A particle is moving in a straight line under constant acceleration of `4 m//s^(2)`. If its velocity at t=0 is 10 m//s, the displacement of particle in the `5^(th)` second of its motion is

Options

A. `28 m`
B. `100 m`
C. `34 m`
D. `50 m`

Answer

To solve the problem of finding the displacement of a particle in the 5th second of its motion under constant acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial velocity (u) = 10 m/s
- Constant acceleration (a) = 4 m/s²
- We need to find the displacement during the 5th second.

2. **Understand the Concept**:
- The displacement in the nth second can be calculated using the formula:
\[ S_n = u + \frac{1}{2} a (2n - 1) \]
- Here, \( S_n \) is the displacement during the nth second, \( u \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the second in which we want to find the displacement.

3. **Calculate Displacement in the 5th Second**:
- For the 5th second (n = 5):
\[ S_5 = u + \frac{1}{2} a (2 \cdot 5 - 1) \]
\[ S_5 = 10 + \frac{1}{2} \cdot 4 \cdot (10 - 1) \]
\[ S_5 = 10 + \frac{1}{2} \cdot 4 \cdot 9 \]
\[ S_5 = 10 + 2 \cdot 9 \]
\[ S_5 = 10 + 18 \]
\[ S_5 = 28 \text{ meters} \]

4. **Conclusion**:
- The displacement of the particle in the 5th second of its motion is **28 meters**.