Answer to Question #22008 in Mechanics | Relativity for ololade

Question

on a 40km bike ride a cyclist rides the first 20km at 20km/h. what is the speed require for the final 20km if the average speed for the trip is to be 10km/h?

Answer 1

Let's denote the average speed for the trip by V (V = 10km/h), the speed for the first 20km by V1 (V1 = 20km/h) and the speed for the last 20km by V2 (V2 is unknown), the first and the second parts of a distance by S1 and S2 (s1 = S2 = 20[km]), and the time for the first and the second parts of a distance by T1 and T2 respectfully. Then

V·(T1 + T2) = V1·T1 + V2·T2 (1)

A cyclist spent

T1 = 20[km]/20[km/h] = 1[h]

riding the first 20km at 20km/h. Also, he spent

T1 = S2/V2

riding the second 20km. So, we can rewrite (1) as

V·(T1 + S2/V2) = V1·T1 + V2·S2/V2,

or

V·T1 + V·S2/V2 = V1·T1 + S2 ==>

V2 = V·S2 / ( V1·T1 + S2 - V·T1 ) =

= 10[km/h]·20[km] / (20[km/h]·1[h] + 20[km] - 10[km/h]·1[h] ) =

= 10[km/h]·20[km] / 30[km] = 20/3 [km/h].

Therefore, the speed required for the final 20km if the average speed for the trip to be 10km/h is 20/3 [km/h].