A particle has an initial velocity of `9m//s` due east and a constant acceleration of `2m//s^(2)` due west. The distance coverd by the particle in the fifth second of its motion is :

Options

A. `0`
B. `0.5 m`
C. ` 2m `
D. None of these

Explanation

To solve the problem of finding the distance covered by the particle in the fifth second of its motion, we can break it down step by step.

### Step 1: Understand the given data
- Initial velocity (u) = 9 m/s (due east)
- Acceleration (a) = -2 m/s² (due west, hence negative)
- We need to find the distance covered in the fifth second.

### Step 2: Calculate the time at which the velocity becomes zero
Using the equation of motion:
\[ v = u + at \]
Setting \( v = 0 \) (when the particle stops):
\[ 0 = 9 - 2t \]
Rearranging gives:
\[ 2t = 9 \]
\[ t = \frac{9}{2} = 4.5 \text{ seconds} \]

### Step 3: Calculate the displacement during the first 4.5 seconds
Using the equation:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting \( t = 4.5 \) seconds:
\[ s = 9 \times 4.5 + \frac{1}{2} \times (-2) \times (4.5)^2 \]
Calculating each term:
1. \( 9 \times 4.5 = 40.5 \)
2. \( \frac{1}{2} \times (-2) \times 20.25 = -20.25 \)

Now, substituting these values:
\[ s = 40.5 - 20.25 = 20.25 \text{ meters} \]

### Step 4: Calculate the distance covered in the next 0.5 seconds (from 4.5 to 5 seconds)
At \( t = 4.5 \) seconds, the velocity is 0, and the particle will start moving in the opposite direction with an acceleration of \( 2 \text{ m/s}^2 \).
Using the equation again for \( t = 0.5 \) seconds:
\[ s = ut + \frac{1}{2}at^2 \]
Here, \( u = 0 \) (initial velocity at \( t = 4.5 \) seconds), \( a = 2 \text{ m/s}^2 \), and \( t = 0.5 \):
\[ s = 0 \times 0.5 + \frac{1}{2} \times 2 \times (0.5)^2 \]
Calculating:
\[ s = 0 + \frac{1}{2} \times 2 \times 0.25 = 0.25 \text{ meters} \]

### Step 5: Calculate the total distance covered in 5 seconds
Total distance covered in 5 seconds:
\[ \text{Total distance} = \text{Distance in first 4.5 seconds} + \text{Distance in next 0.5 seconds} \]
\[ = 20.25 + 0.25 = 20.5 \text{ meters} \]

### Step 6: Calculate the distance covered in the fifth second
To find the distance covered in the fifth second, we need to subtract the distance covered in the first 4 seconds from the total distance covered in 5 seconds.
First, calculate the distance covered in the first 4 seconds:
\[ s_4 = u \cdot t + \frac{1}{2} a t^2 \]
For \( t = 4 \):
\[ s_4 = 9 \times 4 + \frac{1}{2} \times (-2) \times (4)^2 \]
Calculating:
1. \( 9 \times 4 = 36 \)
2. \( \frac{1}{2} \times (-2) \times 16 = -16 \)

Thus,
\[ s_4 = 36 - 16 = 20 \text{ meters} \]

Now, the distance covered in the fifth second:
\[ \text{Distance in fifth second} = s_5 - s_4 = 20.5 - 20 = 0.5 \text{ meters} \]

### Final Answer
The distance covered by the particle in the fifth second of its motion is **0.5 meters**.

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