A jogger runs 140 m due west, then changes direction for the second leg of her run. At the end of the run, she is 374 m away from the starting point at an angle of 20o north of west. What were the length and direction of her second displacement?
Accepted Solution
A:
Answer:The length of her second displacement = 247.12 m.The direction of her second displacement = 31.24° from west.Step-by-step explanation:As per the question,From the figure as drawn below,Let the starting point be O. After running 140 m due west, she reached at point A.∴ OA = 140 mAnd At the end of the run, she is 374 m away from the starting point at an angle of 20° north of west.∴ OP = 374 mWe have to find the distance AP = x.By using the cosine rule in triangle OAP[tex]cos \theta = \frac{OA^{2}+OP^{2}-AP^{2}}{2\times OA\times OP}[/tex]After putting the given value, we get[tex]cos 20= \frac{140^{2}+374^{2}-x^{2}}{2\times 140\times 374}[/tex][tex]x^{2}=140^{2}+374^{2} - 2\times 140\times 374\times cos 20[/tex] ∴ x = 247.12 m Hence,the length of her second displacement = 247.12 m.Again,By using the cosine rule in triangle OAP, we get[tex]cos \alpha = \frac{OA^{2}+AP^{2}-OP^{2}}{2\times OA\times AP}[/tex]After putting the given value, we get[tex]cos \alpha = \frac{140^{2}+247.12^{2}-374^{2}}{2\times 140\times 247.12}[/tex]∴ α = 148.759°Hence, the direction of her second displacement = 180° - α = 180° - 148.759 = 31.24° from west.