If a body starts from rest and travels 120 cm in the 6 second then what is the acceleration

Options

A. `0.20 m//s^(2)`
B. `0.0027 m//s^(2)`
C. `0.218 m//s^(2)`
D. `0.06 m//s^(2)`

Answer

To solve the problem step by step, we will use the second equation of motion.

### Step 1: Understand the Given Information
- The body starts from rest, which means the initial velocity \( u = 0 \).
- The distance traveled \( S = 120 \, \text{cm} = 1.2 \, \text{m} \) (since \( 120 \, \text{cm} = 120 \times 10^{-2} \, \text{m} \)).
- The time taken \( t = 6 \, \text{s} \).

### Step 2: Write the Second Equation of Motion
The second equation of motion is given by:
\[ S = ut + \frac{1}{2} a t^2 \]
Substituting the known values into the equation:
\[ 1.2 = 0 \cdot 6 + \frac{1}{2} a (6^2) \]
This simplifies to:
\[ 1.2 = \frac{1}{2} a (36) \]

### Step 3: Simplify the Equation
Now, we can simplify the equation:
\[ 1.2 = 18a \]

### Step 4: Solve for Acceleration \( a \)
To find \( a \), we will rearrange the equation:
\[ a = \frac{1.2}{18} \]
Calculating this gives:
\[ a = 0.0667 \, \text{m/s}^2 \]

### Step 5: Final Answer
Thus, the acceleration of the body is:
\[ a \approx 0.067 \, \text{m/s}^2 \]