The energy of a system as a function of time `t` is given as `E(t) = A^(2)exp(-alphat)`, `alpha = 0.2 s^(-1)`. The measurement of `A` has an error of `1.25%`. If the error In the measurement of time is `1.50%`, the percentage error in the value of `E(t)` at t = 5 s` is

Explanation

To find the percentage error in the energy \( E(t) = A^2 e^{-\alpha t} \) at \( t = 5 \) seconds, we will follow these steps:

### Step 1: Identify the given values
- The function for energy is \( E(t) = A^2 e^{-\alpha t} \).
- The error in the measurement of \( A \) is \( 1.25\% \).
- The error in the measurement of time \( t \) is \( 1.50\% \).
- The value of \( \alpha = 0.2 \, \text{s}^{-1} \).
- We need to evaluate at \( t = 5 \, \text{s} \).

### Step 2: Determine the percentage error in \( E(t) \)
To find the percentage error in \( E(t) \), we can use the formula for the propagation of errors. The percentage error in a product or quotient is the sum of the percentage errors of the factors involved.

The energy \( E(t) \) can be expressed as:
\[ E(t) = A^2 \cdot e^{-\alpha t} \]
The percentage error in \( E(t) \) can be calculated as:
\[ \text{Percentage Error in } E(t) = 2 \times \text{Percentage Error in } A + \text{Percentage Error in } e^{-\alpha t} \]

### Step 3: Calculate the percentage error in \( e^{-\alpha t} \)
The term \( e^{-\alpha t} \) is a function of time \( t \). The percentage error in \( e^{-\alpha t} \) due to the error in \( t \) can be expressed as:
\[ \text{Percentage Error in } e^{-\alpha t} = -\alpha \times \text{Percentage Error in } t \]
Substituting the values:
\[ \text{Percentage Error in } e^{-\alpha t} = -0.2 \times 5 \times 1.50\% \]
Calculating this gives:
\[ = -1.5\% \]

### Step 4: Combine the errors
Now we can combine the errors:
\[ \text{Percentage Error in } E(t) = 2 \times 1.25\% + (-1.5\%) \]
Calculating this gives:
\[ = 2.5\% - 1.5\% = 1.0\% \]

### Final Answer
The percentage error in the value of \( E(t) \) at \( t = 5 \, \text{s} \) is **1.0%**.

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