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Chapter 3: Problem 70

A woman walks \(250 \mathrm{~m}\) in the direction \(30^{\circ}\) east of north, then \(175 \mathrm{~m}\) directly east. Find (a) the magnitude and (b) the angle of her final displacement from the starting point. (c) Find the distance she walks. (d) Which is greater, that distance or the magnitude of her displacement?

Short Answer

Expert verified
(a) 369.8 m; (b) 54.6掳 east of north; (c) 425 m; (d) The distance walked is greater.

Step by step solution

01

Break down the initial displacement into components

The woman first walks \(250 \mathrm{~m}\) at \(30^{\circ}\) east of north. To break this into components, we identify the north and east directions. The north component is \(250 \cos(30^{\circ})\), and the east component is \(250 \sin(30^{\circ})\).
02

Calculate the components for both displacements

Calculate the components of the first displacement:- North component: \(250 \cos(30^{\circ}) = 250 \times \frac{\sqrt{3}}{2} \approx 216.5 \mathrm{~m}\).- East component: \(250 \sin(30^{\circ}) = 250 \times \frac{1}{2} = 125 \mathrm{~m}\).The second displacement is directly east, adding 175 m to the east component.
03

Determine the total displacement components

Add the components from both displacements to determine net displacement:- Total North component: \(216.5 \mathrm{~m}\).- Total East component: \(125 + 175 = 300 \mathrm{~m}\).
04

Compute the magnitude of the final displacement

Use the Pythagorean theorem to find the total displacement magnitude: \[ d = \sqrt{(216.5)^2 + (300)^2} \approx \sqrt{46800.25 + 90000} \approx \sqrt{136800.25} \approx 369.8 \mathrm{~m} \]
05

Determine the angle of the final displacement

Find the angle \( \theta \) with respect to north using the tangent function: \[ \tan(\theta) = \frac{\text{East Component}}{\text{North Component}} = \frac{300}{216.5} \]\[ \theta = \tan^{-1}\left(\frac{300}{216.5}\right) \approx 54.6^{\circ} \text{ east of north} \]
06

Calculate the total distance walked

Add the distances walked in both segments: \(250 \mathrm{~m} + 175 \mathrm{~m} = 425 \mathrm{~m}\).
07

Compare distances

The distance walked (425 m) is greater than the magnitude of the displacement (369.8 m).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
When solving problems involving vectors, like the path walked by the woman in the exercise, separating a vector into its components is a foundational step.
This concept involves breaking down a vector into perpendicular directions, often using the north and east directions for simplicity.
For example, if a vector is described in terms of an angle from a cardinal direction, the vector can be split as follows:
  • The north (or y-axis) component as the vector's magnitude times the cosine of the angle.
  • The east (or x-axis) component as the vector's magnitude times the sine of the angle.
This breakdown allows for the easier computation of resultant vectors, as vectors in the horizontal and vertical directions can be added or subtracted independently.
Displacement Magnitude
Displacement is a vector that shows the change in position of an object.
The magnitude of this vector indicates the shortest distance from the starting point to the final position, which is not necessarily the same as the total path traveled, as this exercise shows.
To determine the magnitude of a resultant vector, the Pythagorean theorem is usually employed. Given components along the x and y axes, the formula is:\[ d = \sqrt{x^2 + y^2} \]where \(d\) is the resultant displacement magnitude.
This gives an easy way to calculate the overall distance between two points based solely on their component contributions.
Trigonometric Functions
Trigonometry plays a crucial role in resolving vector components and finding angles in physics problem-solving.
Understanding sine, cosine, and tangent functions is essential. These functions relate angles to the ratios of a right triangle's sides:
  • Cosine relates the adjacent side to the hypotenuse.
  • Sine relates the opposite side to the hypotenuse.
  • Tangent compares the opposite side to the adjacent side.
For instance, in this problem, the cosine of the angle was used to find the north component and the sine to find the east component.
Also, the tangent function helped determine the angle of the resultant displacement.
Vector Addition
Vector addition is fundamental when determining the resultant displacement when multiple movements are involved.
By adding vector components separately鈥攏orth with north, east with east鈥攐ne finds the overall displacement vector.
This is achieved by:
  • Summing the north components of each vector to find the total movement north.
  • Summing the east components of each vector to find the total movement east.
The final resultant vector is then represented as a diagonal component when these total components are combined.
This comprehensive approach gives not only the magnitude but also the direction of final displacement.

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Most popular questions from this chapter

Two beetles run across flat sand, starting at the same point. Beetle 1 runs \(0.50 \mathrm{~m}\) due east, then \(0.80 \mathrm{~m}\) at \(30^{\circ}\) north of due east. Beetle 2 also makes two runs; the first is \(1.6 \mathrm{~m}\) at \(40^{\circ}\) east of due north. What must be (a) the magnitude and (b) the direction of its second run if it is to end up at the new location of beetle \(1 ?\)

A person desires to reach a point that is \(3.40 \mathrm{~km}\) from her present location and in a direction that is \(35.0^{\circ}\) north of east. However, she must travel along streets that are oriented either north-south or east-west. What is the minimum distance she could travel to reach her destination?

Consider \(\vec{a}\) in the positive direction of \(x, \vec{b}\) in the positive direction of \(y\), and a scalar \(d\). What is the direction of \(\vec{b} / d\) if \(d\) is (a) positive and (b) negative? What is the magnitude of (c) \(\vec{a} \cdot \vec{b}\) and (d) \(\vec{a} \cdot \vec{b} / d ?\) What is the direction of the vector resulting from (e) \(\vec{a} \times \vec{b}\) and (f) \(\vec{b} \times \vec{a}\) ? (g) What is the magnitude of the vector product in (e)? (h) What is the magnitude of the vector product in (f)? What are (i) the magnitude and (j) the direction of \(\vec{a} \times \vec{b} / d\) if \(d\) is positive?

Vectors \(\vec{A}\) and \(\vec{B}\) lie in an \(x y\) plane. \(\vec{A}\) has magnitude \(8.00\) and angle \(130^{\circ} ; \vec{B}\) has components \(B_{x}=-7.72\) and \(B_{y}=-9.20\). What are the angles between the negative direction of the \(y\) axis and (a) the direction of \(\vec{A}\), (b) the direction of the product \(\vec{A} \times \vec{B}\), and \((\mathrm{c})\) the direction of \(\vec{A} \times(\vec{B}+3.00 \hat{\mathrm{k}}) ?\)

For the following three vectors, what is \(3 \vec{C} \cdot(2 \vec{A} \times \vec{B}) ?\) $$\begin{aligned} &\vec{A}=2.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}}-4.00 \hat{\mathrm{k}} \\ &\vec{B}=-3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}} \quad \vec{C}=7.00 \hat{\mathrm{i}}-8.00 \hat{\mathrm{j}}\end{aligned}$$
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