# Constant Acceleration involving two identical objects

1. May 25, 2009

### coco87

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:

$$R_x_0 = 0m$$ 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 $$R_v = 11.3889 \frac{m}{s}$$

2. Relevant 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 $$x=v_0 t + \frac{1}{2} a t^2 + x_0$$

3. 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!

2. May 25, 2009

### 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).

3. May 25, 2009

### coco87

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