Answer to Question #119187 in Mechanics | Relativity for gly

Question #119187

a boomerang is thrown with the following displacements: 4.5-m to the east, 7.5-m 30° north of west, and 3.0-m 10° south of west. what is the resultant displacement?


1
Expert's answer
2020-06-01T14:26:55-0400

Solution.

"AB=4.5m;"

"BC=7.5m;"

"CD=3m;"

"\\angle B=30^o;"

"\\angle DCB=10^o;"



"AC^2=AB^2+BC^2-2ABBCcos\\angle B;"

"AC^2=4.5^2+7.5^2-2\\sdot4.5\\sdot7.5\\sdot 0.866=18.045;"

"AC=4.25m;"

"\\dfrac{AC}{sin\\angle B}=\\dfrac{AB}{sin \\angle C}\\implies sin\\angle C=\\dfrac{ABsin\\angle B}{AC}" ;


"sin\\angle C=\\dfrac{4.5\\sdot 0.5}{4.25}=0.529;"

"\\angle C=32^o;"

"\\angle ACD=22^o;"

"AD^2=AC^2+CD^2-2ACCDcos\\angle ACD;"

"AD^2=4.25^2+3^2-2\\sdot4.25\\sdot3\\sdot0.9272=3.4189;"

"AD=1.85m;"

"\\dfrac{AD}{sin\\angle ACD}=\\dfrac{CD}{sin \\angle CAD}\\implies sin\\angle CAD=\\dfrac{CDsin\\angle ACD}{AD};"


"sin\\angle CAD=\\dfrac{3\\sdot 0.3746}{1.85}=0.6074;"

"\\angle CAD=37^o;"

"\\angle BAD=118^o-37^o=81^o;"

Answer: "AD=1.85m, \\angle BAD=81^o;" ,


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