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Relative Motion - What is your thought process?

  • Thread starter 03125
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  • #1
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This is not a homework question. I know the answer to this question I'm just not sure how, from reading the question, I would have known what the question is actually asking. If someone could explain what their thought process is after reading the question, and how it is that you know what the question is really asking, I would very much appreciate it.

Homework Statement


Susan, driving North at 60 mph, and Shawn, driving East at 45 mph, are approaching an intersection. What is Shawn's speed relative to Susan's reference frame?


The Attempt at a Solution


The solution is [45^2+60^2]^.5=75mph
.... but I don't really understand why. I do realize that 75 is the hypotenuse of the triangle, I just don't understand how I would have known what this question is asking for, it seems as if there is way too much that the question doesn't tell you. How would I know from looking at this question that it is implied that the vehicles are on a collision course with each other at their exact center of mass?

Any tips for breaking a question down in order to see what it is really asking and implying are appreciated.

I realize that this is kind of a strange question, so any helpful input is appreciated and if it just seems like I'm asking a dumb question, well, I understand that too =)
 

Answers and Replies

  • #2
cepheid
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Welcome to PF 03125

Here's my thought process:

If we're going to view the problem from Susan's reference frame, we're switching to a reference frame in which Susan is stationary and the whole world is moving south past her at 60 mph. Therefore, in this reference frame, in addition to having his 45 mph eastward component of motion relative to the ground, Shawn also has a 60 mph southward component of motion (which is the speed of the ground relative to Susan). These two components must be added (vectorially), and his total speed relative to Susan is the magnitude of that resultant vector (hence the use of Pythagoras: if you resolve his velocity vector into its eastward and southward components, you get a right triangle, and you need to solve for the hypotenuse).

I'm not sure what you mean when you say "they are on a collision course." It's true they are heading towards the same point, but that does not necessarily mean that they will reach it at the same time. Even if they did collide, the street intersection is not actually the point where their centre of mass is located, at least not for the majority of the trip. At any instant, their centre of mass lies somewhere along a direct straight line connecting the two cars.
 
  • #3
gneill
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You are given two velocity vectors, as measured in the "Earth" or "ground" frame of reference. If you want to frame-jump to one of them ("What is Shawn's speed relative to Susan's reference frame?") then from Susan's point of view she is stationary, and she sees Shawn moving with an extra velocity component that is the negative of her velocity with respect to the ground. In other words, from her point of view Shawn has a southbound component velocity component equal to 60 mph (her velocity with respect to the ground).

Shawn's total velocity (from Susan's perspective) is then his original velocity in the ground frame minus her velocity in the ground frame. The speed is calculated by taking the vector magnitude of this velocity, or sqrt(452 + 602).
 

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