A body sliding on a smooth inclined plane requires `4s` to reach the bottom, starting from rest at the at the top. How much time does it take to cover ont-foruth the distance startion from rest at the top?

Options

A. 1 s
B. 4 s
C. 2 s
D. 16 s

Explanation

To solve the problem step by step, let's break it down:

1. **Understanding the Problem**:
- A body is sliding down a smooth inclined plane from rest.
- It takes 4 seconds to reach the bottom of the incline.
- We need to find the time taken to cover one-fourth of the distance from the top.

2. **Using the Equation of Motion**:
- The distance covered by the body when it reaches the bottom (full distance L) can be expressed using the second equation of motion:
\[ L = \frac{1}{2} a t^2 \]
- Here, \( a = g \sin \theta \) (the acceleration down the incline), and \( t = 4 \, \text{s} \).
- Substituting the values, we have:
\[ L = \frac{1}{2} (g \sin \theta) (4^2) = \frac{1}{2} (g \sin \theta) (16) = 8 g \sin \theta \]

3. **Finding Time for One-Fourth Distance**:
- Now, we need to find the time taken to cover one-fourth of the distance, which is \( \frac{L}{4} \).
- Using the same equation of motion:
\[ \frac{L}{4} = \frac{1}{2} a t^2 \]
- Substituting \( a = g \sin \theta \):
\[ \frac{L}{4} = \frac{1}{2} (g \sin \theta) t^2 \]
- Since we know \( L = 8 g \sin \theta \), we can substitute this into the equation:
\[ \frac{8 g \sin \theta}{4} = \frac{1}{2} (g \sin \theta) t^2 \]
- Simplifying gives:
\[ 2 g \sin \theta = \frac{1}{2} (g \sin \theta) t^2 \]

4. **Cancelling out common terms**:
- Dividing both sides by \( g \sin \theta \) (assuming \( g \sin \theta \neq 0 \)):
\[ 2 = \frac{1}{2} t^2 \]
- Multiplying both sides by 2:
\[ 4 = t^2 \]
- Taking the square root:
\[ t = 2 \, \text{s} \]

5. **Conclusion**:
- The time taken to cover one-fourth of the distance from rest at the top is **2 seconds**.