Question

A point moves in a straight line so that its displacement x m at time t sec is given by $x^2=1+t^2$. What is its acceleration in m/sec${}^2$at a time t?
(A) ${\displaystyle \frac{1}{x^3}}$
(B) ${\displaystyle \frac{t}{x^3}}$
(C) ${\displaystyle \frac{1}{x}}-{\displaystyle \frac{t^2}{x^3}}$
(D) ${\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x^2}}$

Answer 1

Hint: The derivative of displacement with respect to time gives velocity and the derivative of velocity with respect to time gives acceleration. So acceleration can be found by differentiating the position vector twice.

Complete step-by-step solution
The displacement of a point at any time t sec is given by the following equation,
$x^2=1+t^2$
Where x is in meters and t is in seconds.
We first need to find the velocity of the point, by taking the derivative of the displacement of the point, which can then be differentiated to find out acceleration.
We take the derivative of displacement of point with respect to time,
${\displaystyle \frac{d(x^2)}{dt}}={\displaystyle \frac{d(1+t^2)}{dt}}$
Since x and t are different variables, we need to use chain rule to find the derivative of the given equation. On the right side of the equation, we can separate the derivatives of the terms separately and add them.
${\displaystyle \frac{d(x^2)}{dx}}{\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d(1)}{dt}}+{\displaystyle \frac{d(t^2)}{dt}}$ …equation (1)
Finding the derivative of separate terms, we obtain,
${\displaystyle \frac{d(x^2)}{dx}}=2x$ [Using the formula ${\displaystyle \frac{d(x^n)}{dx}}=n{x}^{n-1}$]
${\displaystyle \frac{d(1)}{dt}}=0$ [Using the formula ${\displaystyle \frac{d(const)}{dt}}=0$]
${\displaystyle \frac{d(t^2)}{dt}}=2t$ [Using the formula ${\displaystyle \frac{d(t^n)}{dt}}=n{t}^{n-1}$]
Substituting all these values in equation (1), we get,
$2x{\displaystyle \frac{dx}{dt}}=2t$
On simplifying, our equation becomes,
${\displaystyle \frac{dx}{dt}}={\displaystyle \frac{t}{x}}$ …equation (2)
But ${\displaystyle \frac{dx}{dt}}$ is the velocity of the point. So we can obtain the acceleration of the point by differentiating the expression for velocity.
${\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{dx}{dt}}\right)={\displaystyle \frac{d^{2}x}{d{t}^2}}={\displaystyle \frac{dv}{dt}}={\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{t}{x}}\right)$
Using the quotient rule of differentiation to find the derivative of the term on right side, we obtain
${\displaystyle \frac{dv}{dt}}={\displaystyle \frac{x{\displaystyle \frac{dt}{dt}}-t{\displaystyle \frac{dx}{dt}}}{x^2}}$ …equation (3)
Substituting the value of equation (2) in equation (3), we obtain,
Acceleration, a = ${\displaystyle \frac{dy}{dx}}=\left({\displaystyle \frac{x}{x^2}}\right)\times 1-\left({\displaystyle \frac{t}{x^2}}\right)\times \left({\displaystyle \frac{t}{x}}\right)$
Therefore, we obtain that the acceleration is a = $${\displaystyle \frac{1}{x}}-{\displaystyle \frac{t^2}{x^3}}$$.

Hence, option C is correct.

Note: The question can also be solved by taking the square root of the displacement equation to obtain an equation in terms of x. However, the differentiation of that equation would be relatively tougher and more time consuming.