Question 1.5: Displacement of a cross-country skier In this problem we wil...
Displacement of a cross-country skier In this problem we will use graphical vector addition to determine your final location on a cross-country ski trip. Suppose that on a cross-country ski trip, you travel 1.00 km north and then 2.00 km east. (a) How far and in what direction are you from your starting point? (b) What are the magnitude and direction of your resultant displacement?
SET UP Figure 1.13 is a scale diagram. By careful measurement on the diagram, we find that the distance d from the starting point is about 2.2 km and the angle Φ is about 63°. But it’s much more accurate to calculate the result. The vectors in the diagram form a right triangle, and we can find the length of the hypotenuse by using the Pythagorean theorem. The angle can be found by simple trigonometry, from the definition of the tangent function.
SOLVE Part (a): Use the Pythagorean theorem to find the length d of the resultant displacement vector (the distance from the starting point):
d=\sqrt{(1.00 km)^2+(2.00 km)^2}
= 2.24 km (magnitude of displacement vector).
From the definition of the tangent function,
\tan \phi =\frac{opposite side}{adjacent side} = \frac{2.00 km}{1.00 km},
Φ = 63.4° (direction of displacement vector).
Part (b): The magnitude of the resultant displacement is just the distance d we found in part (a), 2.24 km. We can describe the direction as 63.4° east of north or 26.6° north of east. Take your choice!
REFLECT The method we used here works only for vectors that form a right triangle. In the next section, we’ll present a more general method.
Practice Problem: Your friend skis 3.00 km east and then 1.50 km north. (a) How far and in what direction is he from the starting point? (b) What are the magnitude and direction of his resultant displacement? Answers: (a) 3.35 km, 26.6°; (b) 3.35 km, 26.6° north of east.
