On Friday, we went over the sheet from the text book. The first question asked us to create a series of arbitrary vector quantities (two vectors) and identify the vector that could represent Vpg, which is the velocity with perspective to the ground.
The first one is correct because the vector with perspective to the ground is the longest side AND it meets a tail and a tail and the other end meets an arrow with an arrow. The second one is incorrect because every side is an arrow and a tail.
The second question deals with an airplane and a passenger walking on the plane. When you're dealing with questions that ask for a maximum and minimum velocity, it is collinear. Because the object is either moving with the velocity or moving in the opposite direction.
MAXIMUM VELOCITY
The first one is correct because the vector with perspective to the ground is the longest side AND it meets a tail and a tail and the other end meets an arrow with an arrow. The second one is incorrect because every side is an arrow and a tail.
The second question deals with an airplane and a passenger walking on the plane. When you're dealing with questions that ask for a maximum and minimum velocity, it is collinear. Because the object is either moving with the velocity or moving in the opposite direction.
MAXIMUM VELOCITY
100 km/h + 1.5 km/h = 101.5 km/hIt would be 100 km/h + 1.5 km/h because with respective to the ground the plane is already traveling at 100 km/h and since the passenger is walking in the same direction at a speed of 1.5km/h that is added to the planes speed which gives us a total of 101.5 km/h.
MINIMUM VELOCITY
100 km/h - 1.5 km/h = 98.5 km/hThe minimum speed with respect to the ground would be 98.5 km/h because since the plane is heading one direction, in this case to the East, and the passenger is walking 1.5 km/h in the opposite direction [W] then we subtract the velocities to get the minimum velocity.
The third question dealt with two cars, Car A with a ground speed of 80 km/h and Car B with a ground speed of 45 km/h.
a) Both cars are heading in the same direction. A behind B in the reference frame of A.
The third question dealt with two cars, Car A with a ground speed of 80 km/h and Car B with a ground speed of 45 km/h.
a) Both cars are heading in the same direction. A behind B in the reference frame of A.
45 km/h - 80 km/h = -35km/h
The next question deals with a boat that wishes to travel east. If there is a current of 10 km/h flowing north and the boat is capable of traveling at 30 km/h, find the heading and velocity of the boat as seen by a person on shore.
We would subtract because since Car A is gaining speed on Car B then it seems as though Car B is slower because Car A is going at 80 km/h which is faster than 45 km/h.
b) Use the reference frame of B for part a
It would be 80 km/h - 45 km/h = 35 km/h because Car A is faster and since it's behind Car B that means it looks like it's gaining speed on Car B
c) Both cars are heading towards each other. Take the reference frame of A.
It would be 45 km/h + 80 km/h = 125 km/h. Because both cars are heading towards one another, it makes it look as though the other car is coming at you faster cause your in motion and they are too coming towards you.
d) Use B's reference frame for part c
It's 125 km/h, its the same thing as part c. Same reasoning and logic.
b) Use the reference frame of B for part a
It would be 80 km/h - 45 km/h = 35 km/h because Car A is faster and since it's behind Car B that means it looks like it's gaining speed on Car B
c) Both cars are heading towards each other. Take the reference frame of A.
It would be 45 km/h + 80 km/h = 125 km/h. Because both cars are heading towards one another, it makes it look as though the other car is coming at you faster cause your in motion and they are too coming towards you.
d) Use B's reference frame for part c
It's 125 km/h, its the same thing as part c. Same reasoning and logic.
The next question deals with a boat that wishes to travel east. If there is a current of 10 km/h flowing north and the boat is capable of traveling at 30 km/h, find the heading and velocity of the boat as seen by a person on shore.
For number 38 you can solve the for the missing side and angle by using the cosine law because it's not a right angled triangle so we cannot use the Pythagorean theorem. We already know two sides and an angle, that's enough information to find the other missing side and angle.
Et Voila! That was Friday's class. And i believe that JESSICA is the scribe for tonight. Once again I'm terribly sorry for not scribing on Thursday or Friday, it totally slipped my mind. So to make up for it I actually tried on this scribe and finished Thursday's scribe as well. Sorry and hope this helps. Oh, and TEST tomorrow, so study and practice kiddy's.
Et Voila! That was Friday's class. And i believe that JESSICA is the scribe for tonight. Once again I'm terribly sorry for not scribing on Thursday or Friday, it totally slipped my mind. So to make up for it I actually tried on this scribe and finished Thursday's scribe as well. Sorry and hope this helps. Oh, and TEST tomorrow, so study and practice kiddy's.
2 comments:
A very thorough post of what we accomplished on Friday. Nicely explained. Thanks a lot!!
thanks ms. k :D
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