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Homework Help: Two moving cars (Kinematics problem)

  1. Jul 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Two cars, A and B are traveling in same direction with velocities vA and vB. When car A is distance d behind car B, the breaks on A are applied, causing a deceleration a. Show that to prevent a collision between A and B, it is necessary that, vA - vB < [itex]\sqrt{2ad}[/itex]

    2. Relevant equations

    1D kinematics

    3. The attempt at a solution

    Suppose at the moment car A starts acceleration, it is at the origin, then car B is at x = d.
    After time t, car A has position vAt - [itex]\frac{1}{2}[/itex]at2
    Then car B has the position vBt + d

    Now, to avoid collision, vBt + d > vAt - [itex]\frac{1}{2}[/itex]at2
    → vA - vB < [itex]\frac{1}{2}[/itex]at + [itex]\frac{d}{t}[/itex]

    Now I have to remove t from this equation.
    Any help please?

    Thanks in advance.
  2. jcsd
  3. Jul 1, 2014 #2


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    Gold Member

    If the car A is behind the car B, then how can it cause a collision when the brakes of A are applied?
  4. Jul 1, 2014 #3
    Well, I think there can be a collision if car A was moving with a much higher velocity than car B. So even after being behind and applying break, it might hit car B.
    Last edited: Jul 1, 2014
  5. Jul 1, 2014 #4


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    Try to solve for the time at which the collision will occur.
  6. Jul 1, 2014 #5
    I can solve the time of collision by taking this inequality as an equality. Is that what you mean?
  7. Jul 1, 2014 #6
    Yes, solve for the time of collision and figure out what condition must be satisfied in order for a solution to exist. If a solution does not exist then the collision doesn't happen.
    Last edited: Jul 1, 2014
  8. Jul 1, 2014 #7


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    Oh, that. :shy:
    Yeah, it is possible. :shy:
  9. Jul 1, 2014 #8

    We have, vBt + d = vAt - [itex]\frac{1}{2}[/itex]at2
    → [itex]\frac{1}{2}[/itex]at2 + (vB - vA)t + d = 0
    Solving for t,
    t = [itex]\frac{-(v_B - v_A) \pm \sqrt{(v_B - v_A)^2 - 4.\frac{1}{2}ad}}{2.\frac{1}{2}at^2}[/itex]
    If we want a solution not to exist then (vB - vA)2 - 2ad < 0
    → (vB - vA)2 < 2ad

    Now we know if x ≤ y then √x ≤ √y if both x, y ≥ 0
    As there is a chance of two cars to collide we must assume vA > vB

    So to take square root on both side we rearrange the equation as
    (vA - vB)2 < 2ad
    → vA - vB < √(2ad)
  10. Jul 1, 2014 #9
    At first, I also thought how can there be a collision?
    But later I also had this Oh! I see moment. :tongue2:
  11. Jul 1, 2014 #10


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    Good job with the solution. Note that there is another approach that is arguably easier.

    Nothing in the situation depends on the speed of the cars relative to the road. All that matters is their velocity, distance and acceleration relative to each other.

    So make your job easy -- assume that the front car is stationary. Now the equation is simple.
  12. Jul 1, 2014 #11
    I hadn't thought this way but now I see that can be another approach. Thank you.
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