As a grade 12 student learning calculus, these videos have always helped me and I appreciate you for making these videos! I noticed that what I found most helpful was you writing numbered steps such as the ones you did at 3:52 and another in 7.1. So, thank you so much!
Hi Ms Havrot are you able to explain question 3 on page 369. I was able to calculate resultant magnitude being 304.14 using Pythagorean theorem but I want to see how you would find the direction.
The direction is determined by using your right angle triangle and the two given sides (Opposite and Adjacent) and then using the tangent ratio. So tan-1(50/300) = approximately 9.5 degrees So the direction becomes west 9.5 degrees South Make sure to sketch it so you understand and add NSEW
Hi Ms. Havrot, thank you so much for the lesson. I am struggling with a problem right now, and can't figure out how to do it. This is the question: "A sailboat is traveling in the ocean at 65km/h N25E. Due to wind, the sailboat ends up traveling 52km/h N1 10W. What is the velocity of the wind?" I was wondering if you could please help me with it?
I’m wondering if you meant N110W or N10W If you make a sketch you would draw the N25E as one direction vector with a magnitude of 65km/hr and the resultant will be your 52 km/hr drawn at N 110W or is that N10W? Then joining these two will be your wind velocity.
@@mshavrotscanadianuniversit6234Thank you so much for taking the time to reply also I'm sorry for the typo, it is N 10 W. What about direction? I struggling to find the degree for direction
Have you been able to figure it out? It helps to draw compass points for each to determine the angles. I’m out of town right now but if you are still struggling let me know. I need to find some paper and a pencil!
@@mshavrotscanadianuniversit6234 Yes, I got 37.3 km/h W 11.9 S. But I am not sure if it's right. Also thank you so much Ms. Havrot, I really appreciate your effort.
I have a question but it’s not from the book Example 1: A canoeist who can paddle at a speed of 5km/h in still water wishes to cross a river 400m wise that has a current of 2km/h. If he steers the canoe in a direction perpendicular to the current, determine the resultant velocity. Determine the point on the opposite bank where the canoe touches. Example 2: Suppose the canoeist in Example 1 wanted to travel straight across the river. Determine the direction he must head and the time it will take him to cross the river. those are 2 examples my teacher gave us
Hey Ms. Havrot!! Could you please help me with this question. I am not sure how to draw vector diagram and solve it Bird A is flying from his home to meet his friend Bird B on the beach, 4 km south of his home. There is a wind coming from the east at a speed of 5 km/h. If Bird A flies at a speed of 7 km/h, what is his resultant velocity so that he ends up exactly on the beach with Bird B? Round your answers to two decimal places and or any angles to the nearest degree.
This is easier than it sounds (as always!). First of all the fact that the other bird is 4 km south has nothing to do with the resultant velocity. (But knowing that you want to fly due south is!) It might be part of a secondary question such as “how long will it take to bird A to travel 4 km?” So, leaving that out of your calculation you have a nice right angle triangle to work with. Remember that where you want to go is the resultant. First draw a line straight down from A to B. That’s your resultant (4 km). Now, because the wind is blowing from the east it means that your bird will have to fly south and east so that that wind blows him straight to bird B. Your hypotenuse is 7 km/hr and if you label the angle at the top angle A then the opposite side is 5 km/hr Using Pythagorean theorem you can evaluate the resultant velocity (~4.69 km/hr) which makes sense as the bird will be slowed down by the wind. Angle A is approx 8.21° so bird A should fly S8.22°E to arrive at the desired location (you were not asked for this but it is a simple sinA = 5/7 calculation. Hope that helps you. It would be much easier to show you the diagram!
The airplane drawing is beautiful!! Modern day Vincent Van Goph
I spent years in art classes 😂
As a grade 12 student learning calculus, these videos have always helped me and I appreciate you for making these videos! I noticed that what I found most helpful was you writing numbered steps such as the ones you did at 3:52 and another in 7.1. So, thank you so much!
Thanks you. That’s a lovely compliment. And thank you for watching ❤️
Hi Ms Havrot are you able to explain question 3 on page 369. I was able to calculate resultant magnitude being 304.14 using Pythagorean theorem but I want to see how you would find the direction.
The direction is determined by using your right angle triangle and the two given sides (Opposite and Adjacent) and then using the tangent ratio. So tan-1(50/300) = approximately 9.5 degrees So the direction becomes west 9.5 degrees South
Make sure to sketch it so you understand and add NSEW
Thanks Ms. Havrot!
THANK YOU SO MUCH
Here you go ... all you need to know about the water!
ua-cam.com/video/trONC30i7mo/v-deo.html
Hi Ms. Havrot, thank you so much for the lesson. I am struggling with a problem right now, and can't figure out how to do it. This is the question: "A sailboat is traveling in the ocean at 65km/h N25E. Due to wind, the sailboat ends up traveling 52km/h N1 10W. What is the velocity of the wind?" I was wondering if you could please help me with it?
I’m wondering if you meant N110W or N10W If you make a sketch you would draw the N25E as one direction vector with a magnitude of 65km/hr and the resultant will be your 52 km/hr drawn at N 110W or is that N10W? Then joining these two will be your wind velocity.
@@mshavrotscanadianuniversit6234Thank you so much for taking the time to reply also I'm sorry for the typo, it is N 10 W. What about direction? I struggling to find the degree for direction
Have you been able to figure it out? It helps to draw compass points for each to determine the angles. I’m out of town right now but if you are still struggling let me know. I need to find some paper and a pencil!
@@mshavrotscanadianuniversit6234 Yes, I got 37.3 km/h W 11.9 S. But I am not sure if it's right. Also thank you so much Ms. Havrot, I really appreciate your effort.
for the chandelier question, why did u draw the resultant vector arrow downward?
Because that's the force of gravity pulling it downward..
@@mshavrotscanadianuniversit6234 but you cant add vectors with that kind of arrow arrangement
The downward force of the chandelier is balanced out by the upward forces of the wires.
Can you also do airplane problems calculating direction of the wind, resultant, or plane speed. And also can you make vector boat problems.
Tell me which questions in the book you would like me to explain.
I have a question but it’s not from the book
Example 1:
A canoeist who can paddle at a speed of 5km/h in still water wishes to cross a river 400m wise that has a current of 2km/h. If he steers the canoe in a direction perpendicular to the current, determine the resultant velocity. Determine the point on the opposite bank where the canoe touches.
Example 2:
Suppose the canoeist in Example 1 wanted to travel straight across the river. Determine the direction he must head and the time it will take him to cross the river.
those are 2 examples my teacher gave us
Hey Ms. Havrot!!
Could you please help me with this question. I am not sure how to draw vector diagram and solve it
Bird A is flying from his home to meet his friend Bird B on the beach, 4 km south of his home. There is a wind coming from the east at a speed of 5 km/h. If Bird A flies at a speed of 7 km/h, what is his resultant velocity so that he ends up exactly on the beach with Bird B? Round your answers to two decimal places and or any angles to the nearest degree.
This is easier than it sounds (as always!). First of all the fact that the other bird is 4 km south has nothing to do with the resultant velocity. (But knowing that you want to fly due south is!) It might be part of a secondary question such as “how long will it take to bird A to travel 4 km?”
So, leaving that out of your calculation you have a nice right angle triangle to work with. Remember that where you want to go is the resultant.
First draw a line straight down from A to B. That’s your resultant (4 km). Now, because the wind is blowing from the east it means that your bird will have to fly south and east so that that wind blows him straight to bird B.
Your hypotenuse is 7 km/hr and if you label the angle at the top angle A then the opposite side is 5 km/hr
Using Pythagorean theorem you can evaluate the resultant velocity (~4.69 km/hr) which makes sense as the bird will be slowed down by the wind.
Angle A is approx 8.21° so bird A should fly S8.22°E to arrive at the desired location (you were not asked for this but it is a simple sinA = 5/7 calculation. Hope that helps you. It would be much easier to show you the diagram!
@@mshavrotscanadianuniversit6234 thank you for your explanation. It makes sense to me now 😊