Question

Beatriz went on a road trip. By the end of the first day, she was 200\text{ km}200 km200, end text from the starting point of her trip, in a direction that is a °300, degree rotation from east. By the end of the second day, she was 150\text{ km}150 km150, start text, space, k, m, end text from where she was at the beginning of that day, in a direction that is a 250\degree250°250, degree rotation from east. What is Beatriz's direction, relative to the starting point of her trip, by the end of the second day?

Answer 1

The direction of Beatriz relative to the starting point of her trip is approximately $278.806^{\circ}$.

How to find the position of Beatriz relative to the starting point of her trip

After a careful reading of the statement, we find that final position ($\vec r$) by the end of the second day is found by means of this vector sum:

$\vec r = \vec r_{1} + \vec r_{2}$ (1)

Where:

• $\vec r_{1}$ - Vector distance of the first day relative to starting point, in kilometers.
• $\vec r_{2}$ - Vector distance of the second day relative to the final point of $\vec r_{1}$, in kilometers.

If we know that $\vec r_{1} = (200\,km \cdot \cos 300^{\circ}, 200\,km\cdot \sin 300^{\circ})$ and $\vec r_{2} = (150\,km\cdot \cos 250^{\circ}, 150\,km\cdot \sin 250^{\circ})$, then final position of Beatriz relative to origin is:

$\vec r = (200\,km\cdot \cos 300^{\circ}, 200\,km\cdot \sin 300^{\circ})+(150\,km \cdot \cos 250^{\circ}, 150\,km\cdot \sin 250^{\circ})$

$\vec r = (48.670, -314.159)\,[km]$

And the direction relative to the starting point ($\theta$), in degrees, is found by following inverse trigonometric relation:

$\theta = \tan^{-1} \frac{r_{y}}{r_{x}}$ (2)

If we know that $r_{x} = 48.670\,km$ and $r_{y} = -314.159\,km$, then the direction of Beatriz relative to the starting point of her trip is:

$\theta = \tan^{-1} \left(\frac{-314.159\,km}{48.670\,km} \right)$

$\theta \approx 278.806^{\circ}$

The direction of Beatriz relative to the starting point of her trip is approximately $278.806^{\circ}$. $\blacksquare$

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