A particle moves along x-axis such that its distance from the origin at the end of time t minutes is given by `x= sqrt(t^(2)+1)-2t`. Its acceleration at `t= sqrt(3)` is

Options

A. `2-2sqrt(3)`
B. `(1)/(2)(sqrt(3)-4)`
C. `(1)/(8)`
D. `(1)/(2)`

Answer

To find the acceleration of the particle at time \( t = \sqrt{3} \), we start with the position function given by:

\[ x(t) = \sqrt{t^2 + 1} - 2t \]

### Step 1: Find the velocity
The velocity \( v(t) \) is the first derivative of the position function \( x(t) \) with respect to time \( t \):

\[ v(t) = \frac{dx}{dt} \]

Using the chain rule and the derivative of \( \sqrt{t^2 + 1} \):

\[ \frac{d}{dt}(\sqrt{t^2 + 1}) = \frac{1}{2\sqrt{t^2 + 1}} \cdot (2t) = \frac{t}{\sqrt{t^2 + 1}} \]

Thus, the velocity becomes:

\[ v(t) = \frac{t}{\sqrt{t^2 + 1}} - 2 \]

### Step 2: Find the acceleration
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):

\[ a(t) = \frac{dv}{dt} \]

Now, we differentiate \( v(t) \):

\[ a(t) = \frac{d}{dt} \left( \frac{t}{\sqrt{t^2 + 1}} - 2 \right) \]

Using the quotient rule for the first term:

\[ \frac{d}{dt} \left( \frac{t}{\sqrt{t^2 + 1}} \right) = \frac{\sqrt{t^2 + 1} \cdot 1 - t \cdot \frac{1}{2\sqrt{t^2 + 1}} \cdot (2t)}{t^2 + 1} \]

This simplifies to:

\[ \frac{\sqrt{t^2 + 1} - \frac{t^2}{\sqrt{t^2 + 1}}}{t^2 + 1} = \frac{(t^2 + 1) - t^2}{(t^2 + 1)\sqrt{t^2 + 1}} = \frac{1}{(t^2 + 1)\sqrt{t^2 + 1}} \]

Thus, the acceleration becomes:

\[ a(t) = \frac{1}{(t^2 + 1)\sqrt{t^2 + 1}} \]

### Step 3: Evaluate the acceleration at \( t = \sqrt{3} \)
Now we substitute \( t = \sqrt{3} \) into the acceleration function:

\[ a(\sqrt{3}) = \frac{1}{((\sqrt{3})^2 + 1)\sqrt{(\sqrt{3})^2 + 1}} = \frac{1}{(3 + 1)\sqrt{3 + 1}} = \frac{1}{4 \cdot 2} = \frac{1}{8} \]

### Final Answer
Thus, the acceleration of the particle at \( t = \sqrt{3} \) is:

\[ \boxed{\frac{1}{8}} \]

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