An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #144 KJ# to # 640KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-14T08:24:39

The average speed is #=434.68ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=144000J#

The final velocity is #=u_2#

#1/2m u_2^2=640000J#

Therefore,

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

and,

#u_2^2=2/4*640000=320000m^2s^-2#

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

The points are #(0,72000)# and #(3,320000)#

The equation of the line is

#v^2-72000=(320000-72000)/3t#

#v^2=82666.67t+72000#

So,

#v=sqrt((82666.67t+72000)#

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

#(3-0)bar v=int_0^3sqrt((82666.67t+72000))dt#

#3 barv=[((82666.67t+72000)^(3/2)/(3/2*82666.67)]_0^3#

#=((82666.67*3+72000)^(3/2)/(124000))-((82666.67*0+72000)^(3/2)/(124000))#

#=320000^(3/2)/124000-72000^(3/2)/124000#

#=1304.03#

So,

#barv=1304.03/3=434.68ms^-1#