An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 64 KJ to 280 KJ over t in [0, 12 s]. What is the average speed of the object?

1 Answer
Narad T.
Oct 23, 2017

The average speed is =288.0ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=4kg

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=64000J

The final kinetic energy is 1/2m u_2^2=280000J

Therefore,

u_1^2=2/4*64000=32000m^2s^-2

and,

u_2^2=2/4*280000=140000m^2s^-2

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

The points are (0,32000) and (12,140000)

The equation of the line is

v^2-32000=(140000-32000)/12t

v^2=9000t+32000

So,

v=sqrt((9000t+32000)

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

(12-0)bar v=int_0^12(sqrt(9000t+32000))dt

12 barv=[((9000t+32000)^(3/2)/(3/2*9000))]_0^12

=((9000*12+32000)^(3/2)/(13500))-((9000*0+32000)^(3/2)/(13500))

=140000^(3/2)/13500-32000^(3/2)/13500

=3456.2

So,

barv=3456.2/12=288.0ms^-1

The average speed is =288.0ms^-1

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