Answer to Question #178867 in Physics for eyoba

Question

A runner jogs 12km north then turns & runs 16km east in 3 hour.

A) what is his displacement

B) what is his average speed

C) what is his average velovity including the direction

Answer 1

(A) We can find the displacement of the runner from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(16\ km)^2+(12\ km)^2}=20\ km.$$

We can find the direction of displacement from the geometry:

$$\theta=cos^{-1}(\dfrac{d_x}{d})=36.9^{\circ}\ N\ of\ E.$$

(B) Average speed of the runner can be found as follows:

$$v_{avg}=\dfrac{Total\ distance}{Total\ time}=\dfrac{d_1+d_2}{t},$$ $$v_{avg}=\dfrac{12\ km+16\ km}{3\ h}=9.33\ \dfrac{km}{h}.$$

(C) Average velocity of the runner can be found as follows:

$$\vec{v}_{avg}=\dfrac{Total\ displacement}{Total\ time},$$ $$\vec{v}_{avg}=\dfrac{20\ km\ [36.9^{\circ}\ N\ of\ E]}{3\ h}=6.67\ \dfrac{m}{s}\ [36.9^{\circ}\ N\ of\ E].$$