1. An Olympic athlete is doing a warm exercise in preparation for Winter Olympics, he walks 8 km East, then 5 km South and finally 6 km West. Find the final    displacement of the athlete.

University Physics Volume 1
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Chapter2: Vectors
Section: Chapter Questions
Problem 29P: A delivery man starts at the post office, chives 40 km north, then 20 km west, then 60 km northeast,...
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Directions: Answer given 1, 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.

1. An Olympic athlete is doing a warm exercise in preparation for Winter Olympics, he walks 8 km East, then 5 km South and finally 6 km West. Find the final    displacement of the athlete. 

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?

2. 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? 

Sample Problem: A spare fisherman happens to catch fish at the Cuenco
Island, known for its good diving site and fish watching. He swims 2 m East,
turns 3 m 400 North of East and finally moves 2.5 m North. What is the total
displacement of the spare fisherman?
Given: d1=2 m East; d2=3 m, 400 N of E; d3=2.5 m North;
dr=?
Solutions: a.
Vector
dx
dy
2 m East
3 m, 400 N of E
2 m
3 m (cos 40°)
2.31 m
3 m (sin 40°) =
1.92 m
2.5 m North
2.50 m
dx = 4.31 m
Edy = 4. 42 m
b. Use the Pythagorean Theorem to
solve for the total displacement.
dr = /(Edx)? + (Edy)²
dr =
V(4.31 m)2 + (4.42 m)2
dr = V18.58 m² + 19.54 m²
dr = v38.12 m²
dr = 6.17 m
c. To solve for the direction, 0
Edy - 4.42 m
4.31 m
tan 0
1.03
0 = tan-1 = 460
dr = 6.17 m, 46º N of E
Transcribed Image Text:Sample Problem: A spare fisherman happens to catch fish at the Cuenco Island, known for its good diving site and fish watching. He swims 2 m East, turns 3 m 400 North of East and finally moves 2.5 m North. What is the total displacement of the spare fisherman? Given: d1=2 m East; d2=3 m, 400 N of E; d3=2.5 m North; dr=? Solutions: a. Vector dx dy 2 m East 3 m, 400 N of E 2 m 3 m (cos 40°) 2.31 m 3 m (sin 40°) = 1.92 m 2.5 m North 2.50 m dx = 4.31 m Edy = 4. 42 m b. Use the Pythagorean Theorem to solve for the total displacement. dr = /(Edx)? + (Edy)² dr = V(4.31 m)2 + (4.42 m)2 dr = V18.58 m² + 19.54 m² dr = v38.12 m² dr = 6.17 m c. To solve for the direction, 0 Edy - 4.42 m 4.31 m tan 0 1.03 0 = tan-1 = 460 dr = 6.17 m, 46º N of E
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