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

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