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

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

Step-by-step 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 = (90)/(v)`

`t_2 = (90)/(v - 1)`

so we will have

`t_1 + t_2 = 19 hr`

`(90)/(v) + (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