An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 48 KJ over t in [0, 9 s]. What is the average speed of the object?

1 Answer
Narad T.
May 5, 2017

The average speed is =202.1ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =3kg

The initial velocity is =u_1

1/2m u_1^2=75000J

The final velocity is =u_2

1/2m u_2^2=48000J

Therefore,

u_1^2=2/3*75000=50000m^2s^-2

and,

u_2^2=2/3*48000=32000m^2s^-2

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

The points are (0,50000) and (9,32000)

The equation of the line is

v^2-50000=(32000-50000)/9t

v^2=-2000t+50000

So,

v=sqrt((-2000t+50000)

We need to calculate the average value of v over t in [0,9]

(9-0)bar v=int_0^9sqrt((-2000t+50000))dt

9 barv=[((-2000t+50000)^(3/2)/(-3/2*2000)]_0^9

=((-2000*9+50000)^(3/2)/(-3000))-((-2000*0+50000)^(3/2)/(-3000))

=50000^(3/2)/3000-32000^(3/2)/3000

=1818.7

So,

barv=1818.7/9=202.1ms^-1

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