An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 254 KJ to 32 KJ over t in [0, 5 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 19, 2017

The average speed is =259.7ms^-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=254000J

The final kinetic energy is 1/2m u_2^2=32000J

Therefore,

u_1^2=2/4*254000=127000m^2s^-2

and,

u_2^2=2/4*32000=16000m^2s^-2

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

The points are (0,127000) and (5,16000)

The equation of the line is

v^2-127000=(16000-127000)/5t

v^2=-22200t+127000

So,

v=sqrt((-22200t+127000)

We need to calculate the average value of v over t in [0,5]

(5-0)bar v=int_0^5sqrt((-22200t+127000))dt

5 barv=[((-22200t+127000)^(3/2)/(-3/2*22200)]_0^5

=((-22200*5+127000)^(3/2)/(-33300))-((-22200*0+127000)^(3/2)/(-33300))

=127000^(3/2)/33300-16000^(3/2)/33300

=1298.4

So,

barv=1298.4/5=259.7ms^-1

The average speed is =259.7ms^-1

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