A particle moves according to the law, `x=acos(pit//2).` . What is the distance covered by it in time interval `t=0` to `t=3` second.

Options

A. 2a
B. 3a
C. 4a
D. a

Answer

To find the distance covered by the particle in the time interval from \( t = 0 \) to \( t = 3 \) seconds, we can follow these steps:

### Step 1: Understand the motion
The motion of the particle is described by the equation:
\[ x = a \cos\left(\frac{\pi t}{2}\right) \]
This indicates that the particle is undergoing simple harmonic motion (SHM).

### Step 2: Determine the time period
The angular frequency \( \omega \) can be identified from the equation:
\[ \omega = \frac{\pi}{2} \]
The time period \( T \) of the motion can be calculated using the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\frac{\pi}{2}} = 4 \text{ seconds} \]

### Step 3: Analyze the motion over one complete cycle
In one complete cycle (4 seconds), the particle moves from:
- Maximum displacement \( +a \) (at \( t = 0 \))
- To the mean position \( 0 \) (at \( t = 2 \) seconds)
- To the minimum displacement \( -a \) (at \( t = 4 \) seconds)

### Step 4: Calculate the distance covered in the first 2 seconds
From \( t = 0 \) to \( t = 2 \) seconds, the particle moves from \( +a \) to \( -a \):
- Distance from \( +a \) to \( 0 \) is \( a \)
- Distance from \( 0 \) to \( -a \) is \( a \)

Thus, the total distance covered in the first 2 seconds is:
\[ a + a = 2a \]

### Step 5: Calculate the distance covered from \( t = 2 \) to \( t = 3 \) seconds
At \( t = 2 \) seconds, the particle is at the mean position \( 0 \). Now we need to find its position at \( t = 3 \) seconds:
\[ x(3) = a \cos\left(\frac{\pi \cdot 3}{2}\right) = a \cos\left(\frac{3\pi}{2}\right) = a \cdot 0 = 0 \]
The particle moves from \( 0 \) to \( 0 \) in this interval, but it actually moves to the maximum position \( +a \) and back to \( 0 \) in the next second.

The distance covered from \( t = 2 \) to \( t = 3 \) seconds is:
- From \( 0 \) to \( +a \) is \( a \)
- From \( +a \) back to \( 0 \) is \( a \)

So, the total distance covered in this interval is:
\[ a \]

### Step 6: Total distance covered
Now, we sum the distances covered in both intervals:
- Distance from \( t = 0 \) to \( t = 2 \) seconds: \( 2a \)
- Distance from \( t = 2 \) to \( t = 3 \) seconds: \( a \)

Thus, the total distance covered from \( t = 0 \) to \( t = 3 \) seconds is:
\[ 2a + a = 3a \]

### Final Answer
The distance covered by the particle in the time interval from \( t = 0 \) to \( t = 3 \) seconds is \( 3a \).

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