An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 64KJ# over #t in [0,12s]#. What is the average speed of the object?

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Answer 1 · 2017-04-24T13:16:43

The average speed is #=123.5ms^-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=64000J#

Therefore,

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

and,

#u_2^2=2/5*64000=25600m^2s^-2#

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

The points are #(0,6000)# and #(12,25600)#

The equation of the line is

#v^2-6000=(25600-6000)/12t#

#v^2=1633.3t+6000#

So,

#v=sqrt((1633.3t+6000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12sqrt((1633.3t+6000))dt#

#12 barv=[((1633.3t+6000)^(3/2)/(3/2*1633.3)]_0^12#

#=((1633.3*12+6000)^(3/2)/(2450))-((1633.3*0+6000)^(3/2)/(2450))#

#=25600^(3/2)/2450-6000^(3/2)/2450#

#=1482.1#

So,

#barv=1482.1/12=123.5ms^-1#