An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 96 KJ to 240 KJ over t in [0, 9 s]. What is the average speed of the object?

1 Answer
Narad T.
May 13, 2017

The average speed is =332.01ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =3kg

The initial velocity is =u_1

1/2m u_1^2=96000J

The final velocity is =u_2

1/2m u_2^2=240000J

Therefore,

u_1^2=2/3*96000=64000m^2s^-2

and,

u_2^2=2/3*240000=160000m^2s^-2

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

The points are (0,64000) and (9,160000)

The equation of the line is

v^2-64000=(160000-64000)/9t

v^2=10666.67t+64000

So,

v=sqrt((10666.67t+64000)

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

(9-0)bar v=int_0^9sqrt((10666.67t+64000))dt

9 barv=[((10666.67t+64000)^(3/2)/(3/2*10666.67)]_0^9

=((10666.67*9+64000)^(3/2)/(16000))-((10666.67*0+64000)^(3/2)/(16000))

=160000^(3/2)/16000-64000^(3/2)/16000

=2988.07

So,

barv=2988.07/9=332.01ms^-1

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