Answer to Question #104507 in Mechanics | Relativity for havefun7741

Question

[img]https://upload.cc/i1/2020/03/03/UFc9i5.jpg[/img]

where d8 is 4

Answer 1

(i) The average speedis total distance over total time:

$$v_{AD}=\frac{d}{t}=\frac{AB+BC+CD}{t_{AB}+t_{BC}+t_{CD}}=\\ \space\\ =\frac{40+80+60}{5+4+3}=15\text{ m/s}.$$

(ii) We have three displacement vectors: 40 m @ -30°, 80 m @ 0° and 60 @ 50°. Their vector sum will be

$$\vec{r}=r_x\textbf{i}+r_y\textbf{j}=\\ =[40\text{ cos}(-30)^\circ+80\text{ cos}0+60\text{ cos}50^\circ]\textbf{i}+\\ +[40\text{ sin}(-30)^\circ+80\text{ sin}0+60\text{ sin}50^\circ]\textbf{j}=\\ =153\textbf{i}+26\textbf{j}.$$

(iii) The average velocity in unit vector form from A to D:

$$\vec{v}=v_x\textbf{i}+v_y\textbf{j}=\frac{r_x}{t_{AD}}\textbf{i}+\frac{r_y}{t_{AD}}\textbf{j} =12.8\textbf{i}+2.16\textbf{j}.$$

(iv) Since we have our vector $\vec{v}$, the magnitude is

$$v=\sqrt{v_x^2+v_y^2}=12.9\text{ m/s}.$$

The angle will be

$$\theta=\text{atan}\frac{v_y}{v_x}=9.58^\circ.$$