An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 42 KJ over t in [0, 15 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 26, 2017

The average speed is =181.3ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=75000J

The final velocity is =u_2

1/2m u_2^2=42000J

Therefore,

u_1^2=2/3*75000=50000m^2s^-2

and,

u_2^2=2/3*42000=28000m^2s^-2

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

The points are (0,50000) and (15,28000)

The equation of the line is

v^2-50000=(28000-50000)/15t

v^2=-1466.7t+50000

So,

v=sqrt((-1466.7t+50000)

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

(15-0)bar v=int_0^15sqrt((-1466.7t+50000)dt

15 barv=[((-1466.7t+50000)^(3/2)/(3/2*-1466.7)]_0^15

=((-1466.7*15+50000)^(3/2)/(-2200))-((-1466.7*0+50000)^(3/2)/(-2200))

=30000^(3/2)/-2200-50000^(3/2)/-2200

=50000^(3/2)/2200-30000^(3/2)/2200

=2720.1

So,

barv=2720.1/15=181.3ms^-1

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