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You drive a car \(680 \mathrm{~m}\) to the east, then \(340 \mathrm{~m}\) to the north. (a) What is the magnitude of your displacement? (b) Using a sketch, estimate the direction of your displacement. (c) Verify your estimate in part (b) with a numerical calculation of the direction.

Short Answer

Expert verified
(a) The magnitude is approximately 760.53 m. (b) and (c) Around 26.57° north of east.

Step by step solution

01

Identify the components of displacement

From the problem statement, you have two components of displacement. The eastward displacement is \(680 \, \mathrm{m}\). The northward displacement is \(340 \, \mathrm{m}\). We will treat these as the horizontal and vertical components in a right triangle.
02

Calculate the magnitude of the displacement

The magnitude of the displacement can be found using the Pythagorean theorem. The formula is \(c = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the components of displacement.\[c = \sqrt{(680)^2 + (340)^2} = \sqrt{462400 + 115600} = \sqrt{578000}\]Calculating gives \(c \approx 760.53\, \mathrm{m}\).
03

Sketch and estimate the direction

Draw a coordinate system with the positive x-axis pointing east and the positive y-axis pointing north. Place a point at \( (680, 340) \). You will likely estimate that the angle is closer to the x-axis because the eastward component is larger than the northward one.
04

Calculate the direction numerically

Using trigonometry, calculate the angle \( \theta \) with respect to the eastward direction:\[\theta = \tan^{-1} \left(\frac{b}{a}\right) = \tan^{-1} \left(\frac{340}{680}\right) = \tan^{-1}(0.5)\]Using a calculator, \( \theta \approx 26.57^\circ \). This angle is measured counterclockwise from the east.

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Key Concepts

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

Pythagorean theorem
The Pythagorean theorem is a fundamental concept in geometry. It helps us calculate the length of any side of a right triangle, given the lengths of the other two sides. In its most famous form, it states:
  • c² = a² + b²
Here, c represents the hypotenuse, or the longest side opposite the right angle. The sides a and b are the two other sides forming the right angle. We apply this theorem to problems involving distance, like the displacement of an object moving in two perpendicular directions, to find the magnitude of the total displacement. By plugging in the eastward (\(680 \, \mathrm{m}\)) and northward (\(340 \, \mathrm{m}\)) components, we solve for c, yielding the result approximately \(760.53 \, \mathrm{m}\), thus representing the straight-line distance from the starting point.
magnitude calculation
Magnitude calculation refers to finding the overall size or length of a vector, which in physics often represents both direction and quantity. For displacement, which differs from distance by having direction, the magnitude describes how far an object is from its starting point.
To find this in two dimensions, we use the components of displacement along the x-axis and y-axis, equivalent to eastward and northward movements respectively. Applying the Pythagorean theorem, you can calculate the magnitude of this 2D vector.
  • Use the formula: \(c = \sqrt{a^2 + b^2}\)
  • Substitute the known values: \(a = 680 \, \mathrm{m}\), \(b = 340 \, \mathrm{m}\)
  • Solve to find the magnitude: \(c \approx 760.53 \, \mathrm{m}\)
This magnitude tells us the direct distance from the start to the end point, irrespective of the path taken.
direction estimation
Direction estimation involves determining the angle or orientation of a vector relative to a reference line, typically the horizontal axis in geometry problems. After identifying the components of a displacement, such as eastward and northward movement, direction estimation allows us to understand which direction the resultant displacement vector points.
Drawing a sketch helps visualize this. By plotting the movements on a coordinate grid and seeing which component (east or north) is longer, you get a rough idea of the angle. In our example, the car traveled further east (\(680 \, \mathrm{m}\)) than north (\(340 \, \mathrm{m}\)), so the angle was estimated to be closer to the east axis.
Through visualization, you can approximate the angle before confirming it numerically.
trigonometry
Trigonometry is the branch of mathematics dealing with the relationships between the sides and angles of triangles. It's particularly useful for direction calculations in problems involving right triangles, like those we find in physics applications. To calculate an angle when two sides of a triangle are known, such as opposite (northward displacement) and adjacent (eastward displacement) sides, one might use:
  • The tangent function: \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)
In our context, this becomes:
  • \(\theta = \tan^{-1}\left(\frac{340}{680}\right)\)
  • This computes to an angle of about \(26.57^\circ\)
This calculated angle helps better understand how the object moved relative to its starting direction, measured counterclockwise from the east or the positive x-axis.

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Most popular questions from this chapter

Diagram Use a sketch to show that two vectors of unequal magnitude cannot add to zero, but that three vectors of unequal magnitude can.

A projectile is launched with an initial speed of \(12 \mathrm{~m} / \mathrm{s}\). At its highest point its speed is \(6 \mathrm{~m} / \mathrm{s}\). What was the launch angle of the projectile?

A softball is thrown from the origin of an \(x-y\) coordinate system with an initial speed of \(18 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) above the horizontal. (a) Find the \(x\) and \(y\) positions of the softball at the times \(t=0.50 \mathrm{~s}, 1.0 \mathrm{~s}, 1.5 \mathrm{~s}\), and \(2.0 \mathrm{~s}\). (b) Plot the results from part (a) on an \(x-y\) coordinate system, and sketch the parabolic curve that passes through them.

For each of the following quantities, indicate whether it is a scalar or a vector: (a) the time it takes you to run the \(100-m\) dash, (b) your displacement after running the \(100-\mathrm{m}\) dash, (c) your average velocity while running, (d) your average speed while running.

Vector \(\overrightarrow{\mathbf{A}}\) points in the negative \(y\) direction and has a magnitude of \(5 \mathrm{~km}\). Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(15 \mathrm{~km}\) and points in the positive \(x\) direction. Use components to find the magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\), and (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\).

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