An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 36 KJ to 270 KJ over t in [0, 9 s]. What is the average speed of the object?

1 Answer
Narad T.
Aug 18, 2017

The average speed is =310.5ms^-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=36000J

The final kinetic energy is 1/2m u_2^2=270000J

Therefore,

u_1^2=2/3*36000=24000m^2s^-2

and,

u_2^2=2/3*270000=180000m^2s^-2

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

The points are (0,24000) and (9,180000)

The equation of the line is

v^2-24000=(180000-24000)/9t

v^2=17333.3t+24000

So,

v=sqrt((17333.3t+24000)

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

(9-0)bar v=int_0^9(sqrt(17333.3t+24000))dt

9 barv=[((17333.3t+24000)^(3/2)/(3/2*17333.3))]_0^9

=((17333.3*9+24000)^(3/2)/(26000))-((17333.3*0+24000)^(3/2)/(26000))

=180000^(3/2)/26000-24000^(3/2)/26000

=2794.2

So,

barv=2794.2/9=310.5ms^-1

The average speed is =310.5ms^-1

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