Constant Acceleration involving two identical objects

[coco87]

Homework Statement

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:

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

Homework Equations

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

The Attempt at a Solution

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

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

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

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

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

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

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

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

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

(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

V_0=0.7698312727\frac{m}{s}

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!

 

[Doc Al]

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).

 

[coco87]

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).

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.

Thanks for your help!