An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 72KJ# over #t in [0,5s]#. What is the average speed of the object?

1 Answer
May 5, 2017

Answer:

The average speed is #=163.3ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=3kg#

The initial velocity is #=u_1#

#1/2m u_1^2=12000J#

The final velocity is #=u_2#

#1/2m u_2^2=72000J#

Therefore,

#u_1^2=2/3*12000=8000m^2s^-2#

and,

#u_2^2=2/3*72000=48000m^2s^-2#

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

The points are #(0,8000)# and #(5,48000)#

The equation of the line is

#v^2-8000=(48000-8000)/5t#

#v^2=8000t+8000#

So,

#v=sqrt((8000t+8000)#

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

#(5-0)bar v=int_0^4sqrt((8000t+8000))dt#

#5 barv=[((8000t+8000)^(3/2)/(3/2*8000)]_0^5#

#=((8000*5+8000)^(3/2)/(12000))-((8000*0+8000)^(3/2)/(12000))#

#=48000^(3/2)/12000-8000^(3/2)/12000#

#=816.7#

So,

#barv=816.7/5=163.3ms^-1#