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 $n =$ ____.
Its acceleration at any time $t$ is $x^{-n}$ where $n =$ ____.
Correct Answer: 3
Approach Solution - 1
The problem provides the relationship between the displacement \(x\) of a particle and time \(t\) as \(x^2 = 1 + t^2\). We are asked to find the value of \(n\) if the particle's acceleration is given by the expression \(x^{-n}\).
Concept Used:
The solution involves finding the acceleration of the particle, which is the second derivative of displacement with respect to time. The key steps are:
- Velocity (\(v\)): The velocity is the first derivative of displacement with respect to time, \(v = \frac{dx}{dt}\).
- Acceleration (\(a\)): The acceleration is the second derivative of displacement, or the first derivative of velocity, with respect to time, \(a = \frac{d^2x}{dt^2} = \frac{dv}{dt}\).
- Implicit Differentiation: Since the displacement \(x\) is not given explicitly as a function of \(t\), we use implicit differentiation to find the derivatives.
- Quotient Rule: The quotient rule for differentiation, \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \), is used to find the derivative of the velocity.
Step-by-Step Solution:
Step 1: Find the velocity (\(v\)) by differentiating the given equation with respect to time \(t\).
The given relation is:
\[ x^2 = 1 + t^2 \]Differentiating both sides with respect to \(t\):
\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(1 + t^2) \]Using the chain rule on the left side:
\[ 2x \frac{dx}{dt} = 2t \]Since \(v = \frac{dx}{dt}\), we have:
\[ 2xv = 2t \]Solving for \(v\):
\[ v = \frac{t}{x} \]Step 2: Find the acceleration (\(a\)) by differentiating the velocity with respect to time \(t\).
Acceleration is \(a = \frac{dv}{dt}\). We use the quotient rule to differentiate \(v = \frac{t}{x}\):
\[ a = \frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x \frac{d(t)}{dt} - t \frac{d(x)}{dt}}{x^2} \]We know that \(\frac{d(t)}{dt} = 1\) and \(\frac{dx}{dt} = v\). So,
\[ a = \frac{x(1) - t(v)}{x^2} = \frac{x - tv}{x^2} \]Step 3: Substitute the expression for velocity \(v\) into the acceleration equation.
Substitute \(v = \frac{t}{x}\) into the equation for acceleration:
\[ a = \frac{x - t\left(\frac{t}{x}\right)}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} \]To simplify the complex fraction, multiply the numerator and denominator by \(x\):
\[ a = \frac{x^2 - t^2}{x^3} \]Step 4: Express the acceleration solely in terms of \(x\).
From the original equation, \(x^2 = 1 + t^2\), we can express \(t^2\) as \(t^2 = x^2 - 1\). Substitute this into the expression for acceleration:
\[ a = \frac{x^2 - (x^2 - 1)}{x^3} \] \[ a = \frac{x^2 - x^2 + 1}{x^3} = \frac{1}{x^3} \]Step 5: Compare the result with the given form to find \(n\).
The calculated acceleration is \(a = \frac{1}{x^3}\), which can be written as:
\[ a = x^{-3} \]We are given that the acceleration is of the form \(x^{-n}\). By comparing the two expressions,
\[ x^{-n} = x^{-3} \]we can conclude that \(n = 3\).
The value of \(n\) is 3.
Approach Solution -2
\[x^2 = 1 + t^2\]
Differentiating with respect to \(t\):
\[2x \frac{dx}{dt} = 2t\]
\[x \cdot v = t \quad \text{(where \(v = \frac{dx}{dt}\))}\]
Differentiating again:
\[x \frac{dv}{dt} + v \frac{dx}{dt} = 1\]
\[x \cdot a + v^2 = 1 \quad \text{(where \(a = \frac{dv}{dt}\))}\]
Simplify:
\[a = \frac{1 - v^2}{x} = \frac{1 - t^2 / x^2}{x}\]
\[a = \frac{1}{x^3} = x^{-3}\]
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