A particle moves rectilinearly. Its displacement x at time t is given by `x^(2) = at^(2) + b` where a and b are constants. Its acceleration at time t is proportional to

Options

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

Explanation

To solve the problem, we need to determine how the acceleration of the particle is related to its displacement \( x \) at time \( t \). The displacement is given by the equation:

\[ x^2 = at^2 + b \]

where \( a \) and \( b \) are constants. We will follow these steps to find the relationship:

### Step 1: Differentiate the displacement equation to find velocity
We start by differentiating the given equation with respect to time \( t \):

\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(at^2 + b) \]

Using the chain rule on the left side, we have:

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

### Step 2: Solve for velocity \( \frac{dx}{dt} \)
Rearranging the equation gives us:

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

### Step 3: Differentiate velocity to find acceleration
Next, we differentiate the velocity \( \frac{dx}{dt} \) to find acceleration \( a \):

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

Using the quotient rule, we have:

\[ \frac{d^2x}{dt^2} = \frac{x \cdot a - at \cdot \frac{dx}{dt}}{x^2} \]

Substituting \( \frac{dx}{dt} = \frac{at}{x} \) into the equation:

\[ \frac{d^2x}{dt^2} = \frac{ax - at \cdot \frac{at}{x}}{x^2} \]

### Step 4: Simplify the expression for acceleration
This simplifies to:

\[ \frac{d^2x}{dt^2} = \frac{ax - \frac{a^2t^2}{x}}{x^2} \]

Now substituting \( x^2 = at^2 + b \):

\[ \frac{d^2x}{dt^2} = \frac{ax - \frac{a^2t^2}{x}}{(at^2 + b)} \]

### Step 5: Identify the proportionality of acceleration
From the above expression, we can see that the acceleration \( a \) is proportional to:

\[ \frac{1}{x} - \frac{a^2t^2}{x^3} \]

Thus, we conclude that the acceleration \( a \) is proportional to \( \frac{1}{x} \) and \( \frac{1}{x^2} \).

### Final Answer
The acceleration at time \( t \) is proportional to:

\[ \frac{1}{x} - \frac{1}{x^2} \]