An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 16KJ# over #t in [0,8s]#. What is the average speed of the object?

1 Answer
Jun 5, 2017

Answer:

The average speed is #=176.4ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #=5kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=150000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=16000J#

Therefore,

#u_1^2=2/5*150000=60000m^2s^-2#

and,

#u_2^2=2/5*16000=6400m^2s^-2#

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

The points are #(0,60000)# and #(8,6400)#

The equation of the line is

#v^2-60000=(6400-60000)/8t#

#v^2=-6700t+60000#

So,

#v=sqrt((-6700t+60000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8sqrt((-6700t+60000))dt#

#8 barv=[((-6700t+60000)^(3/2)/(-3/2*6700)]_0^8#

#=((-6700*8+60000)^(3/2)/(-10050))-((-6700*0+60000)^(3/2)/(-10050))#

#=60000^(3/2)/10050-6400^(3/2)/10050#

#=1411.4#

So,

#barv=1411.4/8=176.4ms^-1#

The average speed is #=176.4ms^-1#