An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #42 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

1 Answer
Mar 26, 2017

Answer:

The average speed is #=181.3ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=75000J#

The final velocity is #=u_2#

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

Therefore,

#u_1^2=2/3*75000=50000m^2s^-2#

and,

#u_2^2=2/3*42000=28000m^2s^-2#

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

The points are #(0,50000)# and #(15,28000)#

The equation of the line is

#v^2-50000=(28000-50000)/15t#

#v^2=-1466.7t+50000#

So,

#v=sqrt((-1466.7t+50000)#

We need to calculate the average value of #v# over #t in [0,15]#

#(15-0)bar v=int_0^15sqrt((-1466.7t+50000)dt#

#15 barv=[((-1466.7t+50000)^(3/2)/(3/2*-1466.7)]_0^15#

#=((-1466.7*15+50000)^(3/2)/(-2200))-((-1466.7*0+50000)^(3/2)/(-2200))#

#=30000^(3/2)/-2200-50000^(3/2)/-2200#

#=50000^(3/2)/2200-30000^(3/2)/2200#

#=2720.1#

So,

#barv=2720.1/15=181.3ms^-1#