An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 42KJ# over #t in [0,8s]#. What is the average speed of the object?

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Answer 1 · 2017-05-14T12:06:08

The average speed is #=105.73ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=5kg#

The initial velocity is #=u_1#

#1/2m u_1^2=15000J#

The final velocity is #=u_2#

#1/2m u_2^2=42000J#

Therefore,

#u_1^2=2/5*15000=6000m^2s^-2#

and,

#u_2^2=2/5*42000=16800m^2s^-2#

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

The points are #(0,6000)# and #(8,16800)#

The equation of the line is

#v^2-6000=(16800-6000)/8t#

#v^2=1350t+6000#

So,

#v=sqrt((1350t+6000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8sqrt((3200t+10666.67))dt#

#8 barv=[((1350t+6000)^(3/2)/(3/2*1350)]_0^8#

#=((1350*8+6000)^(3/2)/(2025))-((1350*0+6000)^(3/2)/(2025))#

#=16800^(3/2)/2025-6000^(3/2)/2025#

#=845.81#

So,

#barv=845.81/8=105.73ms^-1#