An object has a mass of 1 kg. The object's kinetic energy uniformly changes from 48 KJ to 36 KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
Jan 30, 2018

The average speed is =204.8ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=1kg

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=2/1*48000=96000m^2s^-2

and,

u_2^2=2/1*36000=72000m^2s^-2

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

The points are (0,48000) and (3,36000)

The equation of the line is

v^2-48000=(36000-48000)/3t

v^2=-4000t+48000

So,

v=sqrt(-4000t+48000)

We need to calculate the average value of v over t in [0,3]

(3-0)bar v=int_0^3(sqrt(-4000t+48000))dt

3 barv= [(-4000t+48000)^(3/2)/(3/2*-4000)] _( 0) ^ (3)

=((-4000*3+48000)^(3/2)/(-6000))-((-4000*0+48000)^(3/2)/(-6000))

=48000^(3/2)/6000-36000^(3/2)/6000

=614.3

So,

barv=614.3/3=204.8ms^-1

The average speed is =204.8ms^-1

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