The position x of a particle varies with time (t) as x=A ^ 2 -B ^ 3 . The acceleration at time t of the particle will be equal to zero. What is the value of t ?

Correct answer: C. (A)/(3 B) Given that x=A t^ 2 -B t^ 3 velocity = dx dt = 2 ~A t- 3 ~B ^ 2 and acceleration = d dt ( dx dt )= 2 ~A- 6 Bt For acceleration to be zero 2 ~A-6 Bt=0 . t= 2 ~A 6 ~B = A 3 ~B

BITSAT 2010

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

BITSAT 2010 · Physics · Motion in One Dimension

The position x of a particle varies with time (t) as x=A ^ 2 -B ^ 3...

The position

\mathrm{x}

of a particle varies with time

(\mathrm{t})

\mathrm{x}=\mathrm{A} \mathrmt^{2}-\mathrm{B} \mathrmt^{3}

. The acceleration at time

\mathrm{t}

of the particle will be equal to zero. What is the value of

t ?

Options

A $\frac{2 \mathrm{ A}}{3 \mathrm{ B}}$
B $\frac{\mathrm{A}}{\mathrm{B}}$
C $\frac{A}{3 B}$
Dzero

Correct answer

C. $\frac{A}{3 B}$

Step-by-step solution

Given that

x=A t^{2}-B t^{3}

\therefore

\text{ velocity }=\frac{\mathrm{dx}}{\mathrm{dt}}=

\mathrm{~A} \mathrm{t}-

\mathrm{~B} \mathrmt^{2}

and acceleration =

\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=

\mathrm{~A}-

\mathrm{Bt}

For acceleration to be zero

2 \mathrm{~A}-6 \mathrm{Bt}=0

. \therefore \quad

\mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}=\frac{\mathrm{A}}{3 \mathrm{~B}}

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