Question

If a particle moves along a straight line with the law of motion given as ${s^2} = a{t^2} + 2bt + c$ then, the acceleration varies as
A. $\dfrac{1}{{{s^3}}}$
B. $\dfrac{1}{s}$
C. $\dfrac{1}{{{s^4}}}$
D. $\dfrac{1}{{{s^2}}}$

Answer 1

In order to solve this question, we should know that the equation of motion given in question is in the form of distance and time where a, b, c are constants. And acceleration is the rate of change of velocity with respect to time and velocity is the rate of change of distance with respect to time, so from the given equation, we will find first velocity and then derivative of velocity will be our required acceleration of the particle.

Formula used:
velocity is calculated as $v = \dfrac{{ds}}{{dt}}$ derivative of distance with respect to time.
Acceleration is calculated as $a = \dfrac{{dv}}{{dt}}$ derivative of velocity with respect to time.
Derivative formulas:
$\dfrac{d}{{dx}}(a{x^n}) = an{x^{n - 1}}$
where $a$ is constant and $x$ is variable.
Derivative of the constant term is zero, $\dfrac{d}{{dx}}(a) = 0$.

Complete step by step answer:
According to the question we have given the relation between distance s and time t with a, b, c are constants as
${s^2} = a{t^2} + 2bt + c$
Or we can write it as
$s = \sqrt {a{t^2} + 2bt + c}$
Now, differentiate above relation s with respect to time we get,
$\dfrac{{ds}}{{dt}} = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }}(\dfrac{d}{{dt}}(a{t^2} + 2bt + c))$
Using derivative formulas $\dfrac{d}{{dx}}(a{x^n}) = an{x^{n - 1}}$ we get,
$\dfrac{{ds}}{{dt}} = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }}(2at + 2b)$
$\Rightarrow v = \dfrac{{(at + b)}}{s} \to (i)$ $(\dfrac{{ds}}{{dt}} = v)$
Now again differentiating above relation with respect to time,
$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\dfrac{{at + b}}{s})$

Using law of differentiation $\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{vu' - uv'}}{{{v^2}}}$
where u’ and v’ are the derivatives of u and v function with respect to x variable.
So,
$\dfrac{{dv}}{{dt}} = \dfrac{{s(\dfrac{d}{{dt}}(at + b)) - (\dfrac{{ds}}{{dt}})(at + b)}}{{{s^2}}}$
$\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{s(a) - (\dfrac{{(at + b)(at + b)}}{s})}}{{{s^2}}}$ On putting $\Rightarrow \dfrac{{ds}}{{dt}} = \dfrac{{at + b}}{s}$ from equation (i)
$\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{a{s^2} - {{(at + b)}^2}}}{{{s^3}}}$
From given relation put ${s^2} = a{t^2} + 2bt + c$ in numerator only, we get
$\dfrac{{dv}}{{dt}} = \dfrac{{a(a{t^2} + 2bt + c) - {a^2}{t^2} - {b^2} - 2abt}}{{{s^3}}}$
$\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{ac - {b^2}}}{{{s^3}}}$
And $\dfrac{{dv}}{{dt}} = {a_{acceleration}}$
$\therefore {a_{acceleration}} = \dfrac{{ac - {b^2}}}{{{s^3}}}$
From the above relation we can see that, ${a_{acceleration}} \propto \dfrac{1}{{{s^3}}}$.

Hence, the correct option is A.

Note: It should be remembered that, while solving such equations, derivative of all constant terms is always zero and chain rule is used while solving differentiation if a function is written as $F = f(s(t))$ and we need to find derivative of function $F$ with respect to t then we use the chain rule as $\dfrac{{dF}}{{dt}} = \dfrac{{df}}{{ds}}.\dfrac{{ds}}{{dt}}$ where “f” is a function of “s” and “s” is a function of t.