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

Options

A. zero
B. 0.5m
C. 2m
D. None

Answer

To solve the problem of finding the displacement covered by the particle in the fifth second of its motion, we will follow these steps:

### Step 1: Understand the Given Information
- Initial velocity (u) = 9 m/s (due east)
- Acceleration (a) = -2 m/s² (due west, hence negative in our eastward direction)
- We need to find the displacement during the fifth second of motion.

### Step 2: Identify the Time Interval
The fifth second refers to the time interval from t = 4 seconds to t = 5 seconds.

### Step 3: Calculate the Velocity at t = 4 seconds
Using the first equation of motion:
\[ V = u + at \]
Where:
- \( u = 9 \, \text{m/s} \)
- \( a = -2 \, \text{m/s}^2 \)
- \( t = 4 \, \text{s} \)

Substituting the values:
\[ V = 9 + (-2)(4) \]
\[ V = 9 - 8 \]
\[ V = 1 \, \text{m/s} \]

### Step 4: Calculate the Displacement in the Fifth Second
We can use the second equation of motion to find the displacement (S) during the fifth second:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 1 \, \text{m/s} \) (velocity at t = 4 seconds)
- \( a = -2 \, \text{m/s}^2 \)
- \( t = 1 \, \text{s} \) (duration of the fifth second)

Substituting the values:
\[ S = (1)(1) + \frac{1}{2}(-2)(1^2) \]
\[ S = 1 + \frac{1}{2}(-2)(1) \]
\[ S = 1 - 1 \]
\[ S = 0 \]

### Conclusion
The displacement covered by the particle in the fifth second of its motion is **0 meters**.

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