b) I'm applying the definition of average velocity into vectors, thus finding a displacement vector in the time interval and plugging in. Then I use the the standard formulas for magnitude and theta.
1) Δr=(BC)-(AB)=(-980km)j-(-410km)i=(-410km)i+(-980km)j.***correct
2) Δt=1.50hr-0.750hr=0.750hr.:***Nope,
3) average velocity vector=(-547km/h)i+(-1.31e3km/h)j***Nope
***Velocity=(-410km)/.75hr i+(-980km/1.5hr)j.***
(c) Average speed is merely a magnitude, so just add the magnitude of each trip and divide by the time interval.****Nope, add the two distances, then divide by total time. total time=2.25 hrs
Clarification: this is calculus-based physics, so the methodology is not restricted to trigonometry only.
I have two questions, so hopefully I can divide them up nice and evenly.
Also, if you know where I should copy and paste proper vector notation from, please tell me so I can be more clear. Thanks!
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1. A plane flies 410 km east from city A to city B in 45.0 min and then 980 km south from city B to city C in 1.50 h. (Assume the +x axis points east and the +y axis points north.)
(a) In magnitude-angle notation, what is the plane's displacement for the total trip?
(b) In magnitude-angle notation, what is the plane's average velocity for the total trip?
(c) What is the plane's average speed for the trip?
(a) Let's forget about this, I got the correct answer.
(b) I'm applying the definition of average velocity into vectors, thus finding a displacement vector in the time interval and plugging in. Then I use the the standard formulas for magnitude and theta.
1) Δr=(BC)-(AB)=(-980km)j-(-410km)i=(-410km)i+(-980km)j.
2) Δt=1.50hr-0.750hr=0.750hr.
3) average velocity vector=(-547km/h)i+(-1.31e3km/h)j
4) |avg V vector|=√[(-547km/h)^2+(-1.31e3km/h)^2]=1.42e3km/h
5) θ=arctan[(-1.31e3km/h)/(-547km/h)]=67.3°
6) 180°+67.3°=247°
(c) Average speed is merely a magnitude, so just add the magnitude of each trip and divide by the time interval.
1) avg speed=(410km+980km)/0.75hr=1.85e3km/h
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2. An Earth satellite moves in a circular orbit 687 km above the Earth's surface. The period of the motion is 98.3 min.
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?
(a) Using formula of T=2πr/v I solve for v. This yields v=2πr/T.
1) r=(mean earth radius)+(satellite height)=6,371,687m
2) v=(12,743,347m)π/5,898s=6.79e3m/s
(b) Now.. if the speed from (a) were correct, we could input into a=v^2/r for centripetal a. However, my calculated speed is incorrect :<
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Any insights on where I screwed up or should've looked at is greatly appreciated!
2 answers
2. An Earth satellite moves in a circular orbit 687 km above the Earth's surface. The period of the motion is 98.3 min.
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?
(a) Using formula of T=2πr/v I solve for v. This yields v=2πr/T.
1) r=(mean earth radius)+(satellite height)=6,371,687m***NOPE NOPE
height=6.37E6+.687E6...
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?
(a) Using formula of T=2πr/v I solve for v. This yields v=2πr/T.
1) r=(mean earth radius)+(satellite height)=6,371,687m***NOPE NOPE
height=6.37E6+.687E6...
1 answer