A particle moves a distance x in time t according to equation x=(t+5)^ -1 . The acceleration of particle is proportional to

Correct answer: A. (velocity) ^ 3 / 2 Given, distance x=(t+5)^ -1 Differentiating Eq. (i) w.r.t. t , we get dx dt =(v)= -1 (t+5)^2 Again, differentiating Eq. (i) w.r.t. t, we get d^2 x dt^2 =(a)= 2 (t+5)^3 Comparing Eqs. (ii) and (iii), we get (a) (v)^ 3 / 2

NEET 2010

Exam: NEET
Year: 2010
Subject: Physics
Chapter: Motion in One Dimension

NEET 2010 · Physics · Motion in One Dimension

A particle moves a distance x in time t according to equation...

A particle moves a distance

\mathrm{x}

in time

\mathrm{t}

according to equation

x=(t+5)^{-1}

. The acceleration of particle is proportional to

Options

A(velocity) ${ }^{3 / 2}$
B $(\text{ distance })^2$
C(distance $)^{-2}$
D(velocity) ${ }^{2 / 3}$

Correct answer

A. (velocity) ${ }^{3 / 2}$

Step-by-step solution

Given, distance

x=(t+5)^{-1}

Differentiating Eq. (i) w.r.t.

t

, we get

\frac{\mathrm{dx}}{\mathrm{dt}}=(\mathrm{v})=\frac{-1}{(\mathrm{t}+5)^2}

Again, differentiating Eq. (i) w.r.t. t, we get

\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=(\mathrm{a})=\frac{2}{(\mathrm{t}+5)^3}

Comparing Eqs. (ii) and (iii), we get (a)

\propto(v)^{3 / 2}

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