A point mives in a straight line so its displacement `x` mertre at time `t` second is given by `x^(2)=1+t^(2)`. Its aceleration in `ms^(-2)` at time `t` second is .

Options

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

Explanation

To find the acceleration of the point moving in a straight line with the given displacement equation \( x^2 = 1 + t^2 \), we will follow these steps:

### Step 1: Differentiate the displacement equation
Given:
\[ x^2 = 1 + t^2 \]
We can 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:
\[ 2x \frac{dx}{dt} = 2t \]

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

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

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

### Final Result
Thus, the acceleration \( a \) at time \( t \) is:
\[ a = \frac{1}{x^3} \]