A particle moves with constant acceleration along a straight line streaing from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second is

Options

A. `33%`
B. `40%`
C. `66%`
D. `77%`

Explanation

To solve the problem of finding the percentage increase in displacement during the 4th second compared to that in the 3rd second for a particle moving with constant acceleration from rest, we can follow these steps:

### Step 1: Understand the motion parameters
The particle starts from rest, which means the initial velocity \( u = 0 \). The displacement during any nth second can be calculated using the formula:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
where \( s_n \) is the displacement during the nth second, \( a \) is the constant acceleration, and \( n \) is the second in question.

### Step 2: Calculate displacement during the 3rd second
For the 3rd second (\( n = 3 \)):
\[ s_3 = 0 + \frac{1}{2} a (2 \cdot 3 - 1) = \frac{1}{2} a (6 - 1) = \frac{1}{2} a \cdot 5 = \frac{5a}{2} \]

### Step 3: Calculate displacement during the 4th second
For the 4th second (\( n = 4 \)):
\[ s_4 = 0 + \frac{1}{2} a (2 \cdot 4 - 1) = \frac{1}{2} a (8 - 1) = \frac{1}{2} a \cdot 7 = \frac{7a}{2} \]

### Step 4: Find the increase in displacement
The increase in displacement from the 3rd to the 4th second is:
\[ \Delta s = s_4 - s_3 = \frac{7a}{2} - \frac{5a}{2} = \frac{2a}{2} = a \]

### Step 5: Calculate the percentage increase
The percentage increase in displacement from the 3rd second to the 4th second is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta s}{s_3} \right) \times 100 = \left( \frac{a}{\frac{5a}{2}} \right) \times 100 \]
This simplifies to:
\[ = \left( \frac{a \cdot 2}{5a} \right) \times 100 = \left( \frac{2}{5} \right) \times 100 = 40\% \]

### Final Answer
Thus, the percentage increase in displacement during the 4th second compared to that in the 3rd second is **40%**.

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