A particle starting from rest travels a distance x in first 2 seconds and a distance y in next two seconds, then

Options

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

Answer

To solve the problem, we will analyze the motion of the particle in two intervals of time and use the equations of motion.

### Step 1: Analyze the first 2 seconds
The particle starts from rest, which means the initial velocity \( u = 0 \). We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the first 2 seconds (\( t = 2 \) seconds), the distance traveled is \( x \):
\[ x = 0 \cdot 2 + \frac{1}{2} a (2^2) \]
This simplifies to:
\[ x = \frac{1}{2} a \cdot 4 \]
\[ x = 2a \]

### Step 2: Analyze the next 2 seconds
Now, we need to find the distance traveled in the next 2 seconds (from \( t = 2 \) seconds to \( t = 4 \) seconds). The time interval is again 2 seconds. The final velocity at \( t = 2 \) seconds can be found using:
\[ v = u + at \]
Substituting \( u = 0 \), \( a = a \), and \( t = 2 \):
\[ v = 0 + a \cdot 2 = 2a \]
Now, using the same equation of motion for the next 2 seconds, we have:
\[ s = vt + \frac{1}{2} a t^2 \]
For the next 2 seconds (\( t = 2 \)):
\[ y = (2a)(2) + \frac{1}{2} a (2^2) \]
This simplifies to:
\[ y = 4a + \frac{1}{2} a \cdot 4 \]
\[ y = 4a + 2a = 6a \]

### Step 3: Relate \( x \) and \( y \)
From the equations we derived:
- \( x = 2a \)
- \( y = 6a \)

### Step 4: Express \( y \) in terms of \( x \)
To express \( y \) in terms of \( x \), we can substitute \( a \) from the first equation into the second:
\[ a = \frac{x}{2} \]
Substituting this into the equation for \( y \):
\[ y = 6a = 6 \left(\frac{x}{2}\right) = 3x \]

### Final Result
Thus, we have the relationship:
\[ y = 3x \]