A particle moves on a straight line, with uniorm acceleration `-5m//s^(2)`. Distance travelled by it in three consecutive intervals of `1s` are in ratio of `3:1:1`. Then the total distance travelled in three seconds is:

Options

A. `25m`
B. `12.5m`
C. `22.5m`
D. `20m`

Explanation

To solve the problem step by step, we will analyze the motion of the particle under uniform acceleration and use the given ratio of distances traveled in consecutive 1-second intervals.

### Step 1: Understand the Problem
The particle moves with a uniform acceleration of `-5 m/s²` (which indicates it is decelerating). The distances traveled in three consecutive 1-second intervals are in the ratio `3:1:1`. We need to find the total distance traveled in 3 seconds.

### Step 2: Set Up the Distances
Let the distances traveled in the three intervals of 1 second be represented as:
- Distance in the first second: `3x`
- Distance in the second second: `x`
- Distance in the third second: `x`

### Step 3: Total Distance
The total distance traveled in 3 seconds can be expressed as:
\[ \text{Total Distance} = 3x + x + x = 5x \]

### Step 4: Use the Equation of Motion
We will use the equation of motion to find the distance traveled in the first second. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = distance traveled
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 5: Calculate Distance for First Second
For the first second (where \( t = 1 \) s):
- Initial velocity \( u = 0 \) (assuming it starts from rest)
- Acceleration \( a = -5 \, m/s² \)

Plugging in the values:
\[ s = 0 \cdot 1 + \frac{1}{2} \cdot (-5) \cdot (1)^2 \]
\[ s = 0 + \frac{1}{2} \cdot (-5) \cdot 1 = -\frac{5}{2} \, m \]

### Step 6: Determine \( x \)
Since the distance traveled in the first second is \( 3x \), we set:
\[ 3x = -\frac{5}{2} \]
To find \( x \):
\[ x = -\frac{5}{2} \cdot \frac{1}{3} = -\frac{5}{6} \, m \]

### Step 7: Calculate Total Distance
Now, substituting \( x \) back into the total distance formula:
\[ \text{Total Distance} = 5x = 5 \cdot \left(-\frac{5}{6}\right) = -\frac{25}{6} \, m \]

### Step 8: Find the Magnitude
Since distance cannot be negative, we take the absolute value:
\[ \text{Total Distance} = \frac{25}{6} \, m \approx 4.17 \, m \]

### Conclusion
The total distance traveled in 3 seconds is approximately \( 4.17 \, m \).