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, we need to find the acceleration of a particle when its velocity is \(1 \, \text{m/s}\) given the relationship between velocity \(v\) and displacement \(x\) as:

\[ v = \sqrt{49 + x} \]

### Step 1: Differentiate the velocity with respect to time
We know that acceleration \(a\) is defined as the rate of change of velocity with respect to time:

\[ a = \frac{dv}{dt} \]

Using the chain rule, we can express this as:

\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]

### Step 2: Find \(\frac{dv}{dx}\)
First, we need to differentiate \(v\) with respect to \(x\):

\[ v = \sqrt{49 + x} \]

Differentiating both sides with respect to \(x\):

\[ \frac{dv}{dx} = \frac{1}{2\sqrt{49 + x}} \cdot \frac{d(49 + x)}{dx} = \frac{1}{2\sqrt{49 + x}} \cdot 1 = \frac{1}{2\sqrt{49 + x}} \]

### Step 3: Substitute \(\frac{dx}{dt}\)
Since \(v = \frac{dx}{dt}\), we can write:

\[ \frac{dx}{dt} = v \]

### Step 4: Substitute into the acceleration formula
Now substituting \(\frac{dv}{dx}\) and \(\frac{dx}{dt}\) into the acceleration formula:

\[ a = \frac{dv}{dx} \cdot v = \left(\frac{1}{2\sqrt{49 + x}}\right) \cdot v \]

### Step 5: Find \(x\) when \(v = 1 \, \text{m/s}\)
Now, we need to find the value of \(x\) when \(v = 1 \, \text{m/s}\):

\[ 1 = \sqrt{49 + x} \]

Squaring both sides:

\[ 1^2 = 49 + x \implies 1 = 49 + x \implies x = 1 - 49 = -48 \]

### Step 6: Substitute \(x\) back into the acceleration formula
Now substitute \(x = -48\) into the acceleration formula:

\[ a = \frac{1}{2\sqrt{49 + (-48)}} \cdot 1 = \frac{1}{2\sqrt{1}} = \frac{1}{2} \]

Thus, the acceleration when the velocity is \(1 \, \text{m/s}\) is:

\[ a = 0.5 \, \text{m/s}^2 \]

### Final Answer
The acceleration when the velocity is \(1 \, \text{m/s}\) is \(0.5 \, \text{m/s}^2\).

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