Question

Power supplied to a particle of mass 3kg varies with time as P = 3/2t² W. Here t is in seconds. If velocity of particle at t = 0 is v = 0.The velocity of particle at t =3 seconds will be:- a)1m/s b)4m/s c) 2m/s d)3 m/s

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Answer 1

Given :

• Mass of the particle = 3 kg
• Power varying with time t = 3t²/2
• Velocity at t = 0 is zero

To find :

• The velocity of the particle at t = 3 sec

Solution :

The relation between mass , Power and velocity is given by ,

$\dag{\large{\boxed {\bf{P= FV}}}}$

Also ,

$:\implies\bf \: F= ma \longrightarrow \: eq(1)$

Where ,

• m is mass
• a is acceleration
• F is force

and a is also given by ,

$\displaystyle : \implies \bf \: a = \frac{dv}{dt} \longrightarrow \: eq(2)$

From eq(1) and eq(2) ,

$: \implies \bf \: p = (ma) \times v \\ \\ : \implies \bf \: p = \bigg(m \frac{dv}{dt} \bigg) \times v \\ \\ : \implies \bf \: p. dt= (m.dv) \times v$

Integrating on both sides ,

$: \implies \bf \large\int p.dt = \int(mv.dv)$

We have ,

• P = 3t²/2

$: \implies \bf \int \: \frac{3 {t}^{2} }{2} \: dt = m\int (v \: dv) \\ \\ : \implies \bf \frac{3}{2} \int \: t {}^{2} \: tdt = m \int \: (v \: dv )$

$\boxed{\bf{ \int a {}^{t} \: dt = \int \frac{a {}^{t + 1} }{t + 1} }}$

$: \implies \bf \: \frac{3}{2} \bigg(\: \frac{t {}^{2 + 1} }{2 + 1} \bigg) = m \bigg( \frac{v {}^{1 + 1} }{1 + 1} \bigg) \\ \\ : \implies \bf \: \frac{3}{2} \bigg( \frac{t {}^{3} }{3} \bigg) = m \bigg ( \frac{ {v}^{2} }{2} \bigg)$

We have ,

• m = 3 kg

$\displaystyle : \implies \bf \: \frac{3 {t}^{3} }{6} = 3 \bigg( \frac{ {v}^{2} }{2} \bigg)$

We need the velocity at t = 3 sec . so substituting t = 3 sec we get ,

$: \implies \bf \: \frac{3(3) {}^{3} }{6} = \frac{3{v}^{2} }{2} \\ \\ : \implies \bf \: \frac{3(27)}{6} = \frac{3 {v}^{2} }{2} \\ \\ : \implies \bf \: \frac{81}{6} = \frac{3 {v}^{2} }{2} \\ \\ : \implies \bf \: 81 \times 2 = 6 \times 3 {v}^{2} \\ \\ : \implies \bf \: 162 = 18 {v}^{2} \\ \\ : \implies \bf \: \frac{162}{18} = {v}^{2} \\ \\ : \implies \bf \: 9 = {v}^{2}$

$: \implies \bf \: \sqrt{9} = v \\ \\ : \implies \bf \: \sqrt{ {(3)}^{2} } = v \\ \\ : \implies \bf \: v = 3 \: m {s}^{ - 1}$

∴ The velocity of the particle at t = 3 sec is 3 m/s

Hence , Option(d) is the required answer.