Copper resistivity is about 1.68 × 10⁻⁸ Ω·m. Standard 1 oz copper is about 35 µm thick.
For a track 10 mm long and 0.25 mm wide in 1 oz copper:
Cross-sectional area = 0.25 × 10⁻³ m × 35 × 10⁻⁶ m = 8.75 × 10⁻⁹ m²
R = ρL ÷ A = (1.68 × 10⁻⁸ × 0.010) ÷ (8.75 × 10⁻⁹) = 0.0192 Ω
At 1 A that is a voltage drop of 19.2 mV and a power dissipation of 19.2 mW in the track.
Two ways to halve it. Double the width to 0.5 mm, or specify 2 oz copper at 70 µm. Either gives 0.0096 Ω and 9.6 mV. Width is normally the cheaper option, since heavier copper raises board cost and limits how fine the rest of the features can be.
This is why power tracks are wide and signal tracks are not: 19 mV is irrelevant on a signal line and serious on a 3.3 V rail feeding a regulator's feedback point. For current-carrying capacity and temperature rise, use the tables in the manufacturing standard your course specifies rather than a rule of thumb.