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.

Explanation

To solve the problem step by step, we will use the formulas for distance traveled in a given time with uniform acceleration.

### Step 1: Understand the problem
We have a particle moving in a straight line with an initial velocity \( u \) and uniform acceleration \( f \). We need to find the velocity of the particle after \( t \) seconds, given that the sum of the distances traveled in the \( t^{th} \) and \( (t + 1)^{th} \) seconds is \( 100 \, \text{cm} \).

### Step 2: Use the formula for distance traveled in the \( t^{th} \) second
The distance traveled in the \( t^{th} \) second, \( S_t \), is given by the formula:
\[ S_t = u + \frac{f}{2} (2t - 1) \]

### Step 3: Use the formula for distance traveled in the \( (t + 1)^{th} \) second
The distance traveled in the \( (t + 1)^{th} \) second, \( S_{t+1} \), is given by:
\[ S_{t+1} = u + \frac{f}{2} (2(t + 1) - 1) = u + \frac{f}{2} (2t + 2 - 1) = u + \frac{f}{2} (2t + 1) \]

### Step 4: Set up the equation
According to the problem, the sum of the distances traveled in the \( t^{th} \) and \( (t + 1)^{th} \) seconds is \( 100 \, \text{cm} \):
\[ S_t + S_{t+1} = 100 \]
Substituting the expressions we derived:
\[ \left( u + \frac{f}{2} (2t - 1) \right) + \left( u + \frac{f}{2} (2t + 1) \right) = 100 \]

### Step 5: Simplify the equation
Combine like terms:
\[ 2u + \frac{f}{2} (2t - 1 + 2t + 1) = 100 \]
This simplifies to:
\[ 2u + \frac{f}{2} (4t) = 100 \]
\[ 2u + 2tf = 100 \]

### Step 6: Solve for \( u + tf \)
Divide the entire equation by 2:
\[ u + tf = 50 \]

### Step 7: Find the velocity after \( t \) seconds
The velocity \( v \) after \( t \) seconds can be calculated using the formula:
\[ v = u + ft \]
From the previous step, we know:
\[ u + tf = 50 \]
Thus:
\[ v = 50 \, \text{cm/s} \]

### Final Answer
The velocity of the particle after \( t \) seconds is \( 50 \, \text{cm/s} \).

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