University Physics with Modern Physics (14th Edition)

Let's find the distance $d$ of the trip. Note that when the speed is 105 km/h, the time of the trip is $\frac{11}{6}~h$. $d = vt_1 = (105 ~km/h)(\frac{11}{6}~h) = 192.5 ~km$ We can use the distance to find the time of the trip when the average speed is 70 km/h. $t_2 = \frac{d}{v} = \frac{192.5 ~km}{70 ~km/h} = 2.75 ~h$ Note that 2.75 hours is equal to 2 hours and 45 minutes. We can calculate the time difference of the two trips. (2 hours and 45 minutes) - (1 hour and 50 minutes) is equal to 55 minutes. At an average speed of 70 km/h, the trip takes 55 minutes longer.