Question #c5aad

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Answer 1 · 2017-12-07T11:53:35

Space average velocity #=(intvdx)/(intdx)# ......(1) (for one dimensional motion)
Time average velocity #=(intvdt)/(intdt)# .......(2)

Applicable kinematic expressions are

#v=u+at#
#s=ut+1/2at^2#

Given that particle starts from rest with constant acceleration. WE have

#v=at# .....(3)
#x=1/2at^2# ......(4)

From (3) and (4) #v# in terms of #x# is

#v=axxsqrt(2x/a)#
#=>v=sqrt(2ax)# ......(5)

From (1)
Space average velocity #=(int_0^xsqrt(2ax)dx)/(int_0^xdx)=(sqrt(2a)x^(3/2)/(3/2))/x=2/3sqrt(2ax)#

Space average velocity#=2/3v# .......(6)

From (3)
Time average velocity #=(intvdt)/(intdt)=(int_0^tatdt)/(int_0^tdt)=(1/2 at^2)/t=1/2at#

Time average velocity#=1/2v# .......(7)

Required Ratio #="space average velocity"/ "time average velocity"=(2/3v)/(1/2v)=4/3#