A point moves such that its displacement as a function of times is given by `x^(2)=t^(2)+1`. Its acceleration at time t is

Options

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

Explanation

To find the acceleration of the point moving according to the displacement function \( x^2 = t^2 + 1 \), we will follow these steps:

### Step 1: Differentiate the displacement equation
Given the equation:
\[ x^2 = t^2 + 1 \]
We differentiate both sides with respect to time \( t \).

Using implicit differentiation:
\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(t^2 + 1) \]
This gives:
\[ 2x \frac{dx}{dt} = 2t \]

### Step 2: Solve for velocity
From the equation \( 2x \frac{dx}{dt} = 2t \), we can simplify it:
\[ x \frac{dx}{dt} = t \]
Now, solving for \( \frac{dx}{dt} \) (which represents velocity \( v \)):
\[ \frac{dx}{dt} = \frac{t}{x} \]

### Step 3: Differentiate velocity to find acceleration
Now, we need to find acceleration \( a \), which is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{x}\right) \]

Using the quotient rule for differentiation:
\[ \frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x \cdot \frac{dt}{dt} - t \cdot \frac{dx}{dt}}{x^2} \]
This simplifies to:
\[ \frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x - t \cdot \frac{dx}{dt}}{x^2} \]

### Step 4: Substitute \( \frac{dx}{dt} \)
We already found \( \frac{dx}{dt} = \frac{t}{x} \). Substituting this into the acceleration equation:
\[ a = \frac{x - t \cdot \frac{t}{x}}{x^2} \]
This simplifies to:
\[ a = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3} \]

### Step 5: Use the original displacement equation
From the original equation \( x^2 = t^2 + 1 \), we can express \( x^2 - t^2 \):
\[ x^2 - t^2 = 1 \]
Substituting this back into the acceleration equation:
\[ a = \frac{1}{x^3} \]

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