body moves along x-axis with velocity v = 5 sin(2t)m/s starting from ` x_i =-2.5 m`Ratio of displacement to distance traveled by body from ` t= 0s "to" t= 3pi //2 sec ` is :

Options

A. `1:1`
B. 0
C. `1:3`
D. `1:2`

Explanation

To solve the problem, we need to find the ratio of displacement to distance traveled by a body moving along the x-axis with a given velocity function. The velocity function is given as \( v(t) = 5 \sin(2t) \) m/s, and the body starts from an initial position of \( x_i = -2.5 \) m. We will analyze the motion from \( t = 0 \) s to \( t = \frac{3\pi}{2} \) s.

### Step-by-Step Solution:

1. **Identify the Velocity Function**:
The velocity of the body is given by:
\[ v(t) = 5 \sin(2t) \]

2. **Determine the Time Intervals**:
We need to find the points in time when the velocity is zero, as this will help us determine the intervals of motion. The velocity is zero when:
\[ 5 \sin(2t) = 0 \implies \sin(2t) = 0 \]
This occurs at:
\[ 2t = n\pi \quad (n \in \mathbb{Z}) \implies t = \frac{n\pi}{2} \]
For \( n = 0, 1, 2, 3 \), we find:
- \( t = 0 \)
- \( t = \frac{\pi}{2} \)
- \( t = \pi \)
- \( t = \frac{3\pi}{2} \)

3. **Calculate Displacement**:
Displacement can be calculated by integrating the velocity function over the time interval from \( t = 0 \) to \( t = \frac{3\pi}{2} \):
\[ \text{Displacement} = \int_{0}^{\frac{3\pi}{2}} v(t) \, dt = \int_{0}^{\frac{3\pi}{2}} 5 \sin(2t) \, dt \]
To solve this integral:
\[ = 5 \left[-\frac{1}{2} \cos(2t)\right]_{0}^{\frac{3\pi}{2}} = -\frac{5}{2} \left[\cos(3\pi) - \cos(0)\right] \]
\[ = -\frac{5}{2} \left[-1 - 1\right] = -\frac{5}{2} \times (-2) = 5 \text{ m} \]

4. **Calculate Distance Traveled**:
The distance traveled is the total area under the velocity-time graph, considering the absolute value of the velocity. We need to calculate the area for each segment where the velocity is positive and negative:
- From \( t = 0 \) to \( t = \frac{\pi}{2} \) (positive velocity).
- From \( t = \frac{\pi}{2} \) to \( t = \pi \) (negative velocity).
- From \( t = \pi \) to \( t = \frac{3\pi}{2} \) (positive velocity).

The distance can be calculated as:
\[ \text{Distance} = \int_{0}^{\frac{\pi}{2}} 5 \sin(2t) \, dt + \left| \int_{\frac{\pi}{2}}^{\pi} 5 \sin(2t) \, dt \right| + \int_{\pi}^{\frac{3\pi}{2}} 5 \sin(2t) \, dt \]
Each of these integrals evaluates to:
\[ \int 5 \sin(2t) \, dt = -\frac{5}{2} \cos(2t) \]
Evaluating each segment:
- From \( 0 \) to \( \frac{\pi}{2} \): \( 5 \) m (positive area)
- From \( \frac{\pi}{2} \) to \( \pi \): \( 5 \) m (negative area, but we take absolute value)
- From \( \pi \) to \( \frac{3\pi}{2} \): \( 5 \) m (positive area)

Thus, the total distance traveled is:
\[ \text{Distance} = 5 + 5 + 5 = 15 \text{ m} \]

5. **Calculate the Ratio of Displacement to Distance**:
Finally, we find the ratio of displacement to distance:
\[ \text{Ratio} = \frac{\text{Displacement}}{\text{Distance}} = \frac{5}{15} = \frac{1}{3} \]

### Final Answer:
The ratio of displacement to distance traveled by the body from \( t = 0 \) s to \( t = \frac{3\pi}{2} \) s is \( \frac{1}{3} \).