An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 4KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-04-04T07:08:29

The average speed of the object is #=72.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=18000J#

The final velocity is #=u_2#

#1/2m u_2^2=4000J#

Therefore,

#u_1^2=2/4*18000=9000m^2s^-2#

and,

#u_2^2=2/4*4000=2000m^2s^-2#

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

The points are #(0,9000)# and #(9,2000)#

The equation of the line is

#v^2-9000=(2000-9000)/9t#

#v^2=-777.8t+9000#

So,

#v=sqrt((-777.8t+9000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^9sqrt((-777.8t+9000))dt#

#9 barv=[((-777.8t+9000)^(3/2)/(-3/2*777.8)]_0^9#

#=((-777.8*9+9000)^(3/2)/(-1166.7))-((-777.8*0+9000)^(3/2)/(-1166.7))#

#=-2000^(3/2)/1166.7+9000^(3/2)/1166.7#

#=655.2#

So,

#barv=655.2/9=72.8ms^-1#