A particle moves a distance x in time t according to equation `x^(2) = 1 + t^(2)` . The acceleration of the particle is

Options

A. `1/x^3`
B. `1/x -1/x^2`
C. `-t/x^2`
D. `1/x-t^2/x^3`

Explanation

To find the acceleration of the particle given the equation \( x^2 = 1 + t^2 \), we can follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ x^2 = 1 + t^2 \]
To find the velocity, we differentiate both sides with respect to time \( t \):
\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(1 + t^2) \]
Using the chain rule on the left side, we have:
\[ 2x \frac{dx}{dt} = 0 + 2t \]
This simplifies to:
\[ 2x \frac{dx}{dt} = 2t \]
Dividing both sides by 2:
\[ x \frac{dx}{dt} = t \]

### Step 2: Solve for velocity
From the equation \( x \frac{dx}{dt} = t \), we can express the velocity \( v = \frac{dx}{dt} \):
\[ v = \frac{t}{x} \]

### Step 3: Differentiate velocity to find acceleration
Next, we differentiate the velocity \( v \) with respect to time \( t \) to find acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{x}\right) \]
Using the quotient rule:
\[ a = \frac{x \cdot \frac{d}{dt}(t) - t \cdot \frac{d}{dt}(x)}{x^2} \]
This gives:
\[ a = \frac{x \cdot 1 - t \cdot \frac{dx}{dt}}{x^2} \]
Substituting \( \frac{dx}{dt} = v = \frac{t}{x} \):
\[ a = \frac{x - t \cdot \frac{t}{x}}{x^2} \]
Simplifying this:
\[ a = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3} \]

### Step 4: Substitute from the original equation
From the original equation \( x^2 - t^2 = 1 \), we can substitute:
\[ a = \frac{1}{x^3} \]

### Final Result
Thus, the acceleration of the particle is:
\[ a = \frac{1}{x^3} \]