One of the activities in Boracay is parasailing which uses a speed Mang Tata the driver of the speed boat drives 2.5 km due North of Island, then turns onto the open sea and continues in a direction 300 N of E for 1.5 km and finally turns 2.0 km due East. What is the total displacement of Mang Tata? 3. Jay leaves the office, drives 26 km due North, then turns onto a street and continues in a direction 300 N of East for 35 km and finally turns onto the highway due East for 40 km. What is his total displacement from the office?
One of the activities in Boracay is parasailing which uses a speed Mang Tata the driver of the speed boat drives 2.5 km due North of Island, then turns onto the open sea and continues in a direction 300 N of E for 1.5 km and finally turns 2.0 km due East. What is the total displacement of Mang Tata? 3. Jay leaves the office, drives 26 km due North, then turns onto a street and continues in a direction 300 N of East for 35 km and finally turns onto the highway due East for 40 km. What is his total displacement from the office?
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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Directions: Answer given 2 and 3 problems using the component method and please show your work with encoded solutions and follow the images for the format of answers.
2. One of the activities in Boracay is parasailing which uses a speed Mang Tata the driver of the speed boat drives 2.5 km due North of Island, then turns onto the open sea and continues in a direction 300 N of E for 1.5 km and finally turns 2.0 km due East. What is the total displacement of Mang Tata?
3. Jay leaves the office, drives 26 km due North, then turns onto a street and continues in a direction 300 N of East for 35 km and finally turns onto the highway due East for 40 km. What is his total displacement from the office?
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The athlete walks 8 km East, then 5 km South, and finally 6 km West.
we need to find the final displacement of the athlete.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a54982e-0c97-462c-b022-56dccb176771%2F3a2ef15d-2e53-412d-946d-88ea2a43ef76%2Fmtrrk7_processed.png&w=3840&q=75)
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The athlete walks 8 km East, then 5 km South, and finally 6 km West.
we need to find the final displacement of the athlete.
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1. An Olympic athlete is doing a warm exercise
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ghjk |
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Step 2
Vector
dx
dy
8 km East
8 km
5 km South
-5 km
6 km West
-6 km
E dz=2 km
Σά-5 km
Therefore, total displacement will be:
dr = V(E dz)² + ( d,)² =
(2)² + (-5)² km
V29km
= 5. 385 km
The direction will be:
E d,
tan 0 =
E d.
or, 0 = –68. 19°
or, 0 = 68. 19° S of E
Note, the negative sign indicates south.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a54982e-0c97-462c-b022-56dccb176771%2F3a2ef15d-2e53-412d-946d-88ea2a43ef76%2Fw2r1jabh_processed.png&w=3840&q=75)
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1. An Olympic athlete is doing a warm exercise
Сaps
d.
ghjk |
a
Enter
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Step 2
Vector
dx
dy
8 km East
8 km
5 km South
-5 km
6 km West
-6 km
E dz=2 km
Σά-5 km
Therefore, total displacement will be:
dr = V(E dz)² + ( d,)² =
(2)² + (-5)² km
V29km
= 5. 385 km
The direction will be:
E d,
tan 0 =
E d.
or, 0 = –68. 19°
or, 0 = 68. 19° S of E
Note, the negative sign indicates south.
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