A particle starts moving from rest under a constant acceleration. It travels a distance x in the first 10 sec and distance y in the next 10 sec, then

Options

A. y = x
B. y = 2x
C. y = 3x
D. y = 4x

Explanation

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

### Step 1: Understand the motion of the particle
The particle starts from rest, which means its initial velocity \( u = 0 \). It moves under constant acceleration \( a \).

### Step 2: Calculate the distance traveled in the first 10 seconds
Using the equation of motion:
\[ x = ut + \frac{1}{2} a t^2 \]
For the first 10 seconds (\( t = 10 \) seconds):
\[ x = 0 \cdot 10 + \frac{1}{2} a (10)^2 \]
\[ x = \frac{1}{2} a \cdot 100 = 50a \]
Thus, the distance traveled in the first 10 seconds is:
\[ x = 50a \]

### Step 3: Calculate the distance traveled in the next 10 seconds
Now, we need to find the distance traveled in the next 10 seconds (from 10 seconds to 20 seconds). The total time for this segment is 20 seconds. Using the same equation:
\[ \text{Total distance in 20 seconds} = 0 \cdot 20 + \frac{1}{2} a (20)^2 \]
\[ \text{Total distance in 20 seconds} = \frac{1}{2} a \cdot 400 = 200a \]
The total distance covered in the first 20 seconds is \( x + y \):
\[ x + y = 200a \]

### Step 4: Relate the distances \( x \) and \( y \)
From the previous steps, we have:
1. \( x = 50a \)
2. \( x + y = 200a \)

Substituting \( x \) into the second equation:
\[ 50a + y = 200a \]
Solving for \( y \):
\[ y = 200a - 50a = 150a \]

### Step 5: Find the relationship between \( y \) and \( x \)
Now, we can express \( y \) in terms of \( x \):
\[ y = 150a \]
Since \( x = 50a \), we can express \( a \) in terms of \( x \):
\[ a = \frac{x}{50} \]
Substituting this into the equation for \( y \):
\[ y = 150 \left(\frac{x}{50}\right) = 3x \]

### Final Result
Thus, the relationship between \( y \) and \( x \) is:
\[ y = 3x \]