Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Velocity of a car in order to meet another car coming towards it

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Car A and Car B and travelling towards each other from 220m apart. If car A travels at 20km/h, both cars will meet at 44.5m. If car A travels at 40km/h, both cars will meet at 76.6m. What is the initial velocity and constant acceleration of car B?

    2. Relevant equations
    1D Kinematics

    3. The attempt at a solution
    For car B, the formulas just involve distance and velocity since we don't have any other information, V^2 = (V_0)^2 + 2a(x - x_0).

    Since acceleration is constant, it just becomes V^2 = (V_0)^2, which is meaningless because they will be the same value since acceleration is constant.

    If we try some other formula like x = x_0 + (v_0)t + 1/2at^2, we are left with 220 = 175.5t + (v_0)t, which again is meaningless because we don't have time. (Same with every other equation, it is impossible to solve for time also given the information.)

    So the problem doesn't seem to have enough information, yet apparently there is a solution...how do you do this?
  2. jcsd
  3. Sep 14, 2012 #2
    It is unclear what the distance to the meeting point is relative to. The original midpoint?

    Apart from this, just treat the initial velocity and acceleration of B as unknowns. The two conditions given translate into two equations with those unknowns. So the problem is fully determined.
  4. Sep 14, 2012 #3
    So for the first condition of car A going at 20km/h, I got:

    0.0445 - 220 = V_0 * 0.002225 + 1/2a(0.002225)^2 for Car B

    For Car A's 40km/h, I got:

    0.0766 - 220 = V_0 * 0.001915 + 1/2a(0.001915)^2 for Car B

    I tried solving for V_0 with the first equation and got:

    (a(0.00000247531) - 219.9555) / 0.002225 = V_0

    I plugged that into the second equation and got -119,670,000 km/hr. Which is obviously wrong. What did I do incorrectly?
  5. Sep 14, 2012 #4
    Again, it is not clear how the distance to the meeting point is measured. Can you clarify?

    Secondly, convert all units to meters and seconds.
  6. Sep 14, 2012 #5
    When car A is going at 20km/h, both cars meet at 44.5m, when they are a distance 220m apart.

    When car A is going at 40km/h, both cars meet at 76.6m, when they are a distance 220m apart.

    The meeting point are just two givens, not measured
  7. Sep 14, 2012 #6
    I converted everything to km and h first because m/s don't make sense to me. I can always convert to meters and seconds afterwards

    I re-did the algebra and came up with 31,011 km/h, which again seems very unreasonable. Not sure what else to do here
  8. Sep 14, 2012 #7
    From what?

    They don't have to make sense to you. They just need to be consistent. Your equations are not.

    Regardless, before you understand what those distances mean, you won't be able to solve this.
  9. Sep 14, 2012 #8
    From the endpoints of 0m and 220m

    I just use it as a reference to check if my solution is reasonable. What should I be understanding?
  10. Sep 14, 2012 #9
    It does not make sense. When the cars meet, they cannot be at 44.5 m from both the 0 m and 220 m endpoints.
  11. Sep 14, 2012 #10
    The spacial equation of car A in the first case:
    t = Sa/Va
    t = 44.5/(200/36)

    The spacial equation of car B in the first case:
    Sb = - V.t - at^2/2
    Being Sb = 220 - 44.5 and T is the value found above in the spacial equation of car A.
    Acceleration and Speed are negative because the car is moving contrary to car A.

    The spacial equation of car A in the second case:
    t = 76.6/(400/36)

    The spacial equation of car B in the second case:
    220 - 76.6 = - Vt - at^2/2

    Putting the numerical value into the two equations of car B, and approximating the values, we get:
    2v + 7a = -41
    v + 4a = -22

    Solving it, we find:
    V = - 10m/s
    A = - 3m/s^2

    So, the absolute value of both are these: 10m/s and 3m/s^2 (approximately, of course).

    But then again, your exercise text is wrong. It must state clear that they meet at 45.5 from car A's start point (that's what I've assumed). Other else, the solution is totally different.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook