Question

Section 1.8 → The route f by a hiker consists of three di vectors A, B, and vector A is along a tral and is 1550 m in a Vector B is trail, but the hiker uses a compass and knows that the direction is 41.0 east of south Similary, the direction of wectors 250 north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A+B+C-0. Find the magnitudes of vector B and vector C Section 1.8

Answer 1

Given $\vec{A}+\vec{B}+\vec{C}=0$

Consider horizontal Components of all three vectors

$A_{x}+B_{x}-C_{x}=0$

$A*cos25+B*sin41-C*cos25=0$

$1550cos25+Bsin41-Ccos25=0$

$1404.78+B*0.656-C*0.906=0$

$1404.78+B*0.656=C*0.906$

$(1404.78+B*0.656)/0.906=C$ ---------------(1)

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Consider all Vertical Components of all three vectors

$A_{y}+C_{y}-B_{y}=0$

$A*Sin25+C*Sin25-B*Cos41=0$

$1550*Sin25+C*Sin25-B*Cos41=0$

$655.1+C*0.423-B*0.755=0$

From (1)

$655.1+(1404.78+B*0.656)*0.423/0.906-B*0.755=0$

$655.1+1404.78*0.423/0.906+B*0.656*0.423/0.906-B*0.755=0$

$655.1+655.87+B*0.306-B*0.755=0$

$1310.97-B*0.449=0$

$1310.97/0.449=B$

ANSWER: Magnitude of $\vec{B}=2919.755m$

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Put value of B in (1)

$(1404.78+2919.755*0.656)/0.906=C$

$3320.139/0.906=C$

ANSWER: Magnitude of $\vec{C}=3664.613m$

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