An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 32KJ# over #t in [0,6s]#. What is the average speed of the object?

1 Answer
May 7, 2017

Answer:

The average speed is #=163ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=3kg#

The initial velocity is #=u_1#

#1/2m u_1^2=48000J#

The final velocity is #=u_2#

#1/2m u_2^2=32000J#

Therefore,

#u_1^2=2/3*48000=32000m^2s^-2#

and,

#u_2^2=2/3*32000=21333.3m^2s^-2#

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

The points are #(0,32000)# and #(6,21333.3)#

The equation of the line is

#v^2-32000=(21333.3-32000)/6t#

#v^2=-1777.8t+32000#

So,

#v=sqrt((-1777.8t+32000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6sqrt((-1777.8t+32000))dt#

#6 barv=[((-1777.8t+32000)^(3/2)/(-3/2*1777.8)]_0^6#

#=((-1777.8*6+32000)^(3/2)/(-2666.7))-((-1777.8*0+32000)^(3/2)/(-2666.7))#

#=32000^(3/2)/2666.7-21333.3^(3/2)/2666.7#

#=978.1#

So,

#barv=978.1/6=163ms^-1#