To find the total charge $ q $ that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $, we can integrate the current function $ I(t) = 0.02t + 0.01 \, \text{A} $ over this interval.

The formula for the charge $ q $ flowing through the wire is given by the integral of the current with respect to time:

$ q = \int I(t) \, dt $

We integrate the function from $ t = 1 $ to $ t = 2 $:

$ q = \int_1^2 (0.02t + 0.01) \, dt $

To solve this, first find the antiderivative:

$ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_1^2 $

Calculate the expression at the upper and lower bounds:

$ = \left[ 0.01t^2 + 0.01t \right]_1^2 $

Plug in the values:

For $ t = 2 $:

$ = 0.01(2)^2 + 0.01(2) = 0.04 + 0.02 = 0.06 $

For $ t = 1 $:

$ = 0.01(1)^2 + 0.01(1) = 0.01 + 0.01 = 0.02 $

Subtract the result at $ t=1 $ from the result at $ t=2 $:

$ q = 0.06 - 0.02 = 0.04 \, \text{C} $

Therefore, the charge that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $ is $ 0.04 \, \text{C} $.