A particle moving along a straight line has a velocity `v` `m s^(-1)`, when it cleared a distance of x m. These two are connected by the relation v = `sqrt (49 + x)`. When its velocity is `1 m s^(-1)`, its acceleration is

Options

A. `2 m s^(-2)`
B. `7 m s^(-2)`
C. `1 m s^(-2)`
D. `0.5 m s^(-2)`

Explanation

To solve the problem step by step, we start with the given relationship between velocity \( v \) and displacement \( x \):

### Step 1: Understand the relationship
The relationship given is:
\[ v = \sqrt{49 + x} \]

### Step 2: Differentiate the velocity with respect to displacement
To find the acceleration, we will use the formula:
\[ a = v \frac{dv}{dx} \]
First, we need to differentiate \( v \) with respect to \( x \). We can rewrite \( v \) as:
\[ v = (49 + x)^{1/2} \]
Now, we differentiate using the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2}(49 + x)^{-1/2} \cdot \frac{d(49 + x)}{dx} \]
Since the derivative of \( 49 + x \) with respect to \( x \) is \( 1 \), we have:
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{49 + x}} \]

### Step 3: Substitute back into the acceleration formula
Now we substitute \( \frac{dv}{dx} \) back into the acceleration formula:
\[ a = v \cdot \frac{dv}{dx} = v \cdot \frac{1}{2\sqrt{49 + x}} \]

### Step 4: Substitute \( v \) in the acceleration equation
From the relationship \( v = \sqrt{49 + x} \), we can substitute \( v \) into the acceleration formula:
\[ a = \sqrt{49 + x} \cdot \frac{1}{2\sqrt{49 + x}} = \frac{1}{2} \]

### Step 5: Conclusion
Thus, the acceleration \( a \) when the velocity \( v = 1 \, \text{m/s} \) is:
\[ a = 0.5 \, \text{m/s}^2 \]