Moving with a uniform acceleration from the state of rest a body covers 22 m during the 6th sec of its motion. What is the distance covered by if during the first 6 sec?

Options

A. 36 m
B. 48 m
C. 72 m
D. 150 m

Answer

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

### Step 1: Understand the given information
The body starts from rest (initial velocity \( u = 0 \)) and covers a distance of 22 m during the 6th second of its motion.

### Step 2: Use the formula for distance covered in the nth second
The distance covered in the nth second can be calculated using the formula:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
Since the initial velocity \( u = 0 \), the formula simplifies to:
\[ s_n = \frac{1}{2} a (2n - 1) \]
For the 6th second (\( n = 6 \)):
\[ s_6 = \frac{1}{2} a (2 \times 6 - 1) = \frac{1}{2} a (12 - 1) = \frac{11}{2} a \]
Given that \( s_6 = 22 \) m, we can set up the equation:
\[ \frac{11}{2} a = 22 \]

### Step 3: Solve for acceleration \( a \)
Multiply both sides by 2:
\[ 11a = 44 \]
Now, divide by 11:
\[ a = 4 \, \text{m/s}^2 \]

### Step 4: Calculate the distance covered in the first 6 seconds
The total distance covered in the first \( t \) seconds can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Again, since \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]
Substituting \( a = 4 \, \text{m/s}^2 \) and \( t = 6 \) seconds:
\[ s = \frac{1}{2} \times 4 \times (6)^2 \]
Calculating this:
\[ s = 2 \times 36 = 72 \, \text{m} \]

### Final Answer
The distance covered by the body during the first 6 seconds is **72 meters**.

---