The position x of a particle at any instant i | vedclass

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Question

(A) Given the velocity equation: $v = \sqrt{2x + 9}$.Squaring both sides,we get: $v^2 = 2x + 9$.Comparing this with the kinematic equation $v^2 = u^2

Vedclass · Accepted Answer

$1\, 	ext{unit}, 3\, 	ext{unit}$

Vedclass · Answer

(A) Given the velocity equation: $v = \sqrt{2x + 9}$.Squaring both sides,we get: $v^2 = 2x + 9$.Comparing this with the kinematic equation $v^2 = u^2 + 2ax$,where $u$ is the initial velocity and $a$ is the acceleration:At the origin,$x = 0$. Substituting $x = 0$ into the velocity equation: $v = \sqrt{2(0) + 9} = \sqrt{9} = 3\, 	ext{unit}$. Thus,the initial velocity $u = 3\, 	ext{unit}$.Comparing $v^2 = 2x + 9$ with $v^2 = u^2 + 2ax$,we identify $2a = 2$,which gives $a = 1\, 	ext{unit}$.Therefore,the initial acceleration is $1\, 	ext{unit}$ and the initial velocity is $3\, 	ext{unit}$.

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

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