An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 18 KJ to 64KJ over t in [0,12s]. What is the average speed of the object?

1 Answer
Narad T.
Apr 17, 2018

The average speed is =115.3ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=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=18000J

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

Therefore,

u_1^2=2/6*18000=6000m^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 (0,6000) and (12,21333.3)

The equation of the line is

v^2-6000=(21333.3-6000)/12t

v^2=1277.8t+6000

So,

v=sqrt(1277.8t+6000)

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

(12-0)bar v=int_0^12(sqrt(1277.8t+6000))dt

12 barv= [(1277.8t+6000)^(3/2)/(3/2*1277.8)] _( 0) ^ (12)

=((1277.8*12+6000)^(3/2)/(1916.7))-((1277.8*0+6000)^(3/2)/(1916.7))

=21333.3^(3/2)/1916.7-6000^(3/2)/1916.7

=1383.2

So,

barv=1383.2/12=115.3ms^-1

The average speed is =115.3ms^-1

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