`S = 1/2 at^2 ` where a is a constant. If relative error in measurement of t is y, relative error in measurment of S will be

Options

A. `y^2`
B. `sqrt(y)`
C. `2y`
D. `y`

Answer

To find the relative error in the measurement of \( S \) given the equation \( S = \frac{1}{2} a t^2 \) and the relative error in the measurement of \( t \) as \( y \), we can follow these steps:

### Step 1: Understand the given equation
The equation provided is:
\[ S = \frac{1}{2} a t^2 \]
where \( a \) is a constant.

### Step 2: Define relative error
The relative error in the measurement of \( t \) is given as:
\[ \frac{\Delta t}{t} = y \]
where \( \Delta t \) is the absolute error in \( t \).

### Step 3: Differentiate \( S \) with respect to \( t \)
To find the error in \( S \), we can use the formula for the error propagation. The differential of \( S \) with respect to \( t \) is:
\[ \Delta S = \frac{dS}{dt} \Delta t \]
Calculating the derivative:
\[ \frac{dS}{dt} = \frac{d}{dt} \left( \frac{1}{2} a t^2 \right) = a t \]
Thus, we have:
\[ \Delta S = a t \Delta t \]

### Step 4: Find the relative error in \( S \)
Now, we can express the relative error in \( S \):
\[ \frac{\Delta S}{S} = \frac{a t \Delta t}{\frac{1}{2} a t^2} \]
Here, \( a \) cancels out:
\[ \frac{\Delta S}{S} = \frac{2 \Delta t}{t} \]

### Step 5: Substitute the relative error of \( t \)
We know that \( \frac{\Delta t}{t} = y \), so we can substitute this into our equation:
\[ \frac{\Delta S}{S} = 2 \cdot \frac{\Delta t}{t} = 2y \]

### Conclusion
Thus, the relative error in the measurement of \( S \) is:
\[ \frac{\Delta S}{S} = 2y \]

### Final Answer
The relative error in the measurement of \( S \) will be \( 2y \).

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