An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #640 KJ# to # 160 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

3 Answers
Apr 11, 2017

Answer:

The average speed is #=440ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=640000J#

The final velocity is #=u_2#

#1/2m u_2^2=160000J#

Therefore,

#u_1^2=2/4*640000=320000m^2s^-2#

and,

#u_2^2=2/4*160000=80000m^2s^-2#

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

The points are #(0,320000)# and #(12,80000)#

The equation of the line is

#v^2-320000=(80000-320000)/12t#

#v^2=-20000t+320000#

So,

#v=sqrt((-20000t+320000)#

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

#(12-0)bar v=int_0^12sqrt((-20000t+320000))dt#

#12 barv=[((-20000t+320000)^(3/2)/(-3/2*20000)]_0^12#

#=((-20000*12+320000)^(3/2)/(-30000))-((-20000*0+320000)^(3/2)/(-30000))#

#=320000^(3/2)/30000-80000^(3/2)/30000#

#=5279.7#

So,

#barv=5279.7/12=440ms^-1#

Apr 11, 2017

Answer:

average speed is defined as total distance travelled divided by total time taken to cover that distance .

Explanation:

the kinetic energy of a body of mass m moving with speed v is defined as : # 1/2mv^2 #
thus you can calculate velocity of body at the starting and the end of interval of time :
lets say initially the velocity was # v_1 # , # 1/2*4*v_1^2 = 640000 #
# v_1 = 565.68 ms^-1 #
let at the end of interval of time the velocity was # v_2 # , # 1/2 * 4 * v_2^2 = 160000 #
# v_2 = 282.84 ms^-1 #
now , total distance travelled by the body over given time interval can be calculated by using # v_2^2 - v_1^2 = 2*a*s # and # v_2 = v_1 + a*t #
thus, # 282.84 = 565.68 + 12a #
# a= -23.57ms^-2 #
# 282.84^2 - 565.68^2 = 2*-23.57*s #
# s= 5091.12m #
average speed is defined as ratio of total distance travelled and total time taken :
# v= s/t #
# v= 5091.12/12 #
# v= 424.26ms^-1 #

Apr 12, 2017

Let rate of change of kinetic energy be
# (dKE(t))/(dt) = C#
where #C # is constant with time #t#.

Integrating both sides with #t# we get
#int (dKE(t))/(dt) cdot dt = int C cdotdt#

#=>KE= Ccdott + c# ......(1)
where #c# is a constant of integration.

To evaluate #c#, use initial condition at #t=0#
#640xx10^3 = Ccdot0+c#
#=>c=640xx10^3#
We have expression for kinetic energy as
#KE= Ccdott + 640xx10^3#

To evaluate #C#, use final condition at #t=12#
#160 xx10^3= Ccdot12+640xx10^3#
#=>C=-40xx10^3#

Equation (1) becomes
#KE= (-t + 16)xx4xx10^4# ......(2)

If #m# is the mass and #v(t)# is the velocity of body, Kinetic energy is given as
#KE=1/2mv^2#
#=> 1/2mv^2(t) = (-t + 16)xx4xx10^4#
Given is #m=4kg#. Above expression becomes
#=> 1/2xx4v^2(t) = (-t + 16)xx4xx10^4#
#=> v(t)^2 = (-t+16)xx2xx10^4#
# => v= +-sqrt((-t+16)xx2xx10^4)#
# => "speed "|v|= sqrt((-t+16)xx2xx10^4)#

Average speed #="Total Distance traveled"/"Time taken"=(int_0^12|v(t)|*dt)/(12-0)#

#:.# Average speed #= (int_(0)^(12)sqrt((-t+16)xx2xx10^4)*dt)/12#
#= 100sqrt(2)/12[-(16-t)^(3/2)/(3/2)]_0^12 #
#=- 50/9sqrt(2)(4^(3/2) - 16^(3/2))#
#= 439.98 ms^-1#, rounded to two decimal places.