An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 12 KJ to 36KJ over t in [0,6s]. What is the average speed of the object?

1 Answer
Narad T.
Aug 12, 2017

The average speed is =125.1ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =3kg

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=12000J

The final kinetic energy is 1/2m u_2^2=36000J

Therefore,

u_1^2=2/3*12000=8000m^2s^-2

and,

u_2^2=2/3*36000=24000m^2s^-2

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

The points are (0,8000) and (6,24000)

The equation of the line is

v^2-8000=(24000-8000)/6t

v^2=2666.7t+8000

So,

v=sqrt((2666.7t+8000)

We need to calculate the average value of v over t in [0,6]

(6-0)bar v=int_0^6(sqrt(2666.7t+8000))dt

6 barv=[((2666.7t+8000)^(3/2)/(3/2*2666.7))]_0^6

=((2666.7*6+8000)^(3/2)/(4000))-((2666.7*0+8000)^(3/2)/(4000))

=24000^(3/2)/4000-8000^(3/2)/4000

=750.6

So,

barv=750.6/6=125.1ms^-1

The average speed is =125.1ms^-1

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