A particle is moving in a straight line with initial velocity `u` and uniform acceleration `f`. If the sum of the distances travelled in `t^(th) and (t + 1)^(th)` seconds is `100 cm`, then its velocity after `t` seconds, in `cm//s`, is.

Options

A. 20
B. 30
C. 50
D. 80

Explanation

To solve the problem, we need to find the velocity of a particle after `t` seconds, given that the sum of the distances traveled in the `t-th` and `(t + 1)-th` seconds is `100 cm`.

### Step-by-Step Solution:

1. **Understanding the Distance Formula**:
The distance traveled by a particle in the `t-th` second is given by the formula:
\[ s_t = u + \frac{a}{2}(2t - 1) \]
where \( u \) is the initial velocity, \( a \) is the uniform acceleration, and \( t \) is the time in seconds.

2. **Distance in the (t + 1)-th Second**:
For the `(t + 1)-th` second, the distance traveled is:
\[ s_{t+1} = u + \frac{a}{2}(2(t + 1) - 1) = u + \frac{a}{2}(2t + 2 - 1) = u + \frac{a}{2}(2t + 1) \]

3. **Setting Up the Equation**:
According to the problem, the sum of the distances traveled in the `t-th` and `(t + 1)-th` seconds is `100 cm`:
\[ s_t + s_{t+1} = 100 \]
Substituting the expressions for \( s_t \) and \( s_{t+1} \):
\[ \left( u + \frac{a}{2}(2t - 1) \right) + \left( u + \frac{a}{2}(2t + 1) \right) = 100 \]

4. **Simplifying the Equation**:
Combine the terms:
\[ 2u + \frac{a}{2}(2t - 1 + 2t + 1) = 100 \]
Simplifying further:
\[ 2u + \frac{a}{2}(4t) = 100 \]
\[ 2u + 2at = 100 \]

5. **Dividing by 2**:
Divide the entire equation by 2:
\[ u + at = 50 \]

6. **Finding the Final Velocity**:
The final velocity \( v \) after `t` seconds can be expressed using the equation of motion:
\[ v = u + at \]
From our previous step, we have:
\[ v = 50 \]

### Conclusion:
The velocity of the particle after `t` seconds is:
\[ \boxed{50 \text{ cm/s}} \]