A particle moves rectilinearly with a constant acceleration `1 m//s^2`. Its speed after `10` seconds is `5 m//s`. The distance covered by the particle in this duration is.

Options

A. 20 m
B. 25 m
C. 30 m
D. 50 m

Explanation

To solve the problem of finding the distance covered by a particle moving rectilinearly with a constant acceleration of \(1 \, \text{m/s}^2\) and a speed of \(5 \, \text{m/s}\) after \(10\) seconds, we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial speed (\(u\)) = ?
- Final speed (\(v\)) = \(5 \, \text{m/s}\)
- Acceleration (\(a\)) = \(1 \, \text{m/s}^2\)
- Time (\(t\)) = \(10 \, \text{s}\)

2. **Use the Equation of Motion:**
We can use the first equation of motion:
\[ v = u + at \]
Rearranging the equation to find the initial speed (\(u\)):
\[ u = v - at \]

3. **Substitute the Known Values:**
Substitute \(v = 5 \, \text{m/s}\), \(a = 1 \, \text{m/s}^2\), and \(t = 10 \, \text{s}\):
\[ u = 5 - (1 \times 10) = 5 - 10 = -5 \, \text{m/s} \]

4. **Calculate the Distance Covered:**
We can use the second equation of motion to find the distance (\(s\)):
\[ s = ut + \frac{1}{2} a t^2 \]
Substitute \(u = -5 \, \text{m/s}\), \(a = 1 \, \text{m/s}^2\), and \(t = 10 \, \text{s}\):
\[ s = (-5 \times 10) + \frac{1}{2} \times 1 \times (10)^2 \]
\[ s = -50 + \frac{1}{2} \times 1 \times 100 \]
\[ s = -50 + 50 = 0 \, \text{m} \]

5. **Conclusion:**
The distance covered by the particle in this duration is \(0 \, \text{m}\).