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2.Motion in Straight Line
hard
If the velocity of a particle is given by $v = {(180 - 16x)^{1/2}} m/s$, then its acceleration will be.......$ms^{-2}$
A
$0$
B
$8$
C
$-8$
D
$4$
Solution
(c) $v = {(180 – 16x)^{1/2}}$
As $a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}$
$\therefore a = \frac{1}{2}{(180 – 16x)^{ – 1/2}} \times ( – 16)\;\left( {\frac{{dx}}{{dt}}} \right)$
$ = – \;8\;{(180 – 16x)^{ – 1/2}} \times \;v$
$ = – \;8\;{(180 – 16x)^{ – 1/2}} \times \;{(180 – 16x)^{1/2}}$
$ = – \;8\;m/{s^2}$
Standard 11
Physics
