A particle moves according to the equation v-a- x - Find the average velocity in the first s meters of the path-

The correct option is D a√( s ) 2 We know that v_avg is given as #160; V_avg=v_i+v_f2 So, for first s meters, V_avg= a√( s ) #160; ...

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Byju's Answer

Standard XII

Physics

Projectile Time, Height and Range

A particle mo...

Question

A particle moves according to the equation $v = a \sqrt{x}$ . Find the average velocity in the first $s$ meters of the path.

\frac{\sqrt{s}}{a}

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\frac{\sqrt{s}}{2 a}

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\frac{2 a}{\sqrt{s}}

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\frac{a \sqrt{s}}{2}

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Solution

The correct option is D $\frac{a \sqrt{s}}{2}$
We know that $v_{a v g}$ is given as
$V_{a v g} = \frac{v_{i} + v_{f}}{2}$

So, for first $s$ meters,
$V_{a v g} = \frac{a \sqrt{s}}{2}$

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A particle moves according to the equation v=a√x. Find the average velocity in the first s meters of the path.

The correct option is D a√s2We know that vavg is given as Vavg=vi+vf2So, for first s meters,Vavg=a√s2

A particle moves according to the equation $v = a \sqrt{x}$ . Find the average velocity in the first $s$ meters of the path.

The correct option is D $\frac{a \sqrt{s}}{2}$
We know that $v_{a v g}$ is given as
$V_{a v g} = \frac{v_{i} + v_{f}}{2}$

So, for first $s$ meters,
$V_{a v g} = \frac{a \sqrt{s}}{2}$