1. SOLVING VELOCITY VECTOR PROBLEMS

2. Presented by:
IAN RIVAS MAGLINES

3. This is a 60-minute discussion about the techniques used in finding the magnitude and direction (angle) of the Resultant Velocity of vector quantities forming right angles. The methods used are the Parallelogram Method and the Basic Trigonometric Ratios (SOHCAHTOA) which could then be verified using the Pythagorean Theorem. The class also discusses further about the Sine Law and Cosine Law which are used in solving vector quantities which are not forming right angles. The class contains the following:
Introduction
The Basic Trigonometric Ratios
The Parallelogram Method
The Pythagorean Theorem
Problem Sets
The Sine Law and the Cosine Law
Summary
Evaluation/Assignment

4. Introduction

5. The Basic Trigonometric Ratios

6. Sine θ= Opposite/Hypotenuse Cosine θ= Adjacent/Hypotenuse
SOHCAHTOA
Sine θ= Opposite/Hypotenuse
Cosine θ= Adjacent/Hypotenuse
Tangent θ= Opposite/Adjacent

7. The Parallelogram Method
V2 V1 VR

8. The Pythagorean Theorem
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation: c2 = a2 + b2 or, solved for c:

9. If we let a = V1, b = V2 and c = VR
The equation is now:

10. Examples:
1. The compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/hr. If there is a wind of km/hr from west to east, what is the velocity of the airplane relative to the earth?
Given: VP/A = 240 km/hr due north (N)
VA/E = 100 km/hr due east (E)
Find: VP/E or VR

12. VP/A = 240 km/hr due north (N) VA/E = 100 km/hr due east (E)
O= VA/E
A = VP/A
H = VR N O A H
VP/A VR θ E
VA/E

13. Find θ (direction): tan θ = tan θ = 0.42 θ = tan-1(0.42)
A= VP/A = 240 km/hr due north (N) SOHCAHTOA O= VA/E = 100 km/hr due east (E) H = VR
Find θ (direction):
tan θ =
tan θ = 0.42
θ = tan-1(0.42)
θ = 22.78o E of N
Find magnitude of VR:
sin θ =
sin θ VR = VA/E
VR= VA/E/ sin θ
= 100 km/hr/ sin 22.78
= km/hr N O A H
VP/A VR θ E
VA/E
Therefore, the velocity of the airplane relative to the earth is km/hr at 22.78o E of N

14. 2. A motor boat traveling 4 m/s, East encounters a current traveling 3
2. A motor boat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a. What is the resultant velocity of the motor boat? b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? c. What distance downstream does the boat reach the opposite shore?

15. tan θ = (opposite/adjacent) tan θ = (3/4) θ = invtan (3/4)
Let O= VW= 3.0 m/s N
A = VB = 4.0 m/s E
H = VR
Find direction:
tan θ = (opposite/adjacent)
tan θ = (3/4)
θ = invtan (3/4)
θ = 36.9 degrees
Find magnitude of VR:
VR = VB / cos θ
= 4 m/s / cos 36.9
= 5.0 m/s

16. b. ∆t= ∆d / Vav = 80m / 4 m/s = 20 s c. ∆d = (Vav)(∆t) = (3 m/s)(20 s) = 60 m

17. Practice Problems: 1. A person can row a boat at the rate of 8
Practice Problems: 1. A person can row a boat at the rate of 8.0 km/hr in still water. The person heads the boat directly across a stream that flows at the rate of 6.0 km/hr. Find its resultant velocity. 2. In flying an airplane, a pilot wants to attain a velocity with respect to the ground of 485 km/hr eastward. A wind is blowing southward at 42.0 km/hr. What velocity must the pilot maintain with respect to the air to achieve the desired ground velocity?