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Constant Acceleration involving two identical objects

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 250 m. If the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 46.5 m, and if it has a constant velocity of 41.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the acceleration of the green car?

    From this I gather the following:

    [tex]R_x_0 = 0m[/tex] iff [tex]t=0s[/tex]
    [tex]G_x_0 = 250m[/tex] iff [tex]t=0s[/tex]
    [tex]R_x = G_x = 46.5m[/tex] iff [tex]R_v = 23 \frac{km}{h} = 6.3889 \frac{m}{s}[/tex]
    [tex]R_x = G_x = 76.6m[/tex] iff [tex]R_v = 41 \frac{km}{h} = 11.3889 \frac{m}{s}[/tex]
    [tex]R_a = 0 \frac{m}{s^2}[/tex] since the velocity is constant
    [tex]T=7.278260871s[/tex] iff [tex]R_v = 6.3889 \frac{m}{s}[/tex]
    [tex]T=6.725853664s[/tex] iff [tex]R_v = 11.3889 \frac{m}{s}[/tex]

    2. Relevant equations
    Since the problem is dealing with constant acceleration, we choose the following formula:
    [tex]x-x_0=v_0 t + \frac{1}{2} a t^2[/tex]
    and rewrite it such as [tex]x=v_0 t + \frac{1}{2} a t^2 + x_0[/tex]

    3. The attempt at a solution
    Well, I converted the velocities above to [tex]\frac{m}{s}[/tex], and obtained the time at both cross points (where [tex]R_x = G_x[/tex]). I also realized that I need to setup both cars into simultaneous equations to solve for the missing variables ([tex]G_v_0[/tex] and [tex]G_a[/tex]). So, I did the following:

    R1: [tex]46.5=(6.3889)(7.278260871) + \frac{1}{2} 0 (7.278026081)^2 + 0[/tex]
    R1: [tex]\Rightarrow 46.5=(6.3889)(7.278260871)[/tex]

    G1: [tex]46.5=V_0(7.278026081)+\frac{1}{2}a(7.278026081)^2+250[/tex]

    E1: [tex](6.3889)(7.278260871) =V_0(7.278026081)+\frac{1}{2}a(7.278026081)^2+250[/tex] since both equations are equal.

    R2: [tex]76.6=(11.3889)(6.725853664) + \frac{1}{2} 0 (6.725853664)^2 + 0[/tex]
    R2: [tex]\Rightarrow 76.6=(11.3889)(6.725853664)[/tex]

    G2: [tex]76.6=V_0(6.725853664)+\frac{1}{2}a(6.725853664)^2+250[/tex]

    E2: [tex](11.3889)(6.725853664)=V_0(6.725853664)+\frac{1}{2}a(6.725853664)^2+250[/tex] since both equations are equal.

    Now, using both E1 and E2, I solve for [tex]a[/tex] in E2, and then plug it into E1 to obtain [tex]V_0[/tex].

    [tex]a = \frac{(11.3889)(6.725853664)-V_0(6.725853664)-250}{\frac{1}{2}(6.725853664)^2}[/tex]

    [tex](6.3889)(7.278260871) =V_0(7.278026081)+\frac{1}{2}\left ( \frac{(11.3889)(6.725853664)-V_0(6.725853664)-250}{\frac{1}{2}(6.725853664)^2} \right )(7.278026081)^2+250[/tex]

    [tex]V_0=0.7698312727\frac{m}{s}[/tex]

    Now I stopped here because I find it very hard to believe that the car is traveling that slow, so I feel that I've set something up wrong. Have I? Or is that the actual initial velocity, and I'm over analyzing it?

    Thanks!
     
  2. jcsd
  3. May 25, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks like you're over-analyzing things a bit. Your R1/R2 and E1/E2 equations serve no purpose, as far as I can see. (You already found the times.)

    The only equations you need to solve together are G1 and G2. Solving them I get a similar answer for V0 (not exactly the same, but close enough).
     
  4. May 25, 2009 #3
    I went on ahead with what I had and got it wrong. But the second time around (it gives you new values), I concentrated on the green car, and only used the red car to get the time, and I got the correct answer. :surprised

    Thanks for your help! :biggrin:
     
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