An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #144 KJ# to # 120KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2017-07-16T08:11:36

The average speed is #=256.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=144000J#

The final kinetic energy is #1/2m u_2^2=120000J#

Therefore,

#u_1^2=2/4*144000=72000m^2s^-2#

and,

#u_2^2=2/4*120000=60000m^2s^-2#

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

The points are #(0,72000)# and #(6,60000)#

The equation of the line is

#v^2-72000=(60000-72000)/6t#

#v^2=-2000t+72000#

So,

#v=sqrt((-2000t+72000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6sqrt(-2000t+72000))dt#

#6 barv=[((-2000t+72000)^(3/2)/(-3/2*2000)]_0^6#

#=((-2000*6+72000)^(3/2)/(-3000))-((-2000*0+72000)^(3/2)/(-3000))#

#=72000^(3/2)/3000-60000^(3/2)/3000#

#=1540.9#

So,

#barv=1540.9/6=256.8ms^-1#

The average speed is #=256.8ms^-1#