An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 64 KJ to 38 KJ over t in [0, 15 s]. What is the average speed of the object?

1 Answer
Narad T.
Sep 12, 2017

The average speed is =225.2ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =2kg

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=38000J

Therefore,

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

and,

u_2^2=2/2*38000=38000m^2s^-2

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

The points are (0,64000) and (15,38000)

The equation of the line is

v^2-64000=(38000-64000)/15t

v^2=-17333t+64000

So,

v=sqrt((-1733.3t+64000)

We need to calculate the average value of v over t in [0,15]

(15-0)bar v=int_0^15(sqrt(-1733.3t+64000))dt

15 barv=[((-1733.3t+64000)^(3/2)/(-3/2*1733.3))]_0^15

=((-1733.3*15+64000)^(3/2)/(-2600))-((-1733.3*0+64000)^(3/2)/(-2600))

=64000^(3/2)/2600-38000^(3/2)/2600

=3378.2

So,

barv=3378.2/15=225.2ms^-1

The average speed is =225.2ms^-1

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