An object has a mass of #1 kg#. The object's kinetic energy uniformly changes from #48 KJ# to #12 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-13T08:23:23

The average speed is #=240.99ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=1kg#

The initial velocity is #=u_1#

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

The final velocity is #=u_2#

#1/2m u_2^2=12000J#

Therefore,

#u_1^2=2/1*48000=96000m^2s^-2#

and,

#u_2^2=2/1*120000=24000m^2s^-2#

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

The points are #(0,96000)# and #(3,24000)#

The equation of the line is

#v^2-96000=(24000-96000)/3t#

#v^2=-24000t+96000#

So,

#v=sqrt((-24000t+96000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt((-24000t+96000))dt#

#3 barv=[((-24000t+96000)^(3/2)/(-3/2*24000)]_0^3#

#=((-24000*3+96000)^(3/2)/(-36000))-((-24000*0+96000)^(3/2)/(-36000))#

#=96000^(3/2)/36000-24000^(3/2)/36000#

#=722.96#

So,

#barv=722.96/3=240.99ms^-1#