Question

A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total time for the round trip is 19 hr. Find the​ bicyclist's average speed on each part of the trip.

Answer 1

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph