A particle moves along the x - axis according to `x = A[1 + sin omega t]`. What distance does is travel in time interval from `t = 0` to `t = 2.5pi//omega` ?

Options

A. `4A`
B. `6 A`
C. `5 A`
D. `3 A`

Explanation

To solve the problem, we need to determine the distance traveled by the particle in the time interval from \( t = 0 \) to \( t = \frac{2.5\pi}{\omega} \) given the equation of motion \( x = A(1 + \sin(\omega t)) \).

### Step-by-Step Solution:

1. **Identify the Time Period**:
The motion described is periodic, and the time period \( T \) of the sine function is given by:
\[ T = \frac{2\pi}{\omega} \]

2. **Break Down the Time Interval**:
The time interval from \( t = 0 \) to \( t = \frac{2.5\pi}{\omega} \) can be expressed as:
\[ t = \frac{2.5\pi}{\omega} = \frac{2\pi}{\omega} + \frac{0.5\pi}{\omega} = T + \frac{T}{4} \]
This indicates that the particle completes one full cycle (from \( t = 0 \) to \( t = T \)) and then moves an additional \( \frac{T}{4} \) (which is \( \frac{1}{4} \) of the period).

3. **Determine the Positions**:
- At \( t = 0 \):
\[ x(0) = A(1 + \sin(0)) = A(1 + 0) = A \]
- At \( t = T \) (one complete cycle):
\[ x(T) = A(1 + \sin(2\pi)) = A(1 + 0) = A \]
- At \( t = \frac{T}{4} \) (quarter period):
\[ x\left(\frac{T}{4}\right) = A\left(1 + \sin\left(\frac{\pi}{2}\right)\right) = A(1 + 1) = 2A \]

4. **Calculate the Distance Traveled**:
- From \( t = 0 \) to \( t = T \), the particle travels from \( A \) to \( A \) (back to the starting point), covering a distance of \( 0 \).
- From \( t = T \) to \( t = \frac{T}{4} \), the particle moves from \( A \) to \( 2A \) (the maximum position), covering a distance of:
\[ 2A - A = A \]
- Then, it returns back from \( 2A \) to \( A \), covering another distance of \( A \).

5. **Total Distance**:
The total distance traveled in the time interval \( t = 0 \) to \( t = \frac{2.5\pi}{\omega} \) is:
\[ \text{Total Distance} = A + A = 2A \]

### Final Answer:
The distance traveled by the particle in the time interval from \( t = 0 \) to \( t = \frac{2.5\pi}{\omega} \) is \( 2A \).