Question

The particle moves a distance $x$ in time $t$ according to the equation $x={{(t+5)}^{-1}}$ . The acceleration of the particle is proportional to
\[\begin{align}
  & \text{A}\text{. }{{(\text{Velocity})}^{\dfrac{3}{2}}} \\
 & \text{B}\text{. }{{(\text{Distance})}^{2}} \\
 & \text{C}\text{. }{{(\text{Distance})}^{-2}} \\
 & \text{D}\text{. }{{(\text{Velocity})}^{\dfrac{2}{3}}} \\
\end{align}\]

Answer 1

Hint: Use the differential formulae for velocity and acceleration in terms of distance and time. From these equations, compare the value of acceleration with the values calculated for velocity and distance to find the correct answer.

Formula used: The derivative formula:
$\begin{align}
  & y={{x}^{n}} \\
 & \dfrac{dy}{dx}=n{{x}^{n-1}} \\
\end{align}$

Formulae for velocity and acceleration:
$\begin{align}
  & v=\dfrac{dx}{dt} \\
 & a=\dfrac{dv}{dt} \\
 & a=v\dfrac{dv}{dx} \\
\end{align}$

Complete step-by-step answer:
Distance covered by a particle is defined as the length it has travelled to reach a point from an initial point.

Whereas, the velocity of the particle is defined as the ratio of its change in position (distance covered) to the time taken.

For the given particle, we are given its distance x as a function of time t.

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

For any point of time t, the instantaneous velocity then becomes:

$v=\dfrac{dx}{dt}$ . That is the ratio of the differential (very small) amount of distance covered in a differential time interval.

$\begin{align}
  & v=\dfrac{dx}{dt} \\
 & v=-{{(t+5)}^{-2}} \\
\end{align}$

Where, the derivative of a function y with respect to x is given by:

 $\begin{align}
  & y={{x}^{n}} \\
 & \dfrac{dy}{dx}=n{{x}^{n-1}} \\
 & ~ \\
\end{align}$

Acceleration is defined as the change in the velocity of the particle to the time taken.
Similarly, the instantaneous acceleration is given below.

Also, by multiplying and dividing dx in the numerator and the denominator, acceleration in terms of velocity and displacement is expressed as:

\[\begin{align}
  & a=\dfrac{dv}{dt} \\
 & a=\dfrac{dv}{dt}\dfrac{dx}{dx}=\dfrac{dx}{dt}\dfrac{dv}{dx}=v\dfrac{dv}{dx} \\
\end{align}\]

That is, the expression for acceleration becomes:

$a=v\dfrac{dv}{dx}$

Now, putting the expression $x={{(t+5)}^{-1}}$ and $v=-{{(t+5)}^{-2}}$ in the previous expression:

$\begin{align}
  & \text{where, }x={{(t+5)}^{-1}}\text{ and }v=-{{(t+5)}^{-2}} \\
 & a=-{{(t+5)}^{-2}}\dfrac{-d{{(t+5)}^{-2}}}{d{{(t+5)}^{-1}}} \\
 & \text{Taking }p={{(t+5)}^{-1}} \\
 & \text{The equation reduces to:} \\
 & a={{p}^{2}}\dfrac{d{{p}^{2}}}{dp}={{p}^{2}}.2p=2{{p}^{3}} \\
\end{align}$

Notice that,

$\begin{align}
  & x=p \\
 & v=-{{p}^{2}} \\
 & \text{Therefore:} \\
 & p=-{{(v)}^{\dfrac{1}{2}}} \\
\end{align}$

Now, let us find by the expression of acceleration, what is the relation it holds with velocity and displacement.

$\begin{align}
  & a=-2{{p}^{3}} \\
 & \text{That is, }a\propto {{p}^{3}} \\
\end{align}$ … (1)

With displacement:
Since,

$p=x$

Therefore, the proportionality (1) becomes:

$\begin{align}
  & a\propto {{x}^{3}} \\
 & \text{acceleration}\propto {{(\text{displacement})}^{3}} \\
\end{align}$

With velocity:

Since,

$p=-{{(v)}^{\dfrac{1}{2}}}$

Therefore, the proportionality (1) becomes:

$\begin{align}
  & a\propto {{({{v}^{\dfrac{1}{2}}})}^{3}} \\
 & \text{acceleration}\propto {{(\text{velocity})}^{\dfrac{3}{2}}} \\
\end{align}$

Therefore, option A is correct.

Note: Be careful while differentiating an expression. As there might be a function of x inside another one. For example, in this question, if the expression for distance was $x'={{(3t+5)}^{-1}}$ instead of $x={{(t+5)}^{-1}}$. Then, the derivatives would have been different. That is,

$\begin{align}
  & v=\dfrac{dx'}{dt} \\
 & v=-3{{(t+5)}^{-2}} \\
 & \text{Whereas, }v=\dfrac{dx}{dt} \\
 & v=-{{(t+5)}^{-2}} \\
\end{align}$

As,
$\begin{align}
  & y={{[f(x)]}^{n}} \\
 & \dfrac{dy}{dx}=n{{[f(x)]}^{n-1}}\dfrac{df(x)}{dx} \\
 & ~ \\
\end{align}$.