Question

The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as ${x^2} = a{t^2} + 2bt + c$ . The acceleration of the particle varies as:
A) ${x^{ - 3}}$
B) ${x^{3/2}}$
C) ${x^2}$
D) ${x^{ - 2/3}}$

Answer 1

Hint: The change in position with respect to the time is called velocity. This means the derivative of position with respect to time is velocity. Similarly the change in velocity with respect to time is called acceleration. This means the derivative of velocity with respect to time is acceleration.

Complete step by step solution:
Here, in this question it is given that the distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as
${x^2} = a{t^2} + 2bt + c$
On taking square root on both side, we get
\[x = \sqrt {a{t^2} + 2bt + c} \]
Now, On differentiating with respect to time $t$ , we have
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }} \times (2at + 2b)$
As we know that the rate of change of distance with respect to time is the velocity.
So, $ \Rightarrow v = \dfrac{{at + b}}{x}$ ….(i)
On further solving, we get
$v_x = at + b$
Now, again differentiating with respect to distance $x$, we get
$ \Rightarrow \dfrac{{dv}}{{dx}} \times x + v = a \times \dfrac{{dt}}{{dx}}$
As we know that, $\dfrac{{dx}}{{dt}} = v$
So, the above equation becomes,
$ \Rightarrow \left( {\dfrac{{dv}}{{dx}}} \right)x + v = a \times \dfrac{1}{v}$
Now, on multiplying by $v$ on both sides, we get
$ \Rightarrow \left( {v\dfrac{{dv}}{{dx}}} \right)x + {v^2} = a$
Here, in above equation, put $\dfrac{{dx}}{{dt}} = v$, we get
$ \Rightarrow \left( {\dfrac{{dv}}{{dt}}} \right)x + {v^2} = a$
Now we also know that the rate of change of velocity with respect to time is the acceleration.
So, the above equation becomes,
$ \Rightarrow a'x + {v^2} = a$
Here, $a'$ is the acceleration.
$ \Rightarrow a'x = a - {v^2}$
Now put the value of $v$ from the equation (i) in the above equation, we get
$ \Rightarrow a'x = a - {\left( {\dfrac{{at + b}}{x}} \right)^2}$
Now, on further solving, we get
$ \Rightarrow a'x = \dfrac{{a{x^2} - {{(at + b)}^2}}}{{{x^2}}}$
Now from the question put the value of ${x^2}$ in the above equation, we get
$ \Rightarrow a'x = \dfrac{{a(a{t^2} + 2bt + c) - {{(at + b)}^2}}}{{{x^2}}}$
On opening the whole square and all the brackets, we get
$ \Rightarrow a'x = \dfrac{{ac - {b^2}}}{{{x^2}}}$
On further solving, we get
$ \Rightarrow a' = \dfrac{{ac - {b^2}}}{{{x^3}}}$
Here, $a,b\& c$ all are constants
So, the above equation becomes,
$\Rightarrow$ $a' \propto \dfrac{1}{{{x^3}}}$
Also, we can rewrite it as,
$ \Rightarrow$ $a' \propto {x^{ - 3}}$
Hence, The acceleration of the particle varies according to ${x^{ - 3}}$.

Therefore, the correct option is (A).

Note: Acceleration is the first derivative of velocity of an object with respect to the time. Acceleration is a vector quantity, meaning it has both magnitude and direction. It is also the second derivative of the position with respect to time. The SI unit for acceleration is given as $m/{s^2}$.