A particle moving along a straight line has a velocity $v\ ms^{- 1}$, when it cleared a distance of $y$ metre. These two are connected by the relation $v = \sqrt{49 + y}$. When its velocity is $1$ $ms^{- 1},$ its acceleration (in $ms^{- 2}$)is

Question

Dr. Abubacker · Accepted Answer

(d) $v = \sqrt{49 + y},a = \frac{dv}{dy}.\frac{dy}{dt} = v\frac{dv}{dy}$
$= (49 + y)^{1/2} 	imes \frac{1}{2}(49 + y)^{1/2 - 1}$= $\frac{1}{2}ms^{- 2}$

Q. A particle moving along a straight line has a velocity $v\ ms^{- 1}$, when it cleared a distance of $y$ metre. These two are connected by the relation $v = \sqrt{49 + y}$. When its velocity is $1$ $ms^{- 1},$ its acceleration (in $ms^{- 2}$)is

Explanation

Reviewed by Dr. Abubacker

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