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You are kidnapped by armed political-science majors (who are upset because you told them that political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0\) min, turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point and (b) in what direction relative to your initial direction of travel are you?

Step by step solution

01

Convert speeds and times

Convert minutes and seconds into hours: 2.0 min = 2/60 hours, 4.0 min = 4/60 hours, 60 s = 1/60 hours, 60 s = 1/60 hours, 2.0 min = 2/60 hours, 30 s = 0.5/60 hours.

02

Calculate distances

Use the formula distance = speed × time:1. Distance for 50 km/h for 2 min: \[d_1 = 50 \times \frac{2}{60} = \frac{100}{60} = \frac{5}{3} \text{ km} \]2. Distance for 20 km/h for 4 min: \[d_2 = 20 \times \frac{4}{60} = \frac{80}{60} = \frac{4}{3} \text{ km} \]3. Distance for 20 km/h for 60 s: \[d_3 = 20 \times \frac{1}{60} = \frac{20}{60} = \frac{1}{3} \text{ km} \]4. Distance for 50 km/h for 60 s: \[d_4 = 50 \times \frac{1}{60} = \frac{50}{60} = \frac{5}{6} \text{ km} \]5. Distance for 20 km/h for 2 min: \[d_5 = 20 \times \frac{2}{60} = \frac{40}{60} = \frac{2}{3} \text{ km} \]6. Distance for 50 km/h for 30 s: \[d_6 = 50 \times \frac{0.5}{60} = \frac{25}{60} = \frac{5}{12} \text{ km} \]

03

Determine the direction of each leg

Analyze each leg and the turns to determine final position relative to starting point:1. Start in a certain direction (let's call it North).2. After traveling \[d_1\]] km, turn 90° to the right (East).3. After traveling \[d_2\]], turn 90° to the right (South).4. After traveling \[d_3\]], turn 90° to the left (East).5. After traveling \[d_4\]], turn 90° to the right (South).6. After traveling \[d_5\]], turn 90° to the left (East).7. After traveling \[d_6\]] meters.

04

Employ vector addition

Use vector addition to determine the resultant displacement vector. Assume each direction change results in a new vector component:1. \[North: d_1 = \frac{5}{3}\]2. \[East: d_2 + d_3 + d_5 - d_4 = \frac{4}{3} + \frac{1}{3} + \frac{2}{3} - \frac{5}{6}\]3. \[South: d_6 = \frac{5}{12}\]

05

Calculate the resultant displacement vector

Summing up these vectors,North component: \[ \text{North component} = \frac{5}{3} - (d_3 + d_5 + d_6)\] \[East component = \frac{11}{6} + \frac{5}{12}\]

06

Use the Pythagorean theorem to calculate the resultant

The magnitude of the displacement is \[R= \text{North component^2 + East component^2}\] sqrt{\frac{5}{3}} - \frac{11}{6}(+ \frac{5}{12}) whole squared.

07

Calculating displacement direction using trigonometric functions

Use trigonometric functions to find the angle. Lets call East the positive x axis n North as positive y axis.

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Key Concepts

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

Displacement

Displacement is a vector quantity that represents the shortest distance from the initial to the final position of an object. It considers both the magnitude and direction. In the given problem, displacement helps us determine how far and in what direction we end up after following a specific path.
Instead of just looking at the total distance traveled, displacement takes into account the straight-line distance from start to finish. This means we need to add up all the changes in position, considering their directions, to find the final displacement vector.

Relative Direction

In physics, direction is critical when dealing with problems involving movement and vectors. Relative direction means understanding how each turn or movement alters the overall direction. For instance, a 90° turn to the right adds an eastward component if you were initially heading north.
In the exercise, each turn specified (right or left) alters the path. We have to account for these changes systematically, updating our direction from the turn's reference point. Visualizing the path on a coordinate system helps manage this well, so each segment's direction is easy to keep track of.

Speed and Time Conversion

Speed and time conversions are essential for calculating distances. Speed is generally given in km/h (kilometers per hour), but we often need to convert time from minutes or seconds into hours to match the speed's unit.
For instance, converting 2.0 minutes to hours involves calculating \(\frac{2}{60} \) hours. Similarly, converting 60 seconds to hours means \(\frac{60}{3600} \) hours, simplifying to \(\frac{1}{60} \) hours. Once converted, these times can be multiplied by the speed to find distances. This is done using the formula Distance = Speed × Time.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, used to find the length of the sides of a right-angled triangle. It states that for a right-angled triangle with sides a and b, and hypotenuse c, the relation is \[ a^2 + b^2 = c^2 \].
In displacement problems, we use this theorem to find the resultant vector's magnitude from the components. If a represents the north-south component and b represents the east-west component, the resultant displacement's magnitude is given by: \[ R = \sqrt{a^2 + b^2} \]. This helps us calculate the direct straight-line distance between the starting and ending points.

Trigonometry in Physics

Trigonometry is a branch of mathematics dealing with angles and their relationships within right-angled triangles. In physics problems involving vectors, trigonometry helps determine the direction of the resultant vector.
Using functions like sine, cosine, and tangent, we can find angles relative to a given direction. For instance, if we have our north component as y and east component as x, the direction of the resultant displacement vector (angle \(\theta \)) from the initial axis can be calculated using: \[ \theta = \tan^{-1} \bigg( \frac{y}{x} \bigg)`. \]Knowing this angle and the resultant displacement's magnitude completely describes the final position relative to the starting point.