An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 16KJ# over #t in [0,8s]#. What is the average speed of the object?

2 Answers
Jun 25, 2017

Answer:

#168.553416ms^-1#

Explanation:

#E_k = 1/2mv^2#

150kJ = 150,000J

16kJ = 16,000J

#150,000 = 1/2*4v^2#
#v = sqrt((300,000)/4) = 273.861279ms^-1#

#16,000 = 1/2*4v^2#
#v = sqrt((32,000)/4) = 63.2455532 ms^-1#

Proof 1:
#(63.2455532+273.861279)/2 = 168.553416ms^-1#

Proof 2:

#s=(t(u+v))/2# s = distance, t = time, v = final velocity, u = initial velocity.

#(8(273.861279+63.2455532))/2=1348.42733m#

#"average speed" = ("total distance")/("total time") = 1348.42733/8 = 168.553416ms^-1#

Jun 25, 2017

Answer:

The average speed is #=197.3ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=150000J#

The final kinetic energy is #1/2m u_2^2=16000J#

Therefore,

#u_1^2=2/4*150000=75000m^2s^-2#

and,

#u_2^2=2/4*16000=8000m^2s^-2#

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

The points are #(0,75000)# and #(8,8000)#

The equation of the line is

#v^2-75000=(8000-75000)/8t#

#v^2=-8375t+75000#

So,

#v=sqrt((-8375t+75000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^3sqrt(-8375t+75000))dt#

#8 barv=[((-8375t+75000)^(3/2)/(-3/2*8375)]_0^8#

#=((-8375*8+75000)^(3/2)/(-12562.5))-((-8375*0+75000)^(3/2)/(-12562.5))#

#=75000^(3/2)/12562.5-8000^(3/2)/12562.5#

#=1578.03#

So,

#barv=1578.03/8=197.3ms^-1#

The average speed is #=197.3ms^-1#