Prove the relation, `s_t=u + at - 1/2 a.`

Explanation

To prove the relation \( s_t = u + at - \frac{1}{2}a \), we will follow these steps:

### Step 1: Understand the Displacement Formula
The displacement \( s \) at any time \( t \) for an object moving with initial velocity \( u \) and constant acceleration \( a \) is given by the formula:
\[ s = ut + \frac{1}{2} a t^2 \]

### Step 2: Calculate Displacement at Time \( t \)
Using the formula, the displacement at time \( t \) is:
\[ s_t = ut + \frac{1}{2} a t^2 \]

### Step 3: Calculate Displacement at Time \( t-1 \)
Now, we need to calculate the displacement at time \( t-1 \):
\[ s_{t-1} = u(t-1) + \frac{1}{2} a (t-1)^2 \]
Expanding this, we get:
\[ s_{t-1} = ut - u + \frac{1}{2} a (t^2 - 2t + 1) = ut - u + \frac{1}{2} a t^2 - a t + \frac{1}{2} a \]

### Step 4: Simplify the Displacement at Time \( t-1 \)
Combining the terms, we have:
\[ s_{t-1} = ut - u + \frac{1}{2} a t^2 - at + \frac{1}{2} a \]

### Step 5: Find the Displacement in the \( t^{th} \) Second
The displacement during the \( t^{th} \) second, \( s_t \), is given by the difference between the displacement at time \( t \) and the displacement at time \( t-1 \):
\[ s_t = s_t - s_{t-1} \]
Substituting the expressions we derived:
\[ s_t = \left( ut + \frac{1}{2} a t^2 \right) - \left( ut - u + \frac{1}{2} a t^2 - at + \frac{1}{2} a \right) \]

### Step 6: Simplify the Expression
Now, simplifying this:
\[ s_t = ut + \frac{1}{2} a t^2 - ut + u - \frac{1}{2} a t^2 + at - \frac{1}{2} a \]
The \( ut \) and \( \frac{1}{2} a t^2 \) terms cancel out:
\[ s_t = u + at - \frac{1}{2} a \]

### Conclusion
Thus, we have proved that:
\[ s_t = u + at - \frac{1}{2} a \]

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