An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #128 KJ# to # 32 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2018-01-11T16:23:38

The average speed is #=196.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=128000J#

The final kinetic energy is #1/2m u_2^2=32000J#

Therefore,

#u_1^2=2/4*128000=64000m^2s^-2#

and,

#u_2^2=2/4*32000=16000m^2s^-2#

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

The points are #(0,64000)# and #(5,16000)#

The equation of the line is

#v^2-64000=(16000-64000)/5t#

#v^2=-9600t+64000#

So,

#v=sqrt(-9600t+64000)#

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

#(5-0)bar v=int_0^5(sqrt(-9600t+64000))dt#

#5 barv= (-9600t+64000)^(3/2)/(3/2*-9600)| _( 0) ^ (5) #

#=((-9600*5+64000)^(3/2)/(-14400))-((-9600*0+64000)^(3/2)/(-14400))#

#=64000^(3/2)/14400-16000^(3/2)/14400#

#=983.82#

So,

#barv=983.82/5=196.8ms^-1#

The average speed is #=196.8ms^-1#