An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 24 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

1 Answer
Jun 18, 2017

Answer:

The average speed is #=84.9ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #=12kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=64000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=24000J#

Therefore,

#u_1^2=2/12*64000=10666.7m^2s^-2#

and,

#u_2^2=2/12*24000=4000m^2s^-2#

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

The points are #(0,10666.7)# and #(5,4000)#

The equation of the line is

#v^2-10666.7=(4000-10666.7)/5t#

#v^2=-1333.3t+10666.7#

So,

#v=sqrt((-1333.3t+10666.7)#

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

#(5-0)bar v=int_0^5sqrt((-1333.3t+10666.7))dt#

#5 barv=[((-1333.3t+10666.7)^(3/2)/(-3/2*1333.3)]_0^5#

#=((-1333.3*5+10666.7)^(3/2)/(2000))-((-1333.3*0+10666.7)^(3/2)/(2000))#

#=10666.7^(3/2)/2000-4000^(3/2)/2000#

#=424.3#

So,

#barv=424.3/5=84.9ms^-1#

The average speed is #=84.9ms^-1#