Question

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

Answer 1

Hint: Acceleration is a quantity of vectors which is defined as the rate at which an object increases its velocity. An object accelerates as its velocity increases. So first we will find velocity by differentiating the distance with respect to time. And then by double differentiating we will get the value of acceleration.
Formula used- \[v = \dfrac{{dx}}{{dt}},a = \dfrac{{dv}}{{dt}}\]

Complete Step-by-Step solution:
Given distance $x = {\left( {t + 5} \right)^{ - 1}}$
We know that velocity:
$v = \dfrac{{dx}}{{dt}}$
So, let us differentiate the equation of distance to find the velocity.
\[
  \because v = \dfrac{{dx}}{{dt}} \\
   \Rightarrow v = \dfrac{d}{{dt}}\left[ {{{\left( {t + 5} \right)}^{ - 1}}} \right] \\
   \Rightarrow v = \dfrac{d}{{d\left( {t + 5} \right)}}\left[ {{{\left( {t + 5} \right)}^{ - 1}}} \right] \times \dfrac{d}{{dt}}\left( {t + 5} \right){\text{ }}\left[ {{\text{chain rule is used here}}} \right] \\
   \Rightarrow v = \left( { - 1} \right){\left( {t + 5} \right)^{ - 2}} \times \left( {\dfrac{{d\left( t \right)}}{{dt}} + \dfrac{{d\left( 5 \right)}}{{dt}}} \right){\text{ }}\left[ {\because \dfrac{d}{{dq}}\left[ {{p^n}} \right] = n{p^{n - 1}}} \right] \\
   \Rightarrow v = \left( { - 1} \right){\left( {t + 5} \right)^{ - 2}} \times \left( {1 + 0} \right) \\
   \Rightarrow v = - {\left( {t + 5} \right)^{ - 2}}..........(1) \\
 \]
As we also know that acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is said to be accelerating if it is continuously changing its velocity.
In order to find the acceleration we will differentiate the velocity with respect to time.
\[
  a = \dfrac{{dv}}{{dt}} \\
   \Rightarrow a = \dfrac{d}{{dt}}\left[ { - {{\left( {t + 5} \right)}^{ - 2}}} \right] \\
   \Rightarrow a = \dfrac{d}{{d\left( {t + 5} \right)}}\left[ { - {{\left( {t + 5} \right)}^{ - 2}}} \right] \times \dfrac{d}{{dt}}\left( {t + 5} \right){\text{ }}\left[ {{\text{chain rule is used here}}} \right] \\
   \Rightarrow a = \left( { - 2} \right){\left( {t + 5} \right)^{ - 3}} \times \left( {\dfrac{{d\left( t \right)}}{{dt}} + \dfrac{{d\left( 5 \right)}}{{dt}}} \right){\text{ }}\left[ {\because \dfrac{d}{{dq}}\left[ {{p^n}} \right] = n{p^{n - 1}}} \right] \\
   \Rightarrow a = \left( { - 2} \right){\left( {t + 5} \right)^{ - 3}} \times \left( {1 + 0} \right) \\
   \Rightarrow a = - 2{\left( {t + 5} \right)^{ - 3}}...........(2) \\
 \]
In order to find the relation between acceleration and velocity:
For equation (2), we have:
\[
   \Rightarrow a = - 2{\left( {t + 5} \right)^{ - 3}} \\
   \Rightarrow a \propto {\left( {t + 5} \right)^{ - 3}} \\
   \Rightarrow a \propto {p^3}.........{\text{(3) }}\left[ {{\text{where }}p = {{\left( {t + 5} \right)}^{ - 1}}} \right] \\
 \]
For equation (1), we have:
\[
   \Rightarrow v = - {\left( {t + 5} \right)^{ - 2}} \\
   \Rightarrow v \propto {\left( {t + 5} \right)^{ - 2}} \\
   \Rightarrow v \propto {p^2}{\text{ }}\left[ {{\text{where }}p = {{\left( {t + 5} \right)}^{ - 1}}} \right] \\
   \Rightarrow {v^{\dfrac{3}{2}}} \propto {p^3}.........{\text{(4)}} \\
 \]
From equation (3) and equation (4)
\[
  \dfrac{a}{{{v^{\dfrac{3}{2}}}}} \propto \dfrac{{{p^3}}}{{{p^3}}} \\
  \dfrac{a}{{{v^{\dfrac{3}{2}}}}} \propto 1 \\
  a = {v^{\dfrac{3}{2}}} \\
 \]
Hence, the acceleration of particle is proportional to \[{\left( {{\text{velocity}}} \right)^{\dfrac{3}{2}}}\]
So, the correct answer is option A.

Note- The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is the equivalent of determining the speed and direction of motion of an object. Acceleration is a quantity of vectors which is defined as the rate at which an object increases its velocity. An object accelerates as its velocity is changing. It has direction and magnitude, since velocity is an example of a vector.