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Tough motion question with acceleration, time and distance

  1. Jun 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A car 4m long came up behind a semi-trailer 20m long travelling at a steady rate of 72km/h. The car driver then overtook the truck. If the car pulled out from behind the truck with the front of the car 10m behind the rear of the truck and accelerated at 3.5 m/s2 until the rear of the car was 10m in front of the truck, how far would the car travel and how long would it take?

    2. Relevant equations

    v = u + at

    v2 = u2 + 2ar

    r = ut + 1/2 * at2

    3. The attempt at a solution

    What I've gotten so far is this:
    • The back of the car is 34m behind the front of the truck
    • They are travelling, initially, at 20m/s
    • Therefore, the distance from the initial to final position is 44m (if both were stationary)
    • The acceleration of the car is 3.5 m/s2

    Could someone kindly lead me in the right direction?
     
    Last edited: Jun 6, 2013
  2. jcsd
  3. Jun 6, 2013 #2
    Very well so far.

    Make use of the "if both were stationary" to solve it further. Is there a reference frame where this is true?
     
  4. Jun 6, 2013 #3
    Sorry, I'm not sure what a reference frame is as I'm new to motion as part of physics (only in Year 9). Is it simply the direction of the vector? If so, no direction is given, and the vehicles are travelling in a straight line. I'll have another attempt at the question, concentrating more specifically on the "if both were stationary" comment.
     
  5. Jun 6, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    There is NO "reference frame in which both are stationary" because they have different speeds.
    (Well, initially they have the same speed and the truck maintains that speed so perhaps voko mean "in the reference frame of the truck".)

    In the reference frame of the truck ("relative to the truck"), the truck's speed is 0 (any object's speed is 0 in its own refererncer frame), the car's speed is v= at= 3.5t so the distance traveled is d= 1.75t^2. The distance the rear of the car must travel is 44 m, as you say.

    You could do this in the reference frame of the road- if you don't know "reference frames" that is probably how you would approach it. There the speed of the truck and the initial speed of the car is 20 m/s. In time t, the car will have moved 20t+ 1.75t^2 while the truck will have moved 20t meters. Now the car must move the 44 m above and the distance 20t that the truck moved.

    You can, of course, do this in the reference frame of the car but it looks a bit odd. Now the car's speed is 0 and the truck will have a speed of -3.5t and so in time t will have moved -1.75 t^2. That must be equal to the distance the truck must go backward, relative to the car, -44 m. Of course, except that everything is negative, this is exactly the same as "relative to the truck".

    Try doing all three ways and see that you get the same answers.
     
    Last edited: Jun 6, 2013
  6. Jun 6, 2013 #5
    Any motion is always with respect to SOMETHING. Typically one chooses some object as the "origin" and some directions as the "axes". This is a reference frame. I am sure this was told you when the concepts of displacement, velocity and acceleration were introduced.

    The problem you have uses, implicitly, the road as the reference frame. In this frame, both the truck and car are not stationary, which makes analysis somewhat difficult.

    But what if you choose the truck itself as a reference frame?
     
  7. Jun 6, 2013 #6
    A reference frame is just a coordinate system from where you do the calculations. All the speeds and acceleration values are given with reference to the road. The truck is moving at 72 kmph with respect to the road. You can choose the truck as your frame of reference... If this is the case, you can say the road is moving backwards at 72 kmph with respect to the truck... (This is just to make you understand what is a frame of reference)
    Take the origin as the front of the truck, and observe the back of the car (which will be accelerating from rest wrt truck at 3.5m/s2) and proceed with the calculations.
     
  8. Jun 6, 2013 #7
    I haven't read what you guys have just said as I'm about to go to bed, but this is what I've done now:

    Using r = ut + 1/2 * at2 and the quadratic formula,

    1/2 * at2 + ut - r = 0
    t = ( -u±√(u2- 4 * 1/2 * a * (-r)) ) / 3.5
    ∴ (if my calculations are correct...) t = ( ±2√(77) ) / 3.5 ≈ 5.01


    I will continue tomorrow with the information you've just given :)
     
    Last edited: Jun 6, 2013
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