An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 96 KJ to 48 KJ over t in [0, 6 s]. What is the average speed of the object?

1 Answer
Narad T.
Feb 26, 2018

The average speed is =199.7ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=6kg

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=96000J

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

Therefore,

u_1^2=2/6*96000=32000m^2s^-2

and,

u_2^2=2/6*48000=16000m^2s^-2

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

The points are (0,48000) and (6,32000)

The equation of the line is

v^2-48000=(32000-48000)/6t

v^2=-2666.7t+48000

So,

v=sqrt(-2666.7t+48000)

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

(6-0)bar v=int_0^6(sqrt(-2666.7t+48000))dt

6 barv= [(-2666.7t+48000)^(3/2)/(-3/2*2666.7)] _( 0) ^ (6)

=((-2666.7*6+48000)^(3/2)/(-4000))-((-2666.7*0+48000)^(3/2)/(-4000))

=48000^(3/2)/4000-32000^(3/2)/4000

=1197.98

So,

barv=1197.98/6=199.7ms^-1

The average speed is =199.7ms^-1

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