An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 48 KJ to 36KJ over t in [0,6s]. What is the average speed of the object?

1 Answer
Narad T.
Dec 30, 2017

The average speed is =167.2ms^-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=48000J

The final kinetic energy is 1/2m u_2^2=36000J

Therefore,

u_1^2=2/3*48000=32000m^2s^-2

and,

u_2^2=2/3*36000=24000m^2s^-2

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

The points are (0,32000) and (6,24000)

The equation of the line is

v^2-32000=(24000-32000)/6t

v^2=-1333.3t+32000

So,

v=sqrt(-1333.3t+32000)

We need to calculate the average value of v over t in [0,6]

(6-0)bar v=int_0^6(sqrt(-1333.3t+32000))dt

6 barv= (-1333.3t+32000)^(3/2)/(3/2*-1333.3)| _( 0) ^ (6)

=((-1333.3*6+32000)^(3/2)/(-2000))-((-1333.3*0+32000)^(3/2)/(-2000))

=32000^(3/2)/2000-24000^(3/2)/2000

=10003.1

So,

barv=1003.1/6=167.2ms^-1

The average speed is =167.2ms^-1

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