An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 64 KJ to 160 KJ over t in [0, 5 s]. What is the average speed of the object?

1 Answer
Narad T.
May 14, 2017

The average speed is =91.79ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =12kg

The initial velocity is =u_1

1/2m u_1^2=64000J

The final velocity is =u_2

1/2m u_2^2=160000J

Therefore,

u_1^2=2/12*64000=10666.67m^2s^-2

and,

u_2^2=2/12*160000=26666.67m^2s^-2

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

The points are (0,10666.67) and (5,26666.67)

The equation of the line is

v^2-10666.67=(26666.67-10666.67)/5t

v^2=3200t+10666.67

So,

v=sqrt((3200t+10666.67)

We need to calculate the average value of v over t in [0,5]

(5-0)bar v=int_0^5sqrt((3200t+10666.67))dt

5 barv=[((3200t+10666.67)^(3/2)/(3/2*3200)]_0^5

=((3200*5+10666.67)^(3/2)/(4800))-((3200*0+10666.67)^(3/2)/(4800))

=26666.67^(3/2)/4800-10666.67^(3/2)/4800

=458.96

So,

barv=458.96/5=91.79ms^-1

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