An object has a mass of #1 kg#. The object's kinetic energy uniformly changes from #64 KJ# to #36 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-08T10:08:37

The average speed is #=315.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=1kg#

The initial velocity is #=u_1#

#1/2m u_1^2=64000J#

The final velocity is #=u_2#

#1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/1*64000=128000m^2s^-2#

and,

#u_2^2=2/1*36000=72000m^2s^-2#

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

The points are #(0,128000)# and #(3,72000)#

The equation of the line is

#v^2-128000=(72000-128000)/3t#

#v^2=-18666.7t+128000#

So,

#v=sqrt((-18666.7t+128000)#

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

#(3-0)bar v=int_0^3sqrt((-18666.7t+128000))dt#

#3 barv=[((-18666.7t+128000)^(3/2)/(-3/2*18666.7)]_0^3#

#=((-18666.7*3+128000)^(3/2)/(-28000))-((-18666.7*0+128000)^(3/2)/(-28000))#

#=128000^(3/2)/28000-72000^(3/2)/28000#

#=945.5#

So,

#barv=945.5/3=315.2ms^-1#