Completely stumped on this one -- Kinematic Conceptual problem

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SUMMARY

The discussion centers on a kinematic problem involving two cars: one stationary at a stoplight and the other approaching at a constant velocity. The key equations used include the position equations for both cars, specifically x = vt for the approaching car and x = x0 + V0xt + 1/2axt^2 for the stationary car accelerating from rest. Participants emphasize the importance of understanding the relationship between displacement, velocity, and acceleration to determine the distance between the two cars at the moment the stationary car begins to move. The conversation highlights the necessity of plotting the position versus time graphs for both cars to visualize their motion and avoid collision.

PREREQUISITES
  • Understanding of kinematic equations, specifically x = vt and x = x0 + V0xt + 1/2axt^2
  • Familiarity with concepts of displacement, velocity, and acceleration
  • Ability to graph functions and interpret motion graphs
  • Basic knowledge of frame of reference in physics
NEXT STEPS
  • Study the derivation and application of kinematic equations in motion problems
  • Learn how to plot position vs. time graphs for different types of motion
  • Explore the concept of relative motion and frames of reference in physics
  • Practice solving collision avoidance problems using kinematic principles
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for insights into teaching motion concepts effectively.

  • #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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