Completely stumped on this one -- Kinematic Conceptual problem

  • #26
haruspex
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When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.
 
  • #27
thanks man - I went through it and I got this:

I thought about it - and I noticed you d is facing the opposite way of s1 and s2, is d negative and thus why you switched the inequality sign? Thanks again that problem kicked my ass but seeing it solved really helps. In math I am used to getting values and solving problems but in something like this where they don't give me any values really just gave me writers block, if that is a thing in math/physics :P. Take care.
I was getting close to end of my board so skipped some steps hence flipped inequality by mistake. d distance and not position or displacement so there's no need to label it as positive or negative. Sorry for not providing the hints and directly stating the answer.
 
  • #28
When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.
I'm sorry. I'm new to forum. It'll take some time to learn all rules
 
  • #29
ruskointhehizzy
When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.
I was going to mention that but it was due yesterday so seeing the answer I feel is okay since I wont get credit for it anyways.

I was getting close to end of my board so skipped some steps hence flipped inequality by mistake. d distance and not position or displacement so there's no need to label it as positive or negative. Sorry for not providing the hints and directly stating the answer.
that's fine man I was able to work through it - I appreciate the solution, it helped me learn in this situation but I learn best when asked questions so I can provide my own solution. I know your intentions are best at heart though, and this was due yesterday so it is totally fine. Thank you again man. No worries on the rules I am new here as well man. Take care!
 
  • #30
Okay I understand that the approaching car will be a straight line because it has constant speed, but I don't understand why X0 is negative? That totally makes sense to graph it I don't know why I didn't think of that. Could you explain to me how you got x = 1/2at^2? I can see how that will be a curve because you are going to have to floor it and you will gain speed at an exponential rate. I also don't understand what you mean by car A is at x = -x0 at t = 0.

Thank you so much. Here is a quick graph I made - is this correct?
I'm just going off an assumption here. Could it be because Xo is prior to your starting point? So anything g behind your initial position is negative?
 
  • #31
Ray Vickson
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I'm just going off an assumption here. Could it be because Xo is prior to your starting point? So anything g behind your initial position is negative?
Of course. Surely you must have seen a "number line" before? East = positive ##x##, west = negative ##x##.

number_line.gif
 

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  • #32
kuruman
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Now that the problem has been marked "solved", I can contribute a quicker way to get to the answer. The required distance ##x_{min.}## will depend on two independent parameters, the constant speed ##v_0## of the trailing car and the maximum acceleration ##a## of the leading car. The leading car must reach speed ##v_0## over the minimum distance ##x_{min.}##. If it moves slower than that, it started accelerating too late; if it moves faster than that, it started accelerating too soon. Therefore, from the "fourth" kinematic equation $$2ax_{min.}=v_0^2 - 0^2 \rightarrow x_{min.}=\frac{v_0^2}{2a}$$
 
  • #33
haruspex
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Now that the problem has been marked "solved", I can contribute a quicker way to get to the answer. The required distance ##x_{min.}## will depend on two independent parameters, the constant speed ##v_0## of the trailing car and the maximum acceleration ##a## of the leading car. The leading car must reach speed ##v_0## over the minimum distance ##x_{min.}##. If it moves slower than that, it started accelerating too late; if it moves faster than that, it started accelerating too soon. Therefore, from the "fourth" kinematic equation $$2ax_{min.}=v_0^2 - 0^2 \rightarrow x_{min.}=\frac{v_0^2}{2a}$$
C.f. post #23.
 
  • #34
kuruman
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C.f. post #23.
Sure, but it's not necessary to run time backwards. In either car's reference frame the other car drops its speed from initial speed ##v_0## to zero over distance ##x_{min.}##.
 
  • #35
haruspex
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Sure, but it's not necessary to run time backwards. In either car's reference frame the other car drops its speed from initial speed ##v_0## to zero over distance ##x_{min.}##.
Yes, it is not necessary to run time backwards; I was just trying to make it exactly like the commonest example of the equation.
 

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