Question

An impatient motorist considers speeding as he travels between two cities. If the trip normally takes 2.80

h at an average speed of 100.0 km/h, how much time, in minutes, will be saved if he exceeds the speed
limit by 10.0 km/h?

Answer 1

T1 = S / V1 T2 = S / V2 where S the distance is the same in both cases

V1 T1 = T2 V2 = (T1 + x) * (V1 + 10)

V1 T1 = V1 T1 + x V1 + 10 T1 + 10 x

x (V1 + 10) = -10 T1

x = 10 T1 / (V1 + 10) where x is the time that would be saved

x = 28 / (100 + 10) = .254 hrs or 15.3 min

Check: 2.8 * 100 = 280 km

2.55 hrs * 110 = 280 km