An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 150 KJ to 64KJ over t in [0,8s]. What is the average speed of the object?

1 Answer
Narad T.
Jul 12, 2017

The average speed is =229.7ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =4kg

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=150000J

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

Therefore,

u_1^2=2/4*150000=75000m^2s^-2

and,

u_2^2=2/4*64000=32000m^2s^-2

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

The points are (0,75000) and (8,32000)

The equation of the line is

v^2-75000=(32000-75000)/8t

v^2=-5375t+75000

So,

v=sqrt((-5375t+75000)

We need to calculate the average value of v over t in [0,8]

(8-0)bar v=int_0^8sqrt(-5375t+75000))dt

8 barv=[((-5375t+75000)^(3/2)/(-3/2*5375)]_0^8

=((-5375*8+75000)^(3/2)/(-8062.5))-((-5375*0+75000)^(3/2)/(-8062.5))

=75000^(3/2)/8062.5-32000^(3/2)/8062.5

=1837.6

So,

barv=1837.6/8=229.7ms^-1

The average speed is =229.7ms^-1

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