An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 15 KJ to 64KJ over t in [0,12s]. What is the average speed of the object?

1 Answer
Narad T.
Jul 19, 2017

The average speed is =107.3ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =6kg

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=15000J

The final kinetic energy is 1/2m u_2^2=64000J

Therefore,

u_1^2=2/6*15000=3000m^2s^-2

and,

u_2^2=2/6*64000=21333.3m^2s^-2

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

The points are (0,3000) and (12,21333.3)

The equation of the line is

v^2-3000=(21333.3-3000)/12t

v^2=1527.8t+3000

So,

v=sqrt((1527.8t+3000)

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

(12-0)bar v=int_0^12sqrt(1527.8t+3000))dt

12 barv=[((1527.8t+3000)^(3/2)/(3/2*1527.8)]_0^12

=((1527.8*12+3000)^(3/2)/(2291.7))-((1527.8*0+3000)^(3/2)/(2291.7))

=21333.6^(3/2)/2291.7-3000^(3/2)/2291.7

=1288

So,

barv=1288/12=107.3ms^-1

The average speed is =107.3ms^-1

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