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anonymous
 3 years ago
PLEASE, HELP been waiting nearly and hour... – Drawing a vector displacement problem, need help!
After being kidnapped and placed in the trunk of a car, you discover you are able to pry open the trunk. The GPS tells you that you're moving 35 mph in a direction 66 north of east at a location 1.33 mi west of the police station. 5 min later, you take another reading that tells you are now 2.1 mi from the police station in a direction 20 degrees east of north and are moving E at a speed of 10 mph. Find the avg. velocity and acceleration during the trip.
anonymous
 3 years ago
PLEASE, HELP been waiting nearly and hour... – Drawing a vector displacement problem, need help! After being kidnapped and placed in the trunk of a car, you discover you are able to pry open the trunk. The GPS tells you that you're moving 35 mph in a direction 66 north of east at a location 1.33 mi west of the police station. 5 min later, you take another reading that tells you are now 2.1 mi from the police station in a direction 20 degrees east of north and are moving E at a speed of 10 mph. Find the avg. velocity and acceleration during the trip.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381583505376:dw The second set of points was meaner to find because it gave them to you in polar form. Remember that polar coordinates in cartesian can be found by x = r cos (theta) y = r sin (theta) But only when theta is measured from quadrant 1. If you used the 20 degrees they gave you, the cos and the sin would be switched around. Remember that the average velocity is the change in position  that's the vector from point A to point B  over time 5 min. The average acceleration is the change in velocities. Whereas you can calculate displacement vector just by making a right triangle using your two points, the change in velocity is most easily done componentwise. Hope this helps :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0* from the origin in quadrant 1 anticlockwise

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hey @AllTehMaffs I'm having trouble finding average velocity. By the way, I switches cos and sin around because I'm using 20 degrees. Could you help me out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How would I use components to find the avg. velocity @AllTehMaffs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Kidnapped? What a horrifying question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi. Sure. For the average velocity you need to find the magnitude of the vector that goes from from the first GPS point to the second. So dw:1381698737663:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay... I used 20° though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That won't change the x,y position, just make sure that where I have a sin 70, you have a cos 20 and vice versa

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how would I use components?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whoops, I definitely forgot my trig identities. scratch what I just said about the swapping. But we'll get back to that in a sec. It's not immediately important.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you see how to get the lengths of the sides of that triangle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381699218098:dw You shouln't need to use sin and cos for the displacement vector. You technically don't know the angles right off the bat :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So to find the deltas I would...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:) Not yet. you already have the points. What's your change in x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, and you find your y the same way, then use Pythagoras to find C

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I contradicted myself by saying we don't need angles, but you're good so far

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This gives the displacement so I would have to divide by time?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, and then that's your average velocity between those points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So then as far as acceleration goes?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's where you need those other angles decided to overheat

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow, don't know how those messages got mashed up :P Sorry, my computer decided to overheat

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So! Acceleration. You have to divide your velocities up into components, too. The first one, anyway; the second is only in 1 direction.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381700640583:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the angle. V1 was 35 mph

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your acceleration in the x direction is \[a_{x}=\frac{ v _{2x}v_{1x} }{ \Delta t }\] and in the y it's \[a_{y}=\frac{ v _{2y}v_{1y} }{ \Delta t }\

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For v2 I would use 10sin20?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:/ \[a_{y}=\frac{ v _{2y}v_{1y} }{ \Delta t }]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not quite. Remember that when the person made the measurement, the car was going straight east.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So all of V2 is in the x direction

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my computer is being crashy. So calculate the change of velocity in the x direction, and then calculate it in the y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How much of V2 is moving in the x direction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381701681185:dw dw:1381701807535:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can't see the v2 stuff on the dawning

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nooooo!! bummer. k dw:1381702127697:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does it make sense why we're not using the 20 degrees for this part of the calculation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not that either. In the problem, what does it say about his velocity when he measures it a second time?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02.1 miles is the distance away from the police station

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"tells you are now 2.1 mi from the police station in a direction 20 degrees east of north and are moving E at a speed of 10 mph"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0moving E at a speed of 10 mph"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Where did you get 1.33?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes!! Now V2y  V1y = ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now (V2x  V1x) / t = ax (V2y  V1y)/t = ay a^2 = (ax)^2 + (ay)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and a is your average acceleration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for acceleration? It has to be positive anyway because you're squaring the term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're finding the magnitude of vector a, which is kinda like a right triangle. It's defined as a = sqrt ((ax)^2 + (ay)^2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When I divide by t it's 5/60 right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ummm , yeah, I think so

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05 min (1 hour/60 min)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I gotta run to work. This problem was mean!! :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok so I get 387. What units should it be?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that'd be miles per hour, so something is really wrong there. Because that's insane. I'll look at it again when I get back

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or, maybe I way over complicated it. It's going 35 miles per hour to start, and it ends up going ten miles per hour

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the magnitude to the deceleration is (1035)/5min

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it might just be that, I don't know. Good mathin'!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When you have a chance just double check for me! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First, units on acceleration here should be miles/(hour^2) (acceleration always has units distance/time^2), so sorry I mistyped that before. Second, I think that the answer is right, and the just 1035 / t is definitely wrong. Not used to working in mph, so the numbers are bigger than I had thought they would be... I got 387miles/h^2 too. It would be negative, since both x and y components are getting smaller, and has a direction. But I don't think you need that for your answer. Hope this was the right method they were looking for! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381728262451:dw
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