If the velocity of a particle is given by v = | vedclass

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Question

(c) $v = {(180 - 16x)^{1/2}}$

As $a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}$

$	herefore a = \frac{1}{2}{(180 - 16x)^{ - 1/2}}

Vedclass · Accepted Answer

$-8$

Vedclass · Answer

(c) $v = {(180 - 16x)^{1/2}}$

As $a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}$

$	herefore a = \frac{1}{2}{(180 - 16x)^{ - 1/2}} 	imes ( - 16)\;\left( {\frac{{dx}}{{dt}}} ight)$

$ = - \;8\;{(180 - 16x)^{ - 1/2}} 	imes \;v$

$ = - \;8\;{(180 - 16x)^{ - 1/2}} 	imes \;{(180 - 16x)^{1/2}}$

$ = - \;8\;m/{s^2}$

If the velocity of a particle is given by $v = {(180 - 16x)^{1/2}} m/s$, then its acceleration will be.......$ms^{-2}$

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