An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 128 KJ to 48 KJ over t in [0, 12 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 1, 2017

The average speed is =207.9ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =4kg

The initial velocity is =u_1

1/2m u_1^2=128000J

The final velocity is =u_2

1/2m u_2^2=48000J

Therefore,

u_1^2=2/4*128000=64000m^2s^-2

and,

u_2^2=2/4*48000=24000m^2s^-2

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

The points are (0,64000) and (12,24000)

The equation of the line is

v^2-64000=(24000-64000)/12t

v^2=-3333.3t+64000

So,

v=sqrt((-3333.3t+6400)

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

(12-0)bar v=int_0^12sqrt((-3333.3t+64000))dt

12 barv=[((-3333.3t+64000)^(3/2)/(3/2*3333.3)]_0^12

=((-3333.3*12+64000)^(3/2)/(5000))-((-3333.3*0+64000)^(3/2)/(5000))

=64000^(3/2)/5000-24000^(3/2)/5000

=2494.6

So,

barv=2494.6/12=207.9ms^-1

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