Question

The position $x$ of a particle varies with time $(t)$ as $x = a{t^2} - b{t^3}$. The
acceleration at time $t$ of the particle will be equal to zero, where $t$ is equal to
(A) $\dfrac{{2a}}{{3b}}$
(B) $\dfrac{a}{b}$
(C) $\dfrac{a}{{3b}}$
(D) $zero$

Answer 1

Hint:Here, the position of the particle is given as a function of time. The velocity will be a function of time and the acceleration will be a function of time as well. Using the given equation of position of the particle, find the acceleration of the particle and equate it to zero, then find $t$.

Complete step by step solution:
The acceleration is the rate of change of velocity with respect to time, whereas the velocity is the rate of change of position with respect to time.

The acceleration is given as the derivative of the velocity with respect to time. Mathematically,
$a = \dfrac{{dv}}{{dt}}$ $...\left( 1 \right)$
The velocity is given as the derivative of the position with respect to time. Mathematically,
$v = \dfrac{{dx}}{{dt}}$ $...\left( 2 \right)$
Substituting the value of velocity $v$ from the equation $\left( 2 \right)$ in the equation $\left( 1
\right)$.
$
a = \dfrac{{d\left( {\dfrac{{dx}}{{dt}}} \right)}}{{dt}} \\
a = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right) \\
a = \dfrac{{{d^2}x}}{{d{t^2}}} \\
$
The acceleration is the double derivative of the position with respect to time.
$
\dfrac{{dx}}{{dt}} = \dfrac{{d\left( {a{t^2} - b{t^3}} \right)}}{{dt}} \\
\dfrac{{dx}}{{dt}} = \dfrac{{d\left( {a{t^2}} \right)}}{{dt}} + \dfrac{{d\left( { - b{t^3}} \right)}}{{dt}}
\\
\dfrac{{dx}}{{dt}} = \left( a \right)\left( {2t} \right) - \left( b \right)\left( {3{t^2}} \right) \\
\dfrac{{dx}}{{dt}} = 2at - 3b{t^2} \\
$
The velocity is given $2at - 3b{t^2}$
Now, let us find the acceleration.
$
a = \dfrac{{dv}}{{dt}} \\

a = \dfrac{d}{{dt}}\left( {2at - 3b{t^2}} \right) \\

a = \dfrac{{d\left( {2at} \right)}}{{dt}} + \dfrac{{d\left( { - 3b{t^2}} \right)}}{{dt}} \\

a = 2a - 6bt \\
$
The acceleration is zero,
$
\therefore a = 0 \\
\therefore 2a - 6bt = 0 \\
\therefore 6bt = 2a \\
\therefore t = \dfrac{a}{{3b}} \\
$
The time at which the acceleration is zero will be $t = \dfrac{a}{{3b}}$.

Hence, the acceleration at time $t$ of the particle will be equal to zero, where $t$ is equal to $\dfrac{a}{{3b}}$.

Note:Keep in mind that the acceleration is the derivative of velocity with respect to time or the double derivative of position with respect to time. While finding derivatives, use the properties of derivatives such as the addition property, the chain rule, the product rule.

Here, a condition is given to you that the acceleration is equal to zero at some moment or at some time $t$. You can find similar questions, you have to apply the given condition appropriately and find the quantity which is asked.