Question

Question 1 1 pts A bicyclist goes on a 20 mile trip. For the first 10 miles of the ride, the bicyclist travels with an average speed of 8 mi/h. (al If the average speed of the bicyclist for the entire trip is 4 mi/h what is his average speed in mi/h) over the second 10 miles of the trip? (b) If instead the average speed of the bicyclist for the entire trio is 12 mi/h, what is his average speed (in mi/h) over the second 10 miles of the trip?

Answer 1

Total distance of the trip S_t = 20 miles Average speed in first 10 miles v_1 = 8 miles/h Time taken for the trip T = S_t /V where V = Average speed of entire trip = 4 T = 20 miles/4 = 5 h therefore Time taken for second 10 miles distance t_2 = T - t_1 = 5 - 1.25 = 3.75 h Average speed of second 10 miles distance V_2 = S_2 /t_2 = 10 miles /3.75 h = 2.666 miles/h Average speed of entire trip V' = 12 So, T' = s_t /V' = 20 miles/12 = 1.6666 h therefore t'_2 = T' - t_1 = 1.6666 h - 1.25 h = 0.41666 h therefore V'_2 = s_2 /t'_2 = 10 miles/0.41666 = 24 miles/h