A particle experiences a constant acceleration for 20 sec after starting from rest. If it travels distance `S_1` in the first 10 sec and a distance `S_2` in the next 10 sec, Then

Options

A. `S_1=S_2`
B. `S_1=(S_2)/(3)`
C. `S_1=(S^2)/(2)`
D. `S_1=(S_2)/(4)`

Explanation

To solve the problem, we will use the equations of motion under constant acceleration. Let's break it down step by step.

### Step 1: Calculate the distance \( S_1 \) in the first 10 seconds

The particle starts from rest, so the initial velocity \( u = 0 \). The formula for distance under constant acceleration is given by:

\[ S = ut + \frac{1}{2} a t^2 \]

Substituting \( u = 0 \) and \( t = 10 \) seconds, we get:

\[ S_1 = 0 \cdot 10 + \frac{1}{2} a (10)^2 = \frac{1}{2} a (100) = 50a \]

### Step 2: Calculate the final velocity \( v \) after the first 10 seconds

Using the formula for final velocity:

\[ v = u + at \]

Again substituting \( u = 0 \) and \( t = 10 \):

\[ v = 0 + a \cdot 10 = 10a \]

### Step 3: Calculate the distance \( S_2 \) in the next 10 seconds

Now, the particle continues to move with an initial velocity \( v = 10a \) and accelerates for another 10 seconds. We will use the same distance formula:

\[ S_2 = vt + \frac{1}{2} a t^2 \]

Substituting \( v = 10a \) and \( t = 10 \):

\[ S_2 = (10a)(10) + \frac{1}{2} a (10)^2 \]

Calculating this gives:

\[ S_2 = 100a + \frac{1}{2} a (100) = 100a + 50a = 150a \]

### Step 4: Relate \( S_1 \) and \( S_2 \)

Now we have:

\[ S_1 = 50a \]
\[ S_2 = 150a \]

To find the relationship between \( S_1 \) and \( S_2 \):

\[ \frac{S_2}{S_1} = \frac{150a}{50a} = 3 \]

Thus, we can express \( S_1 \) in terms of \( S_2 \):

\[ S_1 = \frac{S_2}{3} \]

### Final Result

The relationship between the distances is:

\[ S_1 = \frac{S_2}{3} \]