In This post you will get answer of Q.
If the velocity of a particle is given by $ v={{(180-16x)}^{1/2}},m/s $ . then its acceleration will be
Solution:
$a =frac{ d v }{ d t }=frac{ d x }{ d t } frac{ d v }{ d x }= v frac{ d v }{ d x }$
$v=(180-16 x)^{frac{1}{2}}$
$v^{2}=180-16 x$
$text { Diff. both sides wrt. } x text { , we get: }$
$2 v frac{d v}{d x}=-16$
$v frac{ d v }{ d x }= a =- 8 , m / s ^{2}$
Solution:
$a =frac{ d v }{ d t }=frac{ d x }{ d t } frac{ d v }{ d x }= v frac{ d v }{ d x }$
$v=(180-16 x)^{frac{1}{2}}$
$v^{2}=180-16 x$
$text { Diff. both sides wrt. } x text { , we get: }$
$2 v frac{d v}{d x}=-16$
$v frac{ d v }{ d x }= a =- 8 , m / s ^{2}$