a car is driven east for a distance of 50km,then in a direction 30 degrees east of north for 25km.sketch the vector diagram and determine
(a) the magnitude and (b) the angle of the total displacement from its starting point.
a)
S ⃗ = A ⃗ + B ⃗ S = A 2 + B 2 − 2 A B cos ( 90 ∘ + 30 ∘ ) = 5 0 2 + 2 5 2 − 2 ∗ 25 ∗ 50 cos ( 90 ∘ + 30 ∘ ) = 66 k m \begin{array}{l}
\vec {S} = \vec {A} + \vec {B} \quad S = \sqrt {A ^ {2} + B ^ {2} - 2 A B \cos (9 0 {}^ {\circ} + 3 0 {}^ {\circ})} \\
= \sqrt {5 0 ^ {2} + 2 5 ^ {2} - 2 * 2 5 * 5 0 \cos (9 0 {}^ {\circ} + 3 0 {}^ {\circ})} = 6 6 \mathrm {k m} \\
\end{array} S = A + B S = A 2 + B 2 − 2 A B cos ( 90 ∘ + 30 ∘ ) = 5 0 2 + 2 5 2 − 2 ∗ 25 ∗ 50 cos ( 90 ∘ + 30 ∘ ) = 66 km
b)
B 2 = A 2 + S 2 − 2 A S cos θ B ^ {2} = A ^ {2} + S ^ {2} - 2 A S \cos \theta B 2 = A 2 + S 2 − 2 A S cos θ cos θ = A 2 + S 2 − B 2 2 A S = 5 0 2 + 6 6 2 − 2 5 2 2 ∗ 50 ∗ 66 = 0 , 944 → θ = 20 ∘ \cos \theta = \frac {A ^ {2} + S ^ {2} - B ^ {2}}{2 A S} = \frac {5 0 ^ {2} + 6 6 ^ {2} - 2 5 ^ {2}}{2 * 5 0 * 6 6} = 0, 9 4 4 \quad \rightarrow \theta = 2 0 {}^ {\circ} cos θ = 2 A S A 2 + S 2 − B 2 = 2 ∗ 50 ∗ 66 5 0 2 + 6 6 2 − 2 5 2 = 0 , 944 → θ = 20 ∘
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