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Plane and Car Problem

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Homework Statement


A plane traffic monitor monitors traffic on a major highway in the U.S. by flying around 1500m above the road with a constant speed of 200 Km / h. Using the radar, the pilot discovers a car thought to be traveling above the speed limit and is located 2400m from the plane and in order for the plane to stand directly above the car, itneeds to reduce its speed at a rate of 220Km/h^2. Calculate the speed of the car.

Homework Equations


We can calculate the distance between the car and the plane using the Pythagorean Theorem, that gives us that the distance is 1,87km.

The plane will travel x distance while catching up to the speed of the car and [tex]x=u_p t - 1/2 a t^2 , where a = 220 km/h^2 [/tex]
The car will travela distance x+x' in the same time (at a constant speed lets say) that is [tex]x+x'=u_c t[/tex] where x'=1,5 km
We substitute t from the second equation and get the equation below :
[tex]x = u_p \frac{x+x'}{u_c} - \frac{a(x+x')^2}{2(u_c)^2}[/tex]


The Attempt at a Solution


As I state before, this is my approach to the problem. It is pretty straightforward, but I feel I'm missing something, since I don't know the distance x . I can't find the clue in the problem statement, and I am stuck trying to figure it out. Any suggestions?
 
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Answers and Replies

  • #2
gneill
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As I state before, this is my approach to the problem. It is pretty straightforward, but I feel I'm missing something, since I don't know the distance x . I can't find the clue in the problem statement, and I am stuck trying to figure it out. Any suggestions?
The initial distance from the plane to the car, 2400m, along with the cruising altitude of the plane, should give you the initial horizontal distance betwixt them.

attachment.php?attachmentid=40956&stc=1&d=1321369196.jpg
 

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  • #3
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XD love the pic gneill

Well then this problem has a distinct false logic, that confuses one as to where the plane is with respect to the car. Because I thought that the car was behind the plane, and it seems that this case is also probable since it's not specified.
 
  • #4
gneill
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XD love the pic gneill
Why, thank you!
Well then this problem has a distinct false logic, that confuses one as to where the plane is with respect to the car. Because I thought that the car was behind the plane, and it seems that this case is also probable since it's not specified.
I think that if the car starts out behind the plane, the plane can't end up above the car and going the same speed using only a constant deceleration.
 
  • #5
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Why can't it? It will reach speed lower than that of the car and then the car will reach it gradually.
 
  • #6
gneill
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Why can't it? It will reach speed lower than that of the car and then the car will reach it gradually.
Yes, but then the plane's speed won't match that of the car -- it won't "stand directly above the car", which I interpret to mean hover indefinitely above the car.
 
  • #7
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Well yea I guess it won't indefintely hover above the car while constantly decelerating but that also applies in your scenario. We assume that the plane decelerates as much as needed to reach the speed of the car. This is achieved in an amount of time that depends on the speed of the car, right? Well anyway I'm not sure I am totally agreeing with you on this, but it really is not much of the case since the solution to the problem is your approach :) Maybe I'll ask my professor about this!
 
  • #8
gneill
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Well yea I guess it won't indefintely hover above the car while constantly decelerating but that also applies in your scenario. We assume that the plane decelerates as much as needed to reach the speed of the car. This is achieved in an amount of time that depends on the speed of the car, right? Well anyway I'm not sure I am totally agreeing with you on this, but it really is not much of the case since the solution to the problem is your approach :) Maybe I'll ask my professor about this!
In one case the plane can decelerate to match position and velocity (presumably it then ceases its deceleration and remains above the car). In the other case it can decelerate to match position, but then will be traveling too slowly to remain above the car; it would have to then accelerate to match speeds.
 
  • #9
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In one case the plane can decelerate to match position and velocity (presumably it then ceases its deceleration and remains above the car). In the other case it can decelerate to match position, but then will be traveling too slowly to remain above the car; it would have to then accelerate to match speeds.
Oh I see, you're right. However, we don't know what the pilot wants to do, probably your case though :)
 
  • #10
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Just revisited this problem to write it down, and I encountered the fact that the problem remains: How much has the car moved until the plane hovers above it?
 
  • #11
gneill
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You have the starting and ending velocity of the plane, and its acceleration. That should allow you to determine the time for the maneuver...
 
  • #12
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The ending velocity of the plane is the velocity of the car, I don't know that. I definitely misunderstood before, both cases, the plane being in front of or behind the car produce to me the same problem. That either the car or the plane respectively travel an extra amount that I can't calculate. So the equation for distance covered by the plane is

l + l' = u_0 * t + 1/2 *a*t^2 where I substitute t= l'/u_c due to linear movement with speed u_c of the car , where l is the horizontal distance initially and l' the distance that the car has moved before the plane has reached it.

Really have you solved it? I would die to see the solution, I'm so tired after a whole day of work...
 
  • #13
gneill
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Ignore the distance of the plane's travel and concentrate on its velocity change. What's the formula for the change in velocity given acceleration?
 
  • #14
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It would be v=v_0 + a*t , where v= u_c , right? Here I know not u_c or t. (u_c is the speed of the car)
 
  • #15
gneill
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It would be v=v_0 + a*t , where v= u_c , right? Here I know not u_c or t. (u_c is the speed of the car)
Well, you're given the initial speed of the plane and its acceleration. You calculated the speed of the car in part 1 of the problem. The only variable left is t, so you can solve for it.
 
  • #16
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Well, you're given the initial speed of the plane and its acceleration. You calculated the speed of the car in part 1 of the problem. The only variable left is t, so you can solve for it.
Noo you must have it wrong :P, the speed of the car is what I'm looking for, there was no part 1! Either that or I've gone crazy!
 
  • #17
gneill
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Noo you must have it wrong :P, the speed of the car is what I'm looking for, there was no part 1! Either that or I've gone crazy!
In post #10 you wrote:
Just revisited this problem to write it down, and I encountered the fact that the problem remains: How much has the car moved until the plane hovers above it?
I thought we'd moved on to that! This means you still need to find the speed of the car. Can you write the equations for the motion of the plane and car? Take the plane's initial position to be the origin (so the plane starts at distance 0 and the car at distance xo = 1873.5m).
 
  • #18
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Yes I meant I was having trouble understanding stuff again, really sorry if I confused you man, you're doing me a favour keepin up with me.

That would be
for the car: x = x_o + u_c * t
for the plane x = u_0 * t + 1/2 * a * t^2.

What's the same in both equations is time t.
 
  • #19
gneill
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Okay! So you've got two equations in three unknowns (x, u_c, and t). That means you need one more equation in order to be able to solve the system. If it is stipulated that the speed of the plane and the car are the same when they meet, then

u_o + a*t = u_c

is another equation that holds for the problem. Can you solve the three equations for u_c?
 
  • #20
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We got this system:

[tex]x=x_o + u_c t[/tex]
[tex]x=u_0 t + 1/2 at^2[/tex]
[tex]u_c = u_0 + at [/tex]

3rd equation gives us [tex]t= \frac{u_c - u_0}{a}[/tex]
and we sub in 2nd equation [tex] x= u_0\frac{u_c - u_0}{a} +1/2 a(\frac{u_c -u_0}{a})^2[/tex]

and we equate that with the first equation getting
[tex] x_0 + u_c\frac{u_c - u_0}{a} = u_0\frac{u_c-u_0}{a} + \frac{(u_c-u_0)^2}{2a}[/tex]
or
[tex]2ax_0 + 2u_c(u_c - u_0) = 2u_0(u_c - u_0) + (u_c - u_0)^2[/tex]
and we calculate u_c from here?


Yea that's probably it. However I understand that the method of solving using a system is exactly equal to substituting in various equations, which is exactly what I was trying to do and couldn't produce a result. It's funny how our brain works sometimes...
 
  • #21
gneill
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Yup. Move everything to one side and collect terms. You should end up with a quadratic in uc.

There will be two solutions. Verify them by seeing what time they would give using your [itex]t= \frac{u_c - u_0}{a}[/itex] expression.
 
  • #22
PeterO
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The initial distance from the plane to the car, 2400m, along with the cruising altitude of the plane, should give you the initial horizontal distance betwixt them.

attachment.php?attachmentid=40956&stc=1&d=1321369196.jpg
Nice pic, but I would have interpreted the 2400 m as the horizontal distance from plane to car.

As such I would solve the problem with a velocity time graph.

The velocity of the plane starts at 200 km/h, and slopes down to match the acceleration.

The velocity time graph of the car is a horizontal line - some value lower that 200.

The trianglular area between the two graphs represents the extra distance covered by the plane as it slows to match the car's speed.

That area has to equal 2400 - or perhaps 1873.5

Area = base by height.
base and height related by the gradient of the line → the height can be found, and that is how much the plane slows down to match the car.
 

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