Acceleration of a particle moving along a straight line is a function of velocity as a =2v. At t = 2s, its velocity v = 16 m/s. Its velocity at t = 3s will be

a=2vdvdt=2vdv v−1/2=2dt∫16vv−1/2dv=2∫23dt(v−1/2+1−1/2+1)16v=2(t)23[2v]16v=2(3−2)[v]16v=1v−16=1v=5v = 25 m/s

Acceleration of a particle moving along a straight line is a function of velocity as a =2v. At t = 2s, its velocity v = 16 m/s. Its velocity at t = 3s will be

a=2vdvdt=2vdv v−1/2=2dt∫16vv−1/2dv=2∫23dt(v−1/2+1−1/2+1)16v=2(t)23[2v]16v=2(3−2)[v]16v=1v−16=1v=5v = 25 m/s

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Acceleration of a particle moving along a straight line is a function of velocity as $a = 2 \sqrt{v}$ . At t = 2s, its velocity v = 16 m/s. Its velocity at t = 3s will be

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25 m/s

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Detailed Solution

$a = 2 \sqrt{v}$

$\frac{d v}{d t} = 2 \sqrt{v}$

$d v v^{- 1 / 2} = 2 d t$

$\int_{16}^{v} v^{- 1 / 2} d v = 2 \int_{2}^{3} d t$

${(\frac{v^{(- 1 / 2) + 1}}{(- 1 / 2) + 1})}_{16}^{v} = {2 (t)}_{2}^{3}$

${[2 \sqrt{v}]}_{16}^{v} = 2 (3 - 2)$

${[\sqrt{v}]}_{16}^{v} = 1$

$\sqrt{v} - \sqrt{16} = 1$

$\sqrt{v} = 5$

v = 25 m/s

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Acceleration of a particle moving along a straight line is a function of velocity as a =2v. At t = 2s, its velocity v = 16 m/s. Its velocity at t = 3s will be

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Acceleration of a particle moving along a straight line is a function of velocity as $a = 2 \sqrt{v}$ . At t = 2s, its velocity v = 16 m/s. Its velocity at t = 3s will be