A novice pilot sets a plane's controls, thinking the plane will fly at 2.50 x 10(squared) km/h to the north. If the wind blows at 75km/h toward the southeast, what is the plane's resultant velocity? Use graphical techniques.
Explain everything to me. I just started physics and I'm going to 8th grade (I'm at the Newark Academy physics thing). I don't even know Geometry yet. Tell me everything.
electron1
v1 = 250 km/h
v2 = 75 km/h
Southeast is 45˚ south of east. To determine the south and east components, use the following equations.
South = 75 * sin 45, East = 75 * cos 25
To graph these velocities, let use a scale of 1 cm = 25 km/h.
250/25 = 10, 75/25 = 3
The first vector is 10 cm north. The second vector is 3 cm at an angle of 45˚ south of east. I drew these two vectors. The resultant is the line from the bottom of the 10 cm line to the end of the 3 cm line. When I measured this line, its length is approximately 8.2 cm. Let’s multiply by 25.
v = 8.2 * 25 = 205 km/h
To check this answer, let’s determine the north component.
North = 250 – 75 * sin 45 = 196.9669914 km/h
East = 75 * cos 25 = 67.97308403 km/h
v = √(196.9669914^2 + 67.97308403^2) = 208.3658702 km/h. This is slightly greater than 205 km/h. When you graph vectors, you can’t get an exact number. I hope this is helpful. To determine the angle north of east, use the following equation.
Tan θ = North ÷ East
On my drawing, this angle is 80˚.
Technobuff
Move the decimal place 2 places right, = 250kph.
The plane does that speed relative to the air. But the air is moving in a direction to the southeast with a speed of 75kph.(45 degrees east of south).
East component = (sin 45 x 75) = 53kph.
South component = the same (cos 45 x 75).
North component of plane's speed = (250 - 53) = 197kph.
East component is 53kph. Nothing changes that.
The plane's speed relative to the ground is sqrt.(197^2 + 53^2) = 204kph.
The direction across the ground is arctan (53/204) = 14.6 degrees east of north.
You do the graphics.
?
"Explain everything to me".
That's impractical. There's too much to explain and it needs diagrams. But this should help.
There are 2 velocities (which are vectors) to add:
vpa is the velocity of the plane relative to the air (250km/h north).
vag is the velocity of the air relative to the ground (75km/h south east).
We want the plane's resultant velocity, vpg, which is the velocity of the plane relative to the ground
vpg = vpa + vag
If you look at the symbols, you will see the pattern.
To add the 2 vectors using graphical techniques, follow these steps:
1) Choose a scale e.g 1cm represents 10km/h.
2) Draw vpa 25cm long and pointing north; include the arrow-head: ↑
3) Starting from vpa's arrow-head, draw vag 7.5cm long pointing south east; include the arrowhead ↘
4) Join the tail of vpa to the arrow head of vag; this is vpg.
5) Measure the length of vpg and convert to km/h using the scale.
6) In this problem vpg is somewhere between north and east; measure the angle (x°) relative to north using a protractor; give the direction as 'x° east of north'.
For an explanation of adding vectors this way, watch the video in the link.
Steve
From the origin, draw a vector length 250 units straight down (S). From the tip of that vector draw another of length 75 in a SE direction. Draw a 3rd vector from the origin to the end of the 2nd vector. Measure its length and angle CW from north. This is the vector sum of the plane's airspeed + wind speed.
The resulting triangle is often called a 'wind triangle' and is used extensively in navigation problems.
Morningfox
To start with, 10*squared is just 100. So 2.50 x 10(squared) is the same as 250. And 250 is faster to type, and much more understandable.