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. ` 2s`
C. ` 4 s`
D. ` 16 s`

Explanation

To solve the problem, we need to determine the time it takes for a body sliding down a smooth inclined plane to cover one-fourth of the distance when starting from rest at the top. We know that the body takes 4 seconds to reach the bottom of the incline.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The body is sliding down the incline under the influence of gravity. The motion can be described using the equations of uniformly accelerated motion.

2. **Using the Second Equation of Motion**:
The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance covered,
- \( u \) is the initial velocity (which is 0 since it starts from rest),
- \( a \) is the acceleration (which is \( g \sin \theta \) for the incline),
- \( t \) is the time taken.

3. **Distance for the Entire Incline**:
For the entire distance \( l \) covered in 4 seconds:
\[ l = 0 \cdot 4 + \frac{1}{2} (g \sin \theta) (4^2) \]
Simplifying this gives:
\[ l = \frac{1}{2} (g \sin \theta) (16) = 8g \sin \theta \]

4. **Distance for One-Fourth of the Incline**:
Now, we want to find the time \( t \) taken to cover \( \frac{l}{4} \):
\[ \frac{l}{4} = \frac{1}{2} (g \sin \theta) t^2 \]
Substituting \( l = 8g \sin \theta \):
\[ \frac{8g \sin \theta}{4} = \frac{1}{2} (g \sin \theta) t^2 \]
This simplifies to:
\[ 2g \sin \theta = \frac{1}{2} (g \sin \theta) t^2 \]

5. **Cancelling Common Terms**:
We can cancel \( g \sin \theta \) from both sides (assuming \( g \sin \theta \neq 0 \)):
\[ 2 = \frac{1}{2} t^2 \]

6. **Solving for \( t^2 \)**:
Multiplying both sides by 2 gives:
\[ 4 = t^2 \]
Taking the square root:
\[ t = 2 \text{ seconds} \]

### Final Answer:
The time taken to cover one-fourth of the distance starting from rest at the top is **2 seconds**.

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