The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by `v^2=108-9x^2` (all equation are in CGS units):

Options

A. (a)The motion is uniformly accelerated along the straight line
B. (b)the magnitude of the acceleration at a distance 3 cm from the point is `27(cm)/(s^2)`
C. (c)the motion is simple harmonic about the given fixed point
D. (d)the maximum displacement from the fixed point is 4 cm

Explanation

To solve the problem, we need to analyze the given equation for the speed of a particle moving along a straight line, which is expressed as:

\[ v^2 = 108 - 9x^2 \]

### Step 1: Express the speed \( v \)

From the equation, we can express the speed \( v \) as:

\[ v = \sqrt{108 - 9x^2} \]

### Step 2: Find the acceleration \( a \)

Acceleration \( a \) can be defined as the rate of change of velocity with respect to time. We can use the chain rule to express acceleration in terms of position \( x \):

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

Using the chain rule, we can express \( \frac{dv}{dt} \) as:

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

Thus, we can rewrite acceleration as:

\[ a = v \cdot \frac{dv}{dx} \cdot v = v^2 \cdot \frac{dv}{dx} \]

### Step 3: Differentiate \( v \) with respect to \( x \)

Now we need to differentiate \( v \) with respect to \( x \):

\[ v = 3\sqrt{12 - x^2} \]

Differentiating \( v \):

\[ \frac{dv}{dx} = 3 \cdot \frac{1}{2\sqrt{12 - x^2}} \cdot (-2x) = -\frac{3x}{\sqrt{12 - x^2}} \]

### Step 4: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration equation

Now we substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration equation:

\[ a = v^2 \cdot \frac{dv}{dx} \]

Substituting the values:

\[ a = (3\sqrt{12 - x^2})^2 \cdot \left(-\frac{3x}{\sqrt{12 - x^2}}\right) \]

Calculating \( v^2 \):

\[ v^2 = 9(12 - x^2) = 108 - 9x^2 \]

Thus, we have:

\[ a = (108 - 9x^2) \cdot \left(-\frac{3x}{\sqrt{12 - x^2}}\right) \]

### Step 5: Evaluate acceleration at \( x = 3 \)

Now, we will evaluate the acceleration at \( x = 3 \):

1. Calculate \( v \) at \( x = 3 \):

\[ v = 3\sqrt{12 - 3^2} = 3\sqrt{3} \]

2. Substitute \( x = 3 \) into \( \frac{dv}{dx} \):

\[ \frac{dv}{dx} = -\frac{3(3)}{\sqrt{12 - 3^2}} = -\frac{9}{\sqrt{3}} = -3\sqrt{3} \]

3. Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration equation:

\[ a = (3\sqrt{3})^2 \cdot (-3\sqrt{3}) = 27 \cdot (-3\sqrt{3}) = -81\sqrt{3} \]

### Step 6: Conclusion

The acceleration at \( x = 3 \) is negative, indicating it is directed opposite to the motion. The magnitude of acceleration is:

\[ |a| = 81\sqrt{3} \]