Answer to Question #99504 in Mechanics | Relativity for Promise

Question

a tricycle moves from A to B 250km away at 60km/. It picks another set of passenger to point C 70km away from B at a deduced speed of 40km/h before returning to the park for the day at B in another 1 hour determine (a) the average speed (b) the average velocity

Answer 1

We have a tricyclist who did 250 km at 60 km/h and then made 70 km more at 40 km/h. After that he or she came back to B in an hour, i.e. the speed there was 70 km/h.

The average velocity is the total displacement divided by the time required to cover this displacement:

$$v_{avg}=\frac{\Delta x}{\Delta t}=\frac{\Delta x}{t_{AB}+t_{BC}+t_{CB}}=\\ \space\\=\frac{250}{250/60+70/40+1}=36\text{ km/h}.$$

The average speed is the total distance covered divided by the time required to cover this distance:

$$s_{avg}=\frac{\Sigma x}{\Delta t}=\frac{x_{AB}+x_{BC}+x_{CB}}{t_{AB}+t_{BC}+t_{CB}}=\\ \space\\=\frac{250+70+70}{250/60+70/40+1}=56\text{ km/h}.$$