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Homework Help: Avoiding vehicular rear-end collision

  1. Sep 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A car A travelling at 161 km/hr is 676 m directly behind another car B travelling at 29.0 km/hr.
    What must the deceleration be in order for car A to avoid collision with car B?

    2. Relevant equations
    Not sure

    3. The attempt at a solution
    So the initial velocity of car A is 161 km/hr and we're trying to find the deceleration (or negative acceleration as I'll make it here) of it. Assuming car B has no acceleration, the distance gap between them will be: 676 m + 29.0 km/hr - (161 km/hr + negative acceleration).

    I tried to calculate it manually by finding the distance every second, but it takes too long on an exam and I can't find any formula that fits this criteria of information. How do I find the formula for the exact distance at a given point in time given this information?
  2. jcsd
  3. Sep 13, 2012 #2
    Can you setup the equations of motion for each car? This a constant acceleration problem so the basic constant acceleration kinematic equations will apply.
  4. Sep 13, 2012 #3


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    Set up a coordinate system with car A at the origin and car B 676 units ahead of it on the x axis. This will help you in writing up the equations for their positions as functions of time.
    Can you do the rest?
  5. Sep 13, 2012 #4
    I think the question should be the minimum deceleration since the harder the brake applied the sooner the car attain speed less than the front car.
  6. Sep 13, 2012 #5
    Car A is going at:
    161 km/hr - a*t

    and car B is just going at:
    29.0 km/hr

    so the position of car A is just:
    161 km/hr - a*t + 0 m.

    and car B's position is:
    29.0 km/hr + 676 m.

  7. Sep 13, 2012 #6


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    I think you are along the right lines.
    The position of car A should be [itex] s_{xA}= 161t + \frac{1}{2}at^2 [/itex]
    The position of car B should be [itex] s_{xB} = 29t + 676 [/itex]
  8. Sep 13, 2012 #7
    I'm trying to stay away from the formulas because they mean nothing to me and their derivation by integration also doesn't make sense to me (why it happens or what the integration process means).

    I can take the equations at face value, but I have the memory of a fish and can barely remember d=r*t unless I actually think about the values.
  9. Sep 13, 2012 #8
    For minimum deceleration,

    Last edited: Sep 13, 2012
  10. Sep 13, 2012 #9


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    Are you sure this is correct? What I thought of doing was to set [itex] s_{xB}=s_{xA},[/itex]however this equation contains two unknowns, a and t. I need to find another equation, which at the moment I am still trying to find.
  11. Sep 13, 2012 #10
    You can still do it that way too.
    The third equation is the final speed of the behind car should be equal to the front car to avoid the collision.
  12. Sep 14, 2012 #11


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    So, [itex] v_{fA} = 161 + at [/itex] and [itex] v_{fB} = 29 [/itex]. Setting these equal, rearranging for t and subbing into [itex] s_{xA}=s_{xB} [/itex] yields a negative acceleration of -12.9m/s2?
    Last edited: Sep 14, 2012
  13. Sep 14, 2012 #12


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  14. Sep 14, 2012 #13

    I've done a mistake for above post.

    a=-132/10.24= -12.9m/s2

    As in your post #11
  15. Sep 14, 2012 #14


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    I think you made the same mistake as me first time. The 676 is in metres, while the 132 is in km/hr.
  16. Sep 14, 2012 #15
    Thanks for pointing that out.
  17. Sep 14, 2012 #16
    I don't understand anything you guys did, but using the given 1D Kinematic equations, I got:

    x - x_0 = 1/2(v_0 + v)t

    .676 - 0 = 1/2(29 + v)t
    v = (1.352 - 29t) / t

    for car B

    and for car A:

    v^2 = (v_0)^2 + 2a(x - x_0)

    v^2 = (161)^2 + 2a(0 - .676)
    v = sqrt(25921 - 1.352a)

    So I set those equal to each other, but I still have two unknown variables (a and t), so how do I get those values?
  18. Sep 14, 2012 #17


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    There seems to be a certain amount of unnecessary agony taking place over this question :smile:

    Quite often when there are two or more moving objects to keep track of in a problem it can be handy to perform what is known as a change of reference frame. So far you've been looking at the problem from the point of view of a theoretical observer standing still on the road. That is to say, the coordinate system being used is anchored to the road.

    But what would happen if, instead, you were to look at the problem from the point of view of an observer traveling along with car B? Suppose he's looking out the rear window of car B and sees car A approaching. What initial distance and speed of approach (also called "closing speed") will he observe?
  19. Sep 14, 2012 #18
    I'm not sure what you mean gneill.

    I tried a different approach by solving for 't' first:

    Car B's position will be 26t + .676

    So the final position of car A will be at car B's position, therefore:

    26t + .676 = 1/2(161 + 26)t

    Since the initial velocity of car A is 161, and the final velocity should be the same as car B which is 26.

    Solving for t reveals t = (.676 / 67.5)

    That's the first part, now to find the acceleration:

    The position of car A and car B must be the same:

    26t + .676 = 161t - 1/2at^2

    Plug in t, and a = 13,480.0295858. Obviously this is wrong as the number seems way too large, and it is a positive acceleration which means it's speeding up. Why does the solution come to be something that doesn't make any sense?
  20. Sep 14, 2012 #19


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    Staff: Mentor

    Your solution is fine... but consider what units would be associated with it. You did your calculations assuming units of km and hours. So what units will this acceleration have?

    It might be convenient to consider converting the starting values to meters and seconds at the outset, so these sort of "surprises" don't take you unawares :smile:

    Regarding my previous suggestion, an observer looking out the back of car B will see car A initially at a distance of 676 meters and approaching at speed

    161 - 26 = 135 km/hr

    As far as he's concerned, from his point of view car A decelerates from that speed to zero in that distance. One simple, well known equation of motion can be applied to find that acceleration.
  21. Sep 14, 2012 #20
    13,480.0295858 km/hr^2 = 1.04012574 m/s^2

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