^-2

and,

u_2^2=2/3*48000=32000m^2s^-2

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

The points are (0,50000) and (15,32000)

The equation of the line is

v^2-50000=(32000-50000)/15t

v^2=-1200t+50000

So,

v=sqrt(-1200t+50000)

We need to calculate the average value of v over t in [0,15]

(15-0)bar v=int_0^15(sqrt(-1200t+50000))dt

15 barv= [(-1200t+50000)^(3/2)/(-3/2*1200)] _( 0) ^ (15)

=((-1200*15+50000)^(3/2)/(-1800))-((-1200*0+50000)^(3/2)/(-1800))

=50000^(3/2)/1800-32000^(3/2)/1800

=3031.11

So,

barv=3031.11/15=202.1ms^-1

The average speed is =202.1ms^-1

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