An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 72KJ# over #t in [0,5s]#. What is the average speed of the object?

2 Answers
Jan 13, 2018

Answer:

The average speed is #=141.5ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=4kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=12000J#

The final kinetic energy is #1/2m u_2^2=72000J#

Therefore,

#u_1^2=2/4*12000=6000m^2s^-2#

and,

#u_2^2=2/4*72000=36000m^2s^-2#

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

The points are #(0,6000)# and #(5,36000)#

The equation of the line is

#v^2-6000=(36000-6000)/5t#

#v^2=6000t+6000#

So,

#v=sqrt(6000t+6000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5(sqrt(6000t+6000))dt#

#5 barv= (6000t+6000)^(3/2)/(3/2*6000)| _( 0) ^ (5) #

#=((6000*5+6000)^(3/2)/(9000))-((6000*0+6000)^(3/2)/(9000))#

#=36000^(3/2)/9000-6000^(3/2)/9000#

#=707.31#

So,

#barv=707.31/5=141.5ms^-1#

The average speed is #=141.5ms^-1#

Jan 13, 2018

Answer:

133.63 m/s

Explanation:

Using equation of kinetic energy, i.e E =# (1/2)m(v^2) #
We get initial velocity=u was 77.46 m/sec and final velocity=v was 189.73 m/sec
Now,using v=u+at(all symbols are bearing their conventional meaning) we get a is 22.45 #m/(s^2)#
Hence putting it in #v^2 = u^2 +2as # (all symbols are bearing their conventional meaning)
We get total distance covered in 5 sec is 668.152 meter.
So, average speed = (total distance covered/total time) = #(668.152/5)# i.e 133.63 m/s