An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 48 KJ to 38KJ over t in [0,12s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 13, 2018

The average speed is =169.2ms^-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=38000J

Therefore,

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

and,

u_2^2=2/3*38000=25333.3m^2s^-2

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

The points are (0,32000) and (12,25333.3)

The equation of the line is

v^2-32000=(25333.3-32000)/12t

v^2=-555.6t+32000

So,

v=sqrt(-555.6t+32000)

We need to calculate the average value of v over t in [0,12]

(12-0)bar v=int_0^12(sqrt(-555.6t+32000))dt

12 barv= [(-555.6t+32000)^(3/2)/(-3/2*555.6)] _( 0) ^ (12)

=((-555.6*8+32000)^(3/2)/(-833.3))-((-555.6*0+32000)^(3/2)/(-833.3))

=32000^(3/2)/833.3-25333.3^(3/2)/833.3

=2030.7

So,

barv=2030.7/12=169.2ms^-1

The average speed is =169.2ms^-1

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