Answer to Question #104507 in Mechanics | Relativity for havefun7741

Question #104507
[img]https://upload.cc/i1/2020/03/03/UFc9i5.jpg[/img]

where d8 is 4
1
Expert's answer
2020-03-09T10:52:56-0400


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


"v_{AD}=\\frac{d}{t}=\\frac{AB+BC+CD}{t_{AB}+t_{BC}+t_{CD}}=\\\\\n\\space\\\\\n=\\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}=\\\\\n=[40\\text{ cos}(-30)^\\circ+80\\text{ cos}0+60\\text{ cos}50^\\circ]\\textbf{i}+\\\\\n+[40\\text{ sin}(-30)^\\circ+80\\text{ sin}0+60\\text{ sin}50^\\circ]\\textbf{j}=\\\\\n=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}\n=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."

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