A person is driving in his/her car for some errands. • The first errand he/she drove 2 km East and 6 km North from the home. • He/she then drives 4 km East and 2 km South. What is the magnitude (number) of the displacement of how far the car is located from the starting point to the nearest km? b 6 km 14 km 7 km 10 km
A person is driving in his/her car for some errands. • The first errand he/she drove 2 km East and 6 km North from the home. • He/she then drives 4 km East and 2 km South. What is the magnitude (number) of the displacement of how far the car is located from the starting point to the nearest km? b 6 km 14 km 7 km 10 km
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement
A person is driving in his/her car for some errands.
- The first errand he/she drove 2 km East and 6 km North from the home.
- He/she then drives 4 km East and 2 km South.
### Question
What is the magnitude (number) of the displacement of how far the car is located from the starting point to the nearest km?
### Options
- a) 6 km
- b) 14 km
- c) 7 km
- d) 10 km
### Answer
- The correct answer is (d) 10 km.
### Explanation and Solution
To calculate the displacement, we need to determine the resultant vector from the initial point to the final point.
1. **First Movement:**
- 2 km East
- 6 km North
This results in coordinates \((2, 6)\)
2. **Second Movement:**
- 4 km East
- 2 km South
This results in coordinates \((6, 4)\).
The displacement is the straight-line distance between the starting point (0,0) and the final point (6,4). Using the Pythagorean theorem:
\[ \text{Displacement} = \sqrt{x^2 + y^2} \]
Where \( x = 6 \) km (total distance East) and \( y = 4 \) km (total distance North-South).
\[ \text{Displacement} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21\, \text{km} \]
Upon revisiting the calculations, I realize I made a mistake in the distance approximation with initial misunderstanding. \( \sqrt{52}\approx 7.21 \) which isn't an optoin in the question.
Checking Options More Carefully,
a) 6 km
b) 14 km
c) 7 km
d) 10 km
Again clarifying calculation steps involved.
Correct trajectory without vectors cross-check needed to clear up.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb69b1dbc-ab62-44ef-9b57-71232ebb3671%2Fe2d0dad0-c4f4-44a1-98c6-a2abd32b595c%2Fg9nyhb9_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A person is driving in his/her car for some errands.
- The first errand he/she drove 2 km East and 6 km North from the home.
- He/she then drives 4 km East and 2 km South.
### Question
What is the magnitude (number) of the displacement of how far the car is located from the starting point to the nearest km?
### Options
- a) 6 km
- b) 14 km
- c) 7 km
- d) 10 km
### Answer
- The correct answer is (d) 10 km.
### Explanation and Solution
To calculate the displacement, we need to determine the resultant vector from the initial point to the final point.
1. **First Movement:**
- 2 km East
- 6 km North
This results in coordinates \((2, 6)\)
2. **Second Movement:**
- 4 km East
- 2 km South
This results in coordinates \((6, 4)\).
The displacement is the straight-line distance between the starting point (0,0) and the final point (6,4). Using the Pythagorean theorem:
\[ \text{Displacement} = \sqrt{x^2 + y^2} \]
Where \( x = 6 \) km (total distance East) and \( y = 4 \) km (total distance North-South).
\[ \text{Displacement} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21\, \text{km} \]
Upon revisiting the calculations, I realize I made a mistake in the distance approximation with initial misunderstanding. \( \sqrt{52}\approx 7.21 \) which isn't an optoin in the question.
Checking Options More Carefully,
a) 6 km
b) 14 km
c) 7 km
d) 10 km
Again clarifying calculation steps involved.
Correct trajectory without vectors cross-check needed to clear up.
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