A particle covers 150m in 8th second starting from rest its acceleration is

Options

A. 1) `15ms^(-2)`
B. 2) `20ms^(-2)`
C. 3) `10ms^(-2)`
D. 4) `8ms^(-2)`

Explanation

To find the acceleration of a particle that covers 150 meters in the 8th second while starting from rest, we can use the formula for the distance covered in the nth second of motion:

\[ S_n = u + \frac{a}{2}(2n - 1) \]

Where:
- \( S_n \) is the distance covered in the nth second,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( n \) is the nth second.

### Step-by-step Solution:

1. **Identify the given values**:
- Distance covered in the 8th second, \( S_8 = 150 \, m \)
- Initial velocity, \( u = 0 \, m/s \) (since the particle starts from rest)
- \( n = 8 \)

2. **Substitute the known values into the formula**:
\[ S_8 = u + \frac{a}{2}(2n - 1) \]
Substituting the values:
\[ 150 = 0 + \frac{a}{2}(2 \cdot 8 - 1) \]

3. **Simplify the equation**:
\[ 150 = \frac{a}{2}(16 - 1) \]
\[ 150 = \frac{a}{2} \cdot 15 \]

4. **Multiply both sides by 2 to eliminate the fraction**:
\[ 300 = 15a \]

5. **Solve for acceleration \( a \)**:
\[ a = \frac{300}{15} \]
\[ a = 20 \, m/s^2 \]

### Final Answer:
The acceleration of the particle is \( 20 \, m/s^2 \).

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