The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is `v^(2)=144-9x^(2)`. Select the correct alternatives

Options

A. The magnitude of acceleration at a distance 3 units from the fixed point is 27 units
B. the motion is simple harmonic with `T=(2pi)/(3)` units
C. The maximum displacement from the fixed point is 4 units
D. all are correct

Explanation

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

**Given:**
\[ v^2 = 144 - 9x^2 \]

### Step 1: Identify the form of the equation
The equation resembles the form of the speed in simple harmonic motion (SHM), which is given by:
\[ v^2 = \omega^2 (a^2 - x^2) \]
where \( \omega \) is the angular frequency and \( a \) is the amplitude.

### Step 2: Compare coefficients
From the given equation, we can compare it with the SHM equation:
- The constant term \( 144 \) corresponds to \( \omega^2 a^2 \).
- The coefficient of \( x^2 \) is \( -9 \), which corresponds to \( -\omega^2 \).

Thus, we can write:
1. \( \omega^2 = 9 \)
2. \( \omega^2 a^2 = 144 \)

### Step 3: Solve for \( \omega \) and \( a \)
From \( \omega^2 = 9 \):
\[ \omega = 3 \]

Now substituting \( \omega^2 \) into the second equation:
\[ 9a^2 = 144 \]
\[ a^2 = \frac{144}{9} = 16 \]
\[ a = 4 \]

### Step 4: Calculate the time period \( T \)
The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 3 \):
\[ T = \frac{2\pi}{3} \]

### Step 5: Calculate the acceleration at a distance of 3 units
The acceleration \( a \) in SHM is given by:
\[ a = -\omega^2 x \]
Substituting \( \omega^2 = 9 \) and \( x = 3 \):
\[ a = -9 \times 3 = -27 \]
The magnitude of acceleration is:
\[ |a| = 27 \text{ units} \]

### Conclusion
Now we can summarize the findings:
1. The magnitude of acceleration at a distance of 3 units from the fixed point is 27 units.
2. The motion is SHM with a time period \( T = \frac{2\pi}{3} \).
3. The maximum displacement (amplitude) from the fixed point is 4 units.

Thus, all options provided in the question are correct.