Question

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$.
Its acceleration at any time $t$ is $x^{-n}$ where $n =$ ____.

Answer 1

Correct Answer: 3

Given the particle's displacement \(x\) and time \(t\) are related by \(x^2 = 1 + t^2\). The particle's acceleration is expressed as \(x^{-n}\). Determine the value of \(n\).

Concepts Utilized:

The calculation requires determining the particle's acceleration, defined as the second derivative of displacement with respect to time. The core steps involve:

1. Velocity (\(v\)): Defined as the first derivative of displacement with respect to time: \(v = \frac{dx}{dt}\).
2. Acceleration (\(a\)): Defined as the second derivative of displacement with respect to time, or the first derivative of velocity: \(a = \frac{d^2x}{dt^2} = \frac{dv}{dt}\).
3. Implicit Differentiation: Employed because displacement \(x\) is not explicitly defined as a function of \(t\).
4. Quotient Rule: Applied for differentiating the velocity function: \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \).

Solution Procedure:

Step 1: Calculate the velocity (\(v\)) by differentiating the provided equation with respect to time \(t\).

The governing relation is:

\[x^2 = 1 + t^2\]

Differentiating both sides with respect to \(t\):

\[\frac{d}{dt}(x^2) = \frac{d}{dt}(1 + t^2)\]

Applying the chain rule on the left side yields:

\[2x \frac{dx}{dt} = 2t\]

Substituting \(v = \frac{dx}{dt}\):

\[2xv = 2t\]

Solving for \(v\):

\[v = \frac{t}{x}\]

Step 2: Determine the acceleration (\(a\)) by differentiating the velocity with respect to time \(t\).

Using \(a = \frac{dv}{dt}\) and the quotient rule for \(v = \frac{t}{x}\):

\[a = \frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x \frac{d(t)}{dt} - t \frac{d(x)}{dt}}{x^2}\]

Given that \(\frac{d(t)}{dt} = 1\) and \(\frac{dx}{dt} = v\), the expression simplifies to:

\[a = \frac{x(1) - t(v)}{x^2} = \frac{x - tv}{x^2}\]

Step 3: Substitute the expression for velocity \(v\) into the acceleration equation.

Replacing \(v\) with \(\frac{t}{x}\) in the acceleration equation:

\[a = \frac{x - t\left(\frac{t}{x}\right)}{x^2} = \frac{x - \frac{t^2}{x}}{x^2}\]

Simplifying the compound fraction by multiplying the numerator and denominator by \(x\):

\[a = \frac{x^2 - t^2}{x^3}\]

Step 4: Rewrite the acceleration solely in terms of \(x\).

From the initial equation \(x^2 = 1 + t^2\), it follows that \(t^2 = x^2 - 1\). Substituting this into the acceleration expression:

\[a = \frac{x^2 - (x^2 - 1)}{x^3}\]\[a = \frac{x^2 - x^2 + 1}{x^3} = \frac{1}{x^3}\]

Step 5: Equate the derived acceleration with the given form to solve for \(n\).

The computed acceleration is \(a = \frac{1}{x^3}\), which is equivalent to:

\[a = x^{-3}\]

Comparing this with the provided acceleration form \(x^{-n}\):

\[x^{-n} = x^{-3}\]

This comparison directly yields \(n = 3\).

The value of \(n\) is 3.