Relative Motion Review— Sine and Cosine Law Solution
Summary
TLDRThe video explains how to determine the heading and airspeed of a plane navigating in windy conditions. The scenario involves a pilot flying to a city 600 km away in two hours, with wind blowing at 70 km/h at 10° west of south. Through vector analysis, diagrams, and calculations using trigonometric laws (cosine and sine laws), the airspeed of 278 km/h and a heading of 1.81° south of west are determined. The process includes understanding resultant vectors and correcting for wind direction to achieve the desired flight path.
Takeaways
- 😀 The pilot's goal is to reach a city 600 km away in 2 hours, heading 15 degrees south of west.
- 🌬️ There is a 70 km/h wind blowing at 10 degrees west of south, which affects the plane's trajectory.
- 🛫 The plane's resultant velocity, considering the wind, is 300 km/h, calculated by dividing the distance by the time.
- 🔍 The script uses vector addition to determine the plane's heading and airspeed, considering the wind's effect.
- 📏 The cosine law is applied to find the plane's airspeed by using the known wind speed and resultant velocity.
- 🧮 The calculated airspeed of the plane is approximately 278 km/h, accounting for the wind's influence.
- 📐 The sine law is used to determine the angle between the wind vector and the plane's heading.
- 🔄 The plane's heading is found by subtracting the angle calculated using the sine law from the desired direction.
- 🏁 The final heading of the plane is approximately 1.81 degrees south of west, adjusted for the wind.
- 📘 The script provides a detailed step-by-step approach to solving the problem using trigonometric laws and vector analysis.
Q & A
What is the distance the pilot needs to cover to reach the city?
-The pilot needs to cover a distance of 600 kilometers to reach the city.
In which direction does the pilot need to fly to reach the city?
-The pilot needs to fly in a direction 15 degrees south of west to reach the city.
What is the speed and direction of the wind?
-The wind is blowing at a speed of 70 kilometers per hour at a direction of 10 degrees west of south.
How long does the pilot have to reach the city?
-The pilot has two hours to reach the city.
What is the resultant speed of the plane required to reach the city in the given time?
-The resultant speed of the plane is 300 kilometers per hour, calculated by dividing the distance by the time.
What is the angle between the wind direction and the desired direction?
-The angle between the wind direction (10 degrees west of south) and the desired direction (15 degrees south of west) is 65 degrees.
What is the formula used to calculate the unknown side of the triangle in this scenario?
-The cosine law is used to calculate the unknown side of the triangle, which is represented as 'b^2 = a^2 + c^2 - 2ac cos(B)'.
What is the calculated airspeed of the plane with respect to the air?
-The calculated airspeed of the plane is 278 kilometers per hour.
How is the heading of the plane determined?
-The heading of the plane is determined by subtracting the angle 'a' from the desired direction angle of 15 degrees south of west.
What is the final heading of the plane after accounting for the wind?
-The final heading of the plane after accounting for the wind is 1.81 degrees south of west.
What trigonometric function is used to find the angle 'a' in the triangle?
-The sine function is used to find the angle 'a' in the triangle, using the formula 'sin(a) = opposite side / hypotenuse'.
Outlines
🛫 Introduction to the Vector Problem of Airplane Heading with Wind
In this paragraph, the problem of determining the heading and airspeed of a plane flying towards a city 600 km away in 2 hours is introduced. The wind is blowing at 70 km/h, 10 degrees west of south, and the goal is to find the plane's required heading and airspeed to compensate for the wind. A diagram is referenced to illustrate the vector relationship between the airspeed, wind, and resultant velocity needed to reach the city in the desired direction, which is 15 degrees south of west.
✈️ Analyzing the Angles in the Problem Using Parallel Lines
This section focuses on determining key angles in the vector diagram using parallel lines and alternate interior angles. First, the angle between the resultant velocity and the east-west line is identified as 80 degrees. Then, the smaller angle of 15 degrees south of west is noted. By subtracting these two angles, a 65-degree angle in the triangle is calculated. This angle will be used later to solve for the plane's airspeed using the cosine law.
📐 Solving for Airspeed Using the Cosine Law
In this paragraph, the cosine law is applied to find the magnitude of the airspeed vector. The sides and angles of the triangle are labeled, and the relevant equation for cosine law is set up. Using the given wind speed (70 km/h), the resultant velocity (300 km/h), and the calculated angle (65 degrees), the airspeed of the plane is found to be 278 km/h. This value represents the speed the plane must fly to compensate for the wind.
🎯 Determining the Plane’s Heading Using the Sine Law
This paragraph explains how to calculate the plane's heading using the sine law. With known values of angle B (65 degrees) and side B (278 km/h), along with side A (70 km/h), angle A is found to be approximately 13.19 degrees using the inverse sine function. Finally, the heading is calculated by subtracting this angle from 15 degrees, resulting in a heading of 1.81 degrees south of west. The paragraph concludes with the final solution: the airspeed of 278 km/h and a heading of 1.81 degrees south of west.
Mindmap
Keywords
💡Relative motion
💡Resultant vector
💡Heading
💡Wind speed
💡Cosine law
💡Sine law
💡Airspeed
💡Degrees south of west
💡Vector addition
💡Parallel lines and angles
Highlights
The pilot aims to reach a city 600 kilometers away in two hours, facing a wind blowing at 70 km/h at 10 degrees west of south.
The desired direction to the city is 15 degrees south of west.
The resultant velocity is calculated as 300 km/h, derived from the distance and time required to reach the destination.
The wind vector is described as 80 degrees south of west for simplification in calculations.
Parallel lines and alternate interior angles are used to determine the angles in the vector diagram.
The angle between the wind vector and the desired direction is calculated to be 65 degrees.
The cosine law is applied to find the airspeed of the plane without wind.
The airspeed is calculated to be 278 km/h using the cosine law.
The heading of the plane is determined by subtracting the angle 'a' from the desired direction angle.
The sine law is used to find the angle 'a' based on the known sides and angles in the triangle.
The angle 'a' is found to be approximately 13.19 degrees using the inverse sine function.
The final heading of the plane is calculated to be 1.81 degrees south of west.
The airspeed vector is determined to be 278 km/h at a heading of 1.81 degrees south of west.
The solution involves correcting for the wind's effect on the plane's trajectory.
The problem-solving process is visualized through a diagram that helps in understanding the vector addition.
Transcripts
this is question two from the vector
problems that have sine and cosine lots
of solutions from the review for
relative motion
so a pilot wishes to reach a city 600
kilometers away in the direction 15
degrees south of west
in two hours
if there is a wind that is below of 70
kilometers per hour
blowing at 10 degrees west of south
what must be the heading and air speed
of the plane
so i've drawn a diagram that goes from a
to b
right and it's in a direction 15 degrees
south of west which i've
labeled here
so we want to find the heading and the
air speed of the plane
so since we know that we want to get
from a to b we want to go in a direction
15 degrees south of west
that must mean that this
is the resultant this is what
someone on the ground would see this is
where they would actually end up going
so that must mean that the pilot has to
correct for the wind speed so we want to
find what the direction and the speed
that the plane would go in
if there was no wind
that is the heading and air speed
so that must mean that our heading and
airspeed our airspeed vector
has to add to the wind to find our
resultant in blue here
so the wind is blowing at 10 degrees
west of south so that's going to look
like this
10 degrees west of south and then our
air speed so our original speed
um is going to be
from here to here
so you can see that this is our
air speed
and it adds to the wind speed
and gives us our resultant
okay so i've relabeled the diagram to
make it a little bit neater
this is our wind speed this
70 kilometers per hour here
i've just labeled the
80 degrees south of west instead of 10
degrees west of south just to make it a
little bit neater
okay so let's take a take a look at what
we
know and what we need to know so they
tell us that
the wind speed is 70 kilometers per hour
but we actually also know the resultant
speed and that is because we know
the distance that we have to travel
and how fast we need to get there
so that is our resultant because that is
how fast that we actually want to go
so
our velocity is going to be distance
divided by time
which is going to be 600 kilometers
divided by 2 hours
that is going to give us 300 kilometers
per hour this is our resultant speed
because this is how fast we want to go
okay so let's label that 300 kilometers
per hour
here
okay so now we have to look for some
angles and in order to find angles we
need parallel lines
so
what i see first is this set of parallel
lines so if you take a look at this
east-west line and this east-west line
they're parallel and this
our wind vector is kind of like a
diagonal
and we know
that our alternate interior angles which
would be this
80 degrees here
and this
big angle so from the east-west line all
the way up to that
vector
they are the same so we know that this
big vector is 80 degrees
now let's look at another pair
of parallel lines
let's look at
this east-west line
and
this
oops
this east-west line
okay
so these two are parallel now if we take
a look at
this diagonal
right
we know this angle here the alternate
interior is going to be this angle here
so they're going to be the same so we
know that this angle is 15 degrees
so
if we want
um
if we want
this angle
then
we can
take 80 degrees which is this
this bigger one minus
15 degrees which is this smaller one
and 80 minus 15 that would give us 65
degrees and that is our angle
oops
our angle that we want
in the triangle so this is
65 degrees
in the triangle here
okay so now that we've found this angle
65 degrees
we can find our
air speed which is going to be this
purple line here
okay so i will i've labeled the
angles
so i called this one angle a i'll call
this one angle b and i'll call this one
angle c
okay so
i'll label the sides as well so i'll
call this one
side c since it's opposite to angle c
and i'll call this one side a since it's
opposite angle a
and our last one which is what we're
looking for is side b and we don't know
that yet
okay so if we take a look at what we
know we have this side length
we have this side length
and we have an angle that is in between
them
and since this angle happens to be the
angle that is opposite
to side b which is what we're looking
for we can use the cosine law
so let's set up the cosine law
and
the letters that are used in the
regular format of the cosine law that
doesn't really matter
as long as you have two sides and an
angle in between so
now my cosine law is going to look like
b squared equals a squared plus c
squared minus 2
a c
cosine
b
where this
is the angle
opposite
to the side that we're looking for so
the letter names don't really matter
as long as you have two sides and an
angle between them
okay so let's solve for b
b is going to equal the square root
of
all of this so a is 70
70 squared plus 300 squared which is
going to be our c squared
minus 2 times
a sorry 2 times a
times c
times cosine b which is the angle across
from the side that we don't know so
across from the angle that don't know 65
degrees
and if you plug this entire expression
into your calculator
that should give you
278
kilometers per hour if we're rounding to
three sig figs
okay
so now we know that b is 278
kilometers per hour
and now
we just need to look for the heading so
to look for the heading
we need an angle so you can look for
this angle
or you can look for
this angle
i am going to stick to
this top angle here that i've labeled in
blue
okay so to do that we can find angle a
and then we can subtract angle a from
15. so we do 15 minus angle a that will
give us our heading
so let's do that to find angle a
we can use the sign law
since we have the side a that's across
from it we have another angle
and we have the side that is across from
that angle so let's set it up
we will have
sine a
over a which is the side that's opposite
to it
equals sine b
over b to the side opposite to that so
sine a
we don't know what angle a is but we do
know the side that is opposite to it
right here
a equals 70 kilometers per hour so sine
a divided by 70 equals sine b
which is the angle that we have here 65
degrees
sine 65 divided by
side b which is across from it and we
just solve for side b side b right
that's 278.
all right so let's isolate sine a
so sine a equals 70 times sine 65
divided by 278 so i've just multiplied
the 70 over
and that will give us
that will give us
0.228
i'm not going to round here since we
still need to find angle a and to find
angle a we need to take the inverse sine
of this decimal so
inverse sine of
0.228
and that will give us
13.19 something degrees now be careful
angle a is not our final answer but
instead we need the heading which again
is going to be 15 degrees minus a
so our heading will be
our
heading
is equal to 15 degrees
minus a which is 13.19
something degrees
which will give us
1.81 degrees if we just round to three
sig figs
so our final velocity velocity of the
plane with respect to the air so our air
speed
is going to be
our magnitude which are which we
calculated here
278 kilometers per hour
at
1.81 degrees now what direction is this
well if
this
is west
we are going
to the south of west
so we are going to go
1.81 degrees
south
of
west
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