The energy of a system as a function of time t is given as $E(t)=A^{2} \exp(-\alpha t),$ where $\alpha=0.2 s^{-1}.$ 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

Step 1: State the given values.

The energy equation and the given parameters are:

$$E(t) = A^{2}e^{-\alpha t}$$ $$\alpha = 0.2\,\text{s}^{-1}$$ $$t = 5\,\text{s}$$

The percentage errors in the measurements of $A$ and $t$ are:

$$\frac{\Delta A}{A} \times 100 = 1.25\%$$ $$\frac{\Delta t}{t} \times 100 = 1.50\%$$

Step 2: Relate the errors using logarithmic differentiation.

Take the natural logarithm of both sides of the energy equation.

$$\ln E = 2\ln A - \alpha t$$

Differentiating both sides gives the relation between the fractional changes:

$$\frac{\Delta E}{E} = 2\frac{\Delta A}{A} - \alpha\,\Delta t$$

Step 3: Express the maximum fractional error.

To find the maximum possible error (worst-case scenario), the magnitudes of the errors are added.

$$\left|\frac{\Delta E}{E}\right| = 2\left|\frac{\Delta A}{A}\right| + \alpha\,\Delta t$$

Step 4: Calculate the individual error contributions.

The contribution from the measurement of $A$ is:

$$2\left(\frac{\Delta A}{A}\right) \times 100 = 2(1.25\%) = 2.5\%$$

The contribution from the measurement of $t$ is:

$$\alpha\,\Delta t \times 100 = \alpha t \left(\frac{\Delta t}{t} \times 100\right)$$

Substitute the values $\alpha = 0.2\,\text{s}^{-1}$ and $t = 5\,\text{s}$:

$$\alpha\,\Delta t \times 100 = (0.2)(5)(1.50\%) = 1.5\%$$

Step 5: Calculate the total percentage error in $E(t)$.

Add the two contributions to find the total percentage error in $E(t)$ at $t = 5\,\text{s}$.

$$\frac{\Delta E}{E} \times 100 = 2.5\% + 1.5\% = 4\%$$

Answer: The percentage error in the value of $E(t)$ at $t=5\,\text{s}$ is $4\%$.