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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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.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw: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 component-wise. Hope this helps :)
* from the origin in quadrant 1 anticlockwise
Hey @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?

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How would I use components to find the avg. velocity @AllTehMaffs
Kidnapped? What a horrifying question.
Hi. 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|
Okay... I used 20° though
Does that matter?
That 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
So how would I use components?
whoops, 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.
Can you see how to get the lengths of the sides of that triangle?
Using sin and cos?
|dw: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
So to find the deltas I would...?
:) Not yet. you already have the points. What's your change in x?
2.1sin20-(-1.33)
right, and you find your y the same way, then use Pythagoras to find C
I think I contradicted myself by saying we don't need angles, but you're good so far
This gives the displacement so I would have to divide by time?
yeah, and then that's your average velocity between those points
So then as far as acceleration goes?
That's where you need those other angles decided to overheat
wow, don't know how those messages got mashed up :P Sorry, my computer decided to overheat
So! Acceleration. You have to divide your velocities up into components, too. The first one, anyway; the second is only in 1 direction.
|dw:1381700640583:dw|
Sorry 35 mph?
that's the angle. V1 was 35 mph
yeah
your 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 }\
For v2 I would use 10sin20?
:/ \[a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }]\]
Not quite. Remember that when the person made the measurement, the car was going straight east.
So all of V2 is in the x direction
So what would it be?
my computer is being crashy. So calculate the change of velocity in the x direction, and then calculate it in the y
What's v2x?
I'm not sure
How much of V2 is moving in the x direction?
2.1sin20?
|dw:1381701681185:dw| |dw:1381701807535:dw|
Does that help?
I can't see the v2 stuff on the dawning
nooooo!! bummer. k |dw:1381702127697:dw|
Does it make sense why we're not using the 20 degrees for this part of the calculation?
Yes
So 2.1cos0
?
Not that either. In the problem, what does it say about his velocity when he measures it a second time?
2.1 miles is the distance away from the police station
??
"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"
2.1?
moving E at a speed of 10 mph"
V2 = ?
10
V2x - V1x = ?
Where did you get 1.33?
10-35cos66°
yes!! Now V2y - V1y = ?
0-35sin66
yes!!
now (V2x - V1x) / t = ax (V2y - V1y)/t = ay a^2 = (ax)^2 + (ay)^2
and a is your average acceleration
Should I use | |?
??
Abs value
for acceleration? It has to be positive anyway because you're squaring the term
Why is a squared?
You're finding the magnitude of vector a, which is kinda like a right triangle. It's defined as a = sqrt ((ax)^2 + (ay)^2)
When I divide by t it's 5/60 right?
ummm , yeah, I think so
5 min (1 hour/60 min)
yeah
I gotta run to work. This problem was mean!! :/
Ok so I get 387. What units should it be?
that'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
Ok!
Or, maybe I way over complicated it. It's going 35 miles per hour to start, and it ends up going ten miles per hour
the magnitude to the deceleration is (10-35)/5min
it might just be that, I don't know. Good mathin'!
When you have a chance just double check for me! :)
First, 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 10-35 / 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! :)
|dw:1381728262451:dw|

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