A body is moving in a straight line with uniform acceleration. It covers distances of 10 m and 12 m in third and fourth second respectively, then the initial velocity in m/sec is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Given:
- Distance covered in the 3rd second, \( S_3 = 10 \, \text{m} \)
- Distance covered in the 4th second, \( S_4 = 12 \, \text{m} \)

### Step 1: Use the formula for distance covered in the nth second
The distance covered in the nth second is given by the formula:
\[ S_n = u + \frac{a}{2} (2n - 1) \]
where \( S_n \) is the distance covered in the nth second, \( u \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the second.

### Step 2: Set up equations for the 3rd and 4th second
For the 3rd second:
\[ S_3 = u + \frac{a}{2} (2 \cdot 3 - 1) = u + \frac{a}{2} (6 - 1) = u + \frac{5a}{2} \]
Given \( S_3 = 10 \, \text{m} \):
\[ u + \frac{5a}{2} = 10 \quad \text{(Equation 1)} \]

For the 4th second:
\[ S_4 = u + \frac{a}{2} (2 \cdot 4 - 1) = u + \frac{a}{2} (8 - 1) = u + \frac{7a}{2} \]
Given \( S_4 = 12 \, \text{m} \):
\[ u + \frac{7a}{2} = 12 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \( u + \frac{5a}{2} = 10 \)
2. \( u + \frac{7a}{2} = 12 \)

Subtract Equation 1 from Equation 2:
\[ \left(u + \frac{7a}{2}\right) - \left(u + \frac{5a}{2}\right) = 12 - 10 \]
This simplifies to:
\[ \frac{7a}{2} - \frac{5a}{2} = 2 \]
\[ \frac{2a}{2} = 2 \quad \Rightarrow \quad a = 2 \, \text{m/s}^2 \]

### Step 4: Substitute the value of \( a \) back into one of the equations
Now substitute \( a = 2 \) into Equation 1:
\[ u + \frac{5(2)}{2} = 10 \]
\[ u + 5 = 10 \]
\[ u = 10 - 5 = 5 \, \text{m/s} \]

### Final Answer:
The initial velocity \( u \) is \( 5 \, \text{m/s} \).

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