An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #254 KJ# to # 32 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-19T06:06:09

The average speed is #=259.7ms^-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=254000J#

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

Therefore,

#u_1^2=2/4*254000=127000m^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 #(0,127000)# and #(5,16000)#

The equation of the line is

#v^2-127000=(16000-127000)/5t#

#v^2=-22200t+127000#

So,

#v=sqrt((-22200t+127000)#

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

#(5-0)bar v=int_0^5sqrt((-22200t+127000))dt#

#5 barv=[((-22200t+127000)^(3/2)/(-3/2*22200)]_0^5#

#=((-22200*5+127000)^(3/2)/(-33300))-((-22200*0+127000)^(3/2)/(-33300))#

#=127000^(3/2)/33300-16000^(3/2)/33300#

#=1298.4#

So,

#barv=1298.4/5=259.7ms^-1#

The average speed is #=259.7ms^-1#