An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 75 KJ to 135 KJ over t in [0, 9 s]. What is the average speed of the object?

1 Answer
Narad T.
Oct 29, 2017

The average speed is =204.2ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=5kg

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=75000J

The final kinetic energy is 1/2m u_2^2=135000J

Therefore,

u_1^2=2/5*75000=30000m^2s^-2

and,

u_2^2=2/5*135000=54000m^2s^-2

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

The points are (0,30000) and (9,54000)

The equation of the line is

v^2-30000=(54000-30000)/9t

v^2=2666.7t+30000

So,

v=sqrt((2666.7t+30000)

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

(9-0)bar v=int_0^9(sqrt(2666.7t+30000))dt

9 barv=[((2666.7t+30000)^(3/2)/(3/2*2666.7))]_0^9

=((2666.7*9+30000)^(3/2)/(4000))-((2666.7*0+30000)^(3/2)/(4000))

=54000^(3/2)/4000-30000^(3/2)/4000

=1838.1

So,

barv=1838.1/9=204.2ms^-1

The average speed is =204.2ms^-1

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