An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 36 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

2 Answers
Dec 1, 2017

Answer:

The average speed is #=108.3ms^-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=36000J#

Therefore,

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

and,

#u_2^2=2/4*36000=18000m^2s^-2#

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

The points are #(0,6000)# and #(12,18000)#

The equation of the line is

#v^2-6000=(18000-6000)/12t#

#v^2=1000t+6000#

So,

#v=sqrt(1000t+6000)#

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

#(12-0)bar v=int_0^12(sqrt(1000t+6000))dt#

#12 barv=[1000t+6000)^(3/2)/(3/2*1000))] _( 0) ^ (12) #

#=((1000*12+6000)^(3/2)/(1500))-((1000*0+6000)^(3/2)/(1500))#

#=18000^(3/2)/1500-6000^(3/2)/1500#

#=1300.1#

So,

#barv=1300.1/12=108.3ms^-1#

The average speed is #=108.3ms^-1#

Dec 1, 2017

Kinetic energy #KE# of an object of mass #m# moving with velocity #v# is given by the expression

#KE=1/2mv^2#

Given #m=4kg#
Let initial velocity be #=u" "ms^-1#
The final velocity be #=v" " ms^-1#

Initial kinetic energy

#1/2m u^2=12000J#

Final kinetic energy is

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

As kinetic energy changes uniformly, the graph of #t" vs "KE# is a straight line.

The points on the graph are #(0, 12000)# and #(12, 36000)#

The equation of the line is of the type #y=mx+c#

#1/2m(v(t))^2=(36000-12000)/12t+12000#
or #1/2xx4(v(t))^2=2000t+12000#
#=>(v(t))^2=1000t+6000#
#=>v(t)=sqrt(1000t+6000)#

We know that velocity can be written as #(ds)/dt# where #s# is distance moved

#=>(ds(t))/dt=sqrt(1000t+6000)#

We need to calculate total distance traveled in time #t# from # 0# to#12s# by integrating over the interval

#:.s=int_0^12sqrt(1000t+6000)cdotdt#
#=>s=[(1000t+6000)^(3/2)/(3/2xx1000)] _( 0) ^ (12) #
#=>s=1/1500[(1000xx12+6000)^(3/2)-(1000xx0+6000)^(3/2)]#
#=>s=1/1500(18000^(3/2)-6000^(3/2))#
#=>s=1300.13m#

We know that
#"Average speed"="Total distance traveled"/"Time of travel"#
#=>barv=1300.13/12=108.3ms^-1#, rounded to one decimal place