Answer to Question #88036 in Mechanics | Relativity for Gly

Question

A bicycle rider travels 250m north and 160 m northwest. What is the riders displacement? What distance does the rider cover?

Answer 1

The rider's displacement:

$$\vec{s}=\vec{s_1}+\vec{s_2}$$

$$\vec{ s_1}=(0, 250)m$$

$$\vec{ s_2}=(160 \cos{45}, 160 \sin{45})m$$

$$\vec{s}=(0, 250)+(160 \cos{45}, 160 \sin{45})=(113.1, 363.1)m$$

The magnitude is

$$s=\sqrt{113.1^2+363.1^2}=380 m$$

The direction is

$$a=\arctan{\frac{113.1}{363.1}}=17.3^\circ$$

west of north.

The distance is

$$d=s_1+s_2=250+160=410 m.$$