# Kinematics Car velocity Problem

1. Aug 9, 2015

### hiineko

1. The problem statement, all variables and given/known data
In the figure here, a red car and a green car 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 = 223 m. If the red car has a constant velocity of 20.0 km/h, the cars pass each other at x = 43.1 m. On the other hand, if the red car has a constant velocity of 40.0 km/h, they pass each other atx = 76.9 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

2. Relevant equations

3. The attempt at a solution
eq 1 = 179.9=Vo(7.75)+1/2a(7.75)^2
eq 2 = 146.1=Vo(6.92)+1/2a(6.92)^2

insert 2 to 1

Vo = 146.1-1/2a(6.92)/(6.92)

179.9=(146.1-1/2a(6.92)/(6.92))(7.75)+1/2a(7.75)^2

got a = 5.06 m/s^2
and vo as 3.60

2. Aug 9, 2015

### Staff: Mentor

You could explain where all those numbers in equation 1 and 2 come from, that makes it easier to understand the solution.
Did you plug the result back into those equations to check?

3. Aug 9, 2015

### hiineko

Vr = 20km/h = 5.56m/s x = 43.1
v=d/t -> t=d/v = 43.1/5.56=7.75s
xgreencar - x = 223 - 43.1 = 179.9

179.9=vo(7.75)+1/2a(7.75)^2

same procedure to eq. 2

Sorry but I need a bit help here

4. Aug 9, 2015

### Staff: Mentor

Yes it is right. Well, several units are missing.

Edit: oh right, forgot about the signs.

Last edited: Aug 9, 2015
5. Aug 9, 2015

### insightful

Check signs per original problem statement.

6. Aug 9, 2015

### hiineko

That's one thing that bothers me I need signs how do I know what signs will I use? I just got a positive does it mean I'll use + sign?

7. Aug 9, 2015

### insightful

On the X axis,which way is the green car going?

8. Aug 9, 2015

### hiineko

So 5.06 and 3.61 will be put a - sign on it? If that's what you meant

9. Aug 9, 2015

### insightful

Correct, if velocity and acceleration are in the (defined) negative direction, they will have negative values.

10. Aug 9, 2015

### hiineko

Okay thank you!! Off to my next question in my next post see you there lol kidding Thankyouu