Question

An impatient motorist considers speeding as he travels between two cities. If the trip normally takes 2.8 h at an average speed of 100.0 km/h, how much time will be saved (in minutes) if he exceeds the speed limit by 10.0 km/h?

Answer 1

 If the trip takes 2.8 h at an average speed of 100.0 km/h, then the distance between these cities is:

$$d = 100\dfrac{km}{h}\cdot 2.8h = 280km$$

If the speed is $v = 100km/h + 10km/h = 110km/h$, then the time required to cover this distance is:

$$t = \dfrac{d}{v}\\ t = \dfrac{280km}{110km/h} \approx 2.5h$$

Answer. 2.5 h.