Question

A salesperson leaves the office and drives 26 km due north along a straight highway. A turn is made onto a highway that leads in the direction 30.0° north of east. The driver continues on the highway for a distance of 62 km and then stops. Using graphical methods, what is the total displacement of the salesperson from the office?
Why the answer is 1. Approximately 79 km?

Answer 1

Answer on Question #85254 - Physics - Mechanics | Relativity

A salesperson leaves the office and drives $26\mathrm{km}$ due north along a straight highway. A turn is made onto a highway that leads in the direction $30.0{}^{\circ}$ north of east. The driver continues on the highway for a distance of $62\mathrm{km}$ and then stops. Using graphical methods, what is the total displacement of the salesperson from the office?

Why the answer is 1. Approximately $79\mathrm{km}$ ?

Solution:

Figure out the driven distances (not to scale):

On this figure:

$AB = 26km$

$BC = 62km$

$AC$ - the unknown total displacement of the salesperson from the office.

From the rectangular triangle ADC we get

$AC = \sqrt{(AB + BD)^2 + CD^2}.$

From the rectangular triangle BDC we get

$BD = BC\cdot cos60{}^{\circ},$

$$CD = BC \cdot \sin 60{}^\circ.$$

So

$$AC = \sqrt{(AB + BC \cdot \cos 60{}^\circ)^2 + (BC \cdot \sin 60{}^\circ)^2} = \sqrt{(26 + 62 \cdot 0.5)^2 + (62 \cdot 0.87)^2} = 78.5 \text{ km}$$

Answer: the total displacement of the salesperson from the office is $78.5 \text{ km}$.

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