geometric vectors
geometric vectors
Velocity
MCV4U: Calculus & Vectors
Recall that speed, a measure of how fast something is
travelling, is a scalar quantity.
Velocity is speed with direction, and is a vector quantity.
By using vectors to represent velocities, it is possible to solve
a variety of problems.
Velocities as Vectors
J. Garvin
J. Garvin — Velocities as Vectors
Slide 2/16
Slide 1/16
geometric vectors
geometric vectors
Resultant Velocity Problems
Resultant Velocity Problems
Example
Use the Pythagorean Theorem to find the kayaker’s speed.
p
|~r | = 82 + 52
√
= 89
A kayaker paddles 8 km/h due south across a river that has a
current flowing 5 km/h due east. What is the resulting
velocity of the kayaker?
Use the following diagram, where ~k is the velocity of the
kayaker, ~c is the velocity of the current, and ~r is the resultant
velocity.
≈ 9.4 km/h
Use a trigonometric ratio to find the direction.
5
θ = tan−1
8
≈ 32◦
Therefore, the resulting velocity is approximately 9.4 km/h
S32◦ E.
J. Garvin — Velocities as Vectors
Slide 3/16
J. Garvin — Velocities as Vectors
Slide 4/16
geometric vectors
Resultant Velocity Problems
geometric vectors
Resultant Velocity Problems
Example
An airplane is travelling on a bearing of 330◦ at a constant
speed of 150 km/h. A wind blows on a bearing of 85◦ at 40
km/h. Determine the speed and direction of the airplane
relative to the ground.
~ is the airplane’s
Use the following diagram, where AH
~ is the wind’s velocity, and AR
~ is the resultant
velocity, AW
velocity of the airplane relative to the ground.
J. Garvin — Velocities as Vectors
Slide 5/16
J. Garvin — Velocities as Vectors
Slide 6/16
geometric vectors
geometric vectors
Resultant Velocity Problems
Resultant Velocity Problems
To determine the velocity of the airplane relative to the
~
ground, we need to determine the magnitude of AR.
To determine the direction, use the sine law (or cosine law)
to find the measure of ∠HAR, then add it to the airplane’s
original bearing.
From the given bearings, ∠WAH = 30◦ + 85◦ = 115◦ , and
∠AHR = 180◦ − 115◦ = 65◦ .
~
Use the cosine law to determine |AR|.
q
~ 2 + |AH|
~ 2 − 2(|HR|)(|
~
~ cos(AHR)
|HR|
AH|)
q
= 402 + 1502 − 2(40)(150) cos(65◦ )
~ =
|AR|
≈ 138 km/h
sin(HAR)
sin(AHR)
=
~
~
|HR|
|AR|
sin(HAR)
sin(65◦ )
≈
40
138
40 sin(65◦ )
∠HAR ≈ sin−1
138
≈ 15◦
Therefore, the actual bearing of the airplane is approximately
330◦ + 15◦ , or 345◦ . Its actual speed is 138 km/h.
J. Garvin — Velocities as Vectors
Slide 7/16
J. Garvin — Velocities as Vectors
Slide 8/16
geometric vectors
geometric vectors
Problems Given a Resultant Velocity
Problems Given a Resultant Velocity
Example
~ represents the resultant trip from
In the diagram, TM
~ be the velocity of the plane on
Toronto to Montréal. Let TB
~ be the velocity of the wind.
its new bearing and let TW
A pilot wishes to fly 508 km from Toronto to Montréal, on a
bearing of 78◦ . The airplane has a top speed of 550 km/h.
An 80 km/h wind is blowing on a bearing of 125◦ .
• In what direction should the pilot fly to reach the
destination?
• At what speed will the plane be travelling?
• How long will the trip take?
Since we are given information about the resultant, we need
to work backward!
Note that ∠TMB = ∠MTW = 12◦ + 35◦ = 47◦ .
J. Garvin — Velocities as Vectors
Slide 10/16
J. Garvin — Velocities as Vectors
Slide 9/16
geometric vectors
geometric vectors
Problems Given a Resultant Velocity
Problems Given a Resultant Velocity
We can use the sine law to find the measure of ∠BTM.
Now that we know the bearing, we can calculate the speed of
the airplane using the cosine law.
sin(BTM)
sin(BMT )
=
~
~
|BM|
|TB|
sin(BTM)
sin(47◦ )
≈
80
550
80 sin(47◦ )
∠BTM ≈ sin−1
550
≈ 6.1◦
∠TBM ≈ 180◦ − 47◦ − 6.1◦ ≈ 126.9◦ .
q
~ = |TB|
~ 2 + |BM|
~ 2 − 2(|TB|)(|
~
~ cos(TBM)
|TM|
BM|)
q
≈ 5502 + 802 − 2(550)(80) cos(126.9◦ )
≈ 601 km/h
The pilot must fly at a bearing of approximately
78◦ − 6.1◦ ≈ 71.9◦ .
Since Montréal is 508 km away, a plane travelling at 601
km/h will make the trip in 508/601 ≈ 0.85 hours, or 51
minutes.
J. Garvin — Velocities as Vectors
Slide 11/16
J. Garvin — Velocities as Vectors
Slide 12/16
geometric vectors
geometric vectors
Relative Velocity
Relative Velocity
Relative velocity is what an observer perceives when s/he
perceives her/himself to be stationary.
Example
It is the difference of two velocities.
Relative Velocity
When two objects, A and B, have velocities v~A and v~B , the
velocity of B relative to A is v~rel = v~B − v~A .
A truck is travelling east at 60 km/h, while a car is travelling
south at 80 km/h. What is the relative velocity of the truck
to the car?
Let v~T be the velocity of the truck and v~C the velocity of the
car. We want v~T − v~C , or v~T + (−v~C ).
While −v~C has the opposite direction as v~C , the two vectors
are still at right angles to each other.
J. Garvin — Velocities as Vectors
Slide 14/16
J. Garvin — Velocities as Vectors
Slide 13/16
geometric vectors
Relative Velocity
geometric vectors
Questions?
Using the Pythagorean Theorem:
p
|v~rel | = 802 + 602
= 100 km/h
Using the tangent ratio for the direction:
60
θ = tan−1
80
≈ 37◦
Therefore, the velocity of the truck relative to the car is 100
km/h, at a bearing of approximately N37◦ E.
J. Garvin — Velocities as Vectors
Slide 15/16
J. Garvin — Velocities as Vectors
Slide 16/16