vector velocity

Report
Presented by:
IAN RIVAS MAGLINES
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
Problem Sets
Summary
Evaluation/Assignment
Introduction
The Basic Trigonometric Ratios
SOHCAHTOA
Sine θ= Opposite/Hypotenuse
Cosine θ= Adjacent/Hypotenuse
Tangent θ= Opposite/Adjacent
The Parallelogram Method
V2
V1
VR
V2
V1
VR
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:
If we let a = V1, b = V2 and c = VR
The equation is now:
VR

V 1 V 2
2
2
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
100 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
VP/A = 240 km/hr due north (N)
VA/E = 100 km/hr due east (E)
N
O
O= VA/E
A = VP/A
H = VR
A
H
VP/A
VR
θ
VA/E
E
SOHCAHTOA
A= VP/A = 240 km/hr due north (N)
O= VA/E = 100 km/hr due east (E)
H = VR
Find magnitude of VR:
Find θ (direction):
tan θ =
100 km / hr
N
sin θ =
O
240 km / hr
tan θ = 0.42
θ = tan-1(0.42)
θ = 22.78o E of N
A
H
VP/A
V A/E
VR
sin θ VR = VA/E
VR= VA/E/ sin θ
= 100 km/hr/ sin 22.78
= 256. 41 km/hr
VR
θ
VA/E
E
Therefore, the velocity of the airplane relative to the earth is 256. 41 km/hr at 22.78o E of N
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?
Let O= VW= 3.0 m/s N
A = VB = 4.0 m/s E
H = VR
Find direction:
Find magnitude of VR:
tan θ = (opposite/adjacent)
tan θ = (3/4)
θ = invtan (3/4)
θ = 36.9 degrees
VR = VB / cos θ
= 4 m/s / cos 36.9
= 5.0 m/s
b. ∆t= ∆d / Vav
= 80m / 4 m/s
= 20 s
c. ∆d = (Vav)(∆t)
= (3 m/s)(20 s)
= 60 m
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?

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