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 $="space average velocity"/ "time average velocity"=(2/3v)/(1/2v)=4/3$