Completely stumped on this one -- Kinematic Conceptual problem

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The discussion revolves around a kinematic problem where one car accelerates from a stoplight to avoid being rear-ended by another car approaching at constant velocity. Participants emphasize the need to set up equations for both cars' positions over time, highlighting that the accelerating car's motion is described by a quadratic equation, while the approaching car's motion is linear. Key points include the importance of correctly identifying initial positions and understanding that the distance between the two cars at the moment of acceleration is critical to avoid a collision. There is a focus on visualizing the problem through graphs to clarify the relationship between the two cars' movements. The conversation underscores the necessity of grasping the underlying concepts rather than seeking direct answers.
  • #31
DoctorPhysics said:
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
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
kuruman said:
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
haruspex said:
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
kuruman said:
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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