Question

john is lost in a market. from his motorcycle, he walked 12 meters, 30° east of south while playing his phone. he lifted his face up and saw he is lost, then, he turned and walk 7 meters, 50° south of west. how far is john from his motorcycle and in what direction? use the parallelogram method to find the resultant vector of john's displacements​

Answer 1

The resultant vector of John's displacement is approximately 15.83 meters,

5.43° East of South.

Reasons:

Known parameters are;

Distances walked by John are;

12 meters, 30° East of South

7 meters, 50° South of West

The distances and directions can be represented using the parallelogram

and the resultant found as follows;

From the constructed parallelogram, the resultant is given by the diagonal  

of the parallelogram drawn from the start point.

By cosine rule, we have;

R² = 12² + 7² - 2× 12 × 7×cos(110°)

The resultant, R = √(12² + 7² - 2× 12 × 7×cos(110°)) ≈ 15.83

Therefore, R ≈ 15.83 meters

By sine rule, we have;

$\dfrac{12}{sin(\theta)} = \dfrac{15.83}{sin(110^{\circ})}$

Therefore;

$\theta = arcsine \left(\dfrac{sin(110^{\circ}) \times 12}{15.83} \right) \approx 45.43^{\circ}$

Therefore;

John's resultant direction is 180° - (45.43 + 50)° = 84.57° South of East,

which is 90° - 84.57° = 5.43° East of South.

The resultant vector of John's displacement, R ≈ 15.83 m, 5.43° East of

South.

Which gives; R = 15.83×sin(5.43)·i - 15.83×cos(5.43)·j ≈ 1.5·i - 15.76·j

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