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.

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aristocles aristocles · Answer 1 · 2019-08-08T02:52:36+02:00

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

[tex]v_2 = v - 1[/tex]

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

so we will have

[tex]t_1 = \frac{90}{v}[/tex]

[tex]t_2 = \frac{90}{v - 1}[/tex]

so we will have

[tex]t_1 + t_2 = 19 hr[/tex]

[tex]\frac{90}{v} + \frac{90}{v-1} = 19[/tex]

[tex]90(2v - 1) = 19(v^2 - v)[/tex]

[tex]19 v^2 - 199 v + 90 = 0[/tex]

by solving above equation we have

[tex]v = 10 mph[/tex]

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