An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 160 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-04-24T06:16:14

The average speed is #=234.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=64000J#

The final velocity is #=u_2#

#1/2m u_2^2=160000J#

Therefore,

#u_1^2=2/4*64000=32000m^2s^-2#

and,

#u_2^2=2/4*160000=80000m^2s^-2#

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

The points are #(0,32000)# and #(12,80000)#

The equation of the line is

#v^2-32000=(80000-32000)/12t#

#v^2=4000t+32000#

So,

#v=sqrt((4000t+32000)#

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

#(12-0)bar v=int_0^12sqrt((4000t+32000))dt#

#12 barv=[((4000t+32000)^(3/2)/(3/2*4000)]_0^12#

#=((4000*12+32000)^(3/2)/(6000))-((4000*0+32000)^(3/2)/(6000))#

#=80000^(3/2)/6000-32000^(3/2)/6000#

#=2817.2#

So,

#barv=2817.2/12=234.8ms^-1#