Question

The relation between time $t$ and distance $x$ is $t=a{{x}^{2}}+bx$ where $a$ and $b$ are constants. The acceleration is:
A. $-2a{{v}^{3}}$
B. $2a{{v}^{2}}$
C. $-2a{{v}^{2}}$
D. $2b{{v}^{3}}$

Answer 1

Hint: This problem can be solved by making use of the fact that acceleration is the second derivative with respect to time of the distance. The first derivative of the distance with respect to time is the velocity of the object. By differentiating the given equation with respect to time twice, we can arrive at the answer.

Formula used:
$\dfrac{d\left( \text{distance} \right)}{dt}=\text{velocity}$
$\dfrac{d\left( \text{velocity} \right)}{dt}=\text{acceleration}$
$\dfrac{{{d}^{2}}\left( \text{distance} \right)}{d{{\left( \text{time} \right)}^{2}}}=\text{acceleration}$

Complete step-by-step answer:
The velocity of an object is the derivative of the distance with respect to time. That is,
$\dfrac{d\left( \text{distance} \right)}{dt}=\text{velocity}$--(1)
and the acceleration of an object is the derivative of the velocity with respect to time. That is,
$\dfrac{d\left( \text{velocity} \right)}{dt}=\text{acceleration}$--(2)
Using these equations we can find out the acceleration of the body. We are given that the relation between time $t$ and distance $x$ is $t=a{{x}^{2}}+bx$ where $a$ and $b$ are constants.
Hence, $t=a{{x}^{2}}+bx$. --(3) Now, differentiating (3) with respect to distance, we get,
$\dfrac{dt}{dx}=2ax+b$ $\therefore \dfrac{dx}{dt}=\dfrac{1}{2ax+b}$ ---$\left( \because \dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}} \right)$ Using (1), $\therefore v=\dfrac{1}{2ax+b}$ $\therefore \dfrac{1}{v}=2ax+b$ --(4)
where v is the velocity of the body.
Now, since $\dfrac{d\left( \text{velocity} \right)}{dt}=\text{acceleration}$ we will differentiate (4) with respect to t to get a relation.
$\therefore \dfrac{d\left( \dfrac{1}{v} \right)}{dt}=2a\dfrac{dx}{dt}+0$ --(since and b are constants and $d\left( \text{constant} \right)=0$ )
$\therefore \dfrac{-1}{{{v}^{2}}}\dfrac{dv}{dt}=2av$ --$\left( \because \dfrac{dx}{dt}=v \right)$
$\therefore \text{acceleration = }-2a{{v}^{3}}$ --[from (2)]
Therefore, the required value of acceleration is $-2a{{v}^{3}}$.
Hence, the correct option is A) $-2a{{v}^{3}}$.

Note: Students might try to change the given relation in the question to find out the value of x in terms of t. However, that will only bring in a lot of complications and unnecessary square roots and fractions into the problem. This will make it hard even while differentiating.
They may again be tempted to proceed with and differentiate the value of $v$ and not $\dfrac{1}{v}$ as done above, in the hopes of arriving at the answer directly. However, that will also bring in unnecessary fractions involving $x$ and make the differentiation process very cumbersome and prone to calculation errors.
This question is in fact purposefully set in this way to confuse the students who go for the straightforward method. Students should follow the above process carefully and try to apply it in similar questions to arrive at the answer easily and in a short time.