An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #16 KJ# to # 36KJ# over #t in [0, 6 s]#. What is the average speed of the object?

2 Answers
Apr 3, 2017

Answer:

The average speed is #=160.2ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=2kg#

The initial velocity is #=u_1#

#1/2m u_1^2=16000J#

The final velocity is #=u_2#

#1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/2*16000=16000m^2s^-2#

and,

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

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

The points are #(0,16000)# and #(6,36000)#

The equation of the line is

#v^2-16000=(36000-16000)/6t#

#v^2=3333.3t+16000#

So,

#v=sqrt((3333.3t+16000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^8sqrt((3333.3t+16000))dt#

#6 barv=[((3333.3t+16000)^(3/2)/(3/2*3333.3)]_0^6#

#=((3333.3*6+16000)^(3/2)/(5000))-((3333.3*0+16000)^(3/2)/(5000))#

#=36000^(3/2)/5000-16000^(3/2)/5000#

#=961.3#

So,

#barv=961.3/6=160.2ms^-1#

Apr 3, 2017

Answer:

Average velocity #bar v approx 160.2 " m/s"#

So average speed is #bar (dot s) approx 160.2 " m/s"#.

Explanation:

During the period we can say that:

#(dT)/(dt) = (36,000-16,000)/(6) = (10,000)/3 " Joules / s"#

Integrating:

#T(t) = 1/2 m v(t) ^2 = int (10,000)/3 \ dt = (10,000)/3 t + alpha#

#T(0) = 16,000 implies alpha = 16,000#

And as #m = 2#, we have:

#v(t) = sqrt ((10,000)/3 t + 16,000)#

For average velocity #bar v#:

# bar v = (int_0^6 v(t) \ dt)/(6 - 0) # (simply put, this calculates that constant velocity that would cover the same distance (sense displacement!) in the same time)

#=([ ((10,000 t)/3 + 16,000)^(3/2) ]_0^6 )/(6 cdot 5,000) approx 160.2 " m/s"#

Reality check:

#v = sqrt ((2T)/m)#

#v(0) approx 126.5 " m/s"#

#v(6) approx 190 " m/s"#