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

Options

A. 0.02
B. 0.04
C. 0.03
D. 0.05

Explanation

To solve the problem, we need to find the percentage error in the energy \( E(t) \) given the function \( E(t) = A^2 e^{-\alpha t} \), where \( \alpha = 0.2 \, \text{s}^{-1} \). We are given that the measurement of \( A \) has an error of \( 1.25\% \) and the error in the measurement of time \( t \) is \( 1.50\% \). We will calculate the percentage error in \( E(t) \) at \( t = 5 \, \text{s} \).

### Step-by-Step Solution:

1. **Write the expression for \( E(t) \)**:
\[ E(t) = A^2 e^{-\alpha t} \]

2. **Take the natural logarithm of both sides**:
\[ \ln E = 2 \ln A - \alpha t \]

3. **Differentiate both sides**:
\[ \frac{dE}{E} = 2 \frac{dA}{A} - \alpha dt \]

4. **Identify the errors**:
- The error in \( A \) is given as \( \frac{dA}{A} = 1.25\% = 0.0125 \).
- The error in \( t \) is given as \( \frac{dt}{t} = 1.50\% = 0.015 \).

5. **Substitute the values into the differentiated equation**:
\[ \frac{dE}{E} = 2(0.0125) - 0.2 \cdot (0.015) \]

6. **Calculate the terms**:
- Calculate \( 2 \times 0.0125 = 0.025 \).
- Calculate \( 0.2 \times 0.015 = 0.003 \).

7. **Combine the results**:
\[ \frac{dE}{E} = 0.025 - 0.003 = 0.022 \]

8. **Convert to percentage**:
\[ \text{Percentage error in } E(t) = 0.022 \times 100\% = 2.2\% \]

### Final Result:
The percentage error in the value of \( E(t) \) at \( t = 5 \, \text{s} \) is **2.2%**.