The velocity of any particle is related with its displacement As, `x = sqrt(v+1)`, Calculate acceleration at `x = 5m`.

Options

A. `sqrt6` unit
B. 24 unit
C. 240 unit
D. 25 unit

Answer

To solve the problem, we need to find the acceleration of a particle given the relationship between its displacement \( x \) and its velocity \( v \) as \( x = \sqrt{v + 1} \). We will calculate the acceleration at \( x = 5 \, \text{m} \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ x = \sqrt{v + 1} \]

2. **Square both sides to eliminate the square root:**
\[ x^2 = v + 1 \]

3. **Rearrange the equation to express \( v \) in terms of \( x \):**
\[ v = x^2 - 1 \]

4. **Now, we need to find the acceleration \( a \). Recall that acceleration can be expressed as:**
\[ a = v \frac{dv}{dx} \]
where \( \frac{dv}{dx} \) is the derivative of velocity with respect to displacement.

5. **Differentiate \( v \) with respect to \( x \):**
\[ \frac{dv}{dx} = \frac{d}{dx}(x^2 - 1) = 2x \]

6. **Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:**
\[ a = (x^2 - 1)(2x) \]

7. **Now, we need to find the acceleration at \( x = 5 \, \text{m} \):**
\[ a = (5^2 - 1)(2 \cdot 5) \]
\[ = (25 - 1)(10) \]
\[ = 24 \cdot 10 \]
\[ = 240 \, \text{units} \]

### Final Answer:
The acceleration at \( x = 5 \, \text{m} \) is \( 240 \, \text{units} \).

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