A particle moves in `x-y` plane according to the law `x = 4 sin 6t and y = 4 (1-cos 6t)`. The distance traversed by the particle in 4 second is (`x & y` are in meters)

Options

A. 96 m
B. 48 m
C. 24 m
D. 108 m

Answer

To solve the problem, we need to find the distance traversed by the particle in 4 seconds given the equations of motion in the x and y directions.

### Step-by-Step Solution:

1. **Identify the equations of motion**:
The position of the particle is given by:
\[ x(t) = 4 \sin(6t) \]
\[ y(t) = 4(1 - \cos(6t)) \]

2. **Differentiate the equations to find velocities**:
- The velocity in the x-direction (\(v_x\)) is the derivative of \(x(t)\) with respect to time \(t\):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(4 \sin(6t)) = 4 \cdot 6 \cos(6t) = 24 \cos(6t) \]
- The velocity in the y-direction (\(v_y\)) is the derivative of \(y(t)\) with respect to time \(t\):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(4(1 - \cos(6t))) = 0 + 4 \cdot 6 \sin(6t) = 24 \sin(6t) \]

3. **Calculate the speed of the particle**:
The speed \(v\) is given by the formula:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the expressions for \(v_x\) and \(v_y\):
\[ v = \sqrt{(24 \cos(6t))^2 + (24 \sin(6t))^2} \]
\[ = \sqrt{576 \cos^2(6t) + 576 \sin^2(6t)} \]
\[ = \sqrt{576 (\cos^2(6t) + \sin^2(6t))} \]
Using the identity \(\cos^2(\theta) + \sin^2(\theta) = 1\):
\[ = \sqrt{576} = 24 \text{ m/s} \]

4. **Calculate the distance traveled in 4 seconds**:
The distance \(d\) is given by:
\[ d = v \cdot t \]
Substituting the values we found:
\[ d = 24 \text{ m/s} \cdot 4 \text{ s} = 96 \text{ m} \]

Thus, the distance traversed by the particle in 4 seconds is **96 meters**.