Answer to Question #103441 in Mechanics | Relativity for Rethabile

Question

You are going on a bicycle ride with a friend. You start 3.0 min ahead of her from her carport, pedal at 7.0 m/s for 10 min, then stop and chat with a neighbor for 4.0 min . As you chat, your friend pedals by, oblivious to you. You notice that she's forgotten to bring the lunch basket you plan to share, and you immediately start to pedal back to her house to get it, now moving at 12 m/s . When you are halfway back, your friend catches up with you, having herself remembered about the basket.

1.What was the magnitude of her average velocity for the trip up to that instant? Treat the trip as though it began at the same time for both of you, and you friend maintained zero velocity for three minutes.

2.What was the magnitude of her average velocity between the instant she passed you and the neighbor and the instant she caught up with you?

Answer 1

1)

$$d=7(10)(60)=4200\ m$$

$$t_{back}=\frac{0.5(4200)}{12}=175\ s$$

$$v=\frac{0.5(4200)}{10(60)+4(60)+175}=2.1\frac{m}{s}$$

2)

$$u=\frac{0.5(4200)}{175}=12\frac{m}{s}$$