A person starts from his home and walks the following distances: i) 200 m, due north ii) 180 m, 30.0° W of N ii) 120m, due south. How far is the person from his home? O 236 m O 260 m O 253 m O 90.0 m O 295 m
A person starts from his home and walks the following distances: i) 200 m, due north ii) 180 m, 30.0° W of N ii) 120m, due south. How far is the person from his home? O 236 m O 260 m O 253 m O 90.0 m O 295 m
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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Transcribed Image Text:1.7 The Components of a Vector
A person starts from his home and walks the following distances: i) 200 m, due north i) 180 m, 30.0° W of N iii) 120m, due south.
How far is the person from his home?
O 236 m
O 260 m
O 253 m
O 90.0 m
O 295 m
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1.7 The Components of a Vector
A person starts from his home and walks the following distances: i) 200 m, due north ii) 180 m, 30.0° W of N iii) 120 m, due south.
How far is the person from his home?
O 236 m
O 253 m
O 90.0 m
O 260 m
O 295 m
Submit Answer
Save for Later
W
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Transcribed Image Text:115⁰
in
1187
Ax,
N
ute
#1: i
मेने पे
300
717²7-1200m
X
1
i=200m i K80m iii = 120m
ix=0
We know this will be
necfativ!
Ho
200m
#2 iix (40) : Sino = # -> Sin OH = Ha(iix) -> Sin (30°) 180m = iix
WE GDCD lixe = 90
TORON
iix = -90
1+ 3 iiy (H₂): Cos Ⓒ - + -D Cos (30°) (180m) = 155m | 1xy = 155 m
O
#ul
iiix = 0
#5y2 -120
x= √ (ix +iix +iiix ) = ( ¡y +iiy tiiiy
-D
J₂ -> x=√ (0+ (-90) +0) ³² + (200 + 155 + (-120)?
x= √(-90)² + (235)² -> x= √√-7865+55325
X> √47360 = x=217.6m
= 12.3
S
iix
Deo.
0+ (-90) +
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