# Basic Kinematics Question

1. Oct 29, 2006

### prace

Hello,

I am having an issue with this simple kinematics problem. I have the question visualized, well, at least I think I do. Here is the question:

Two cars are traveling along a straight road. Car A maintains a constand speed of 80 km/h; car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. How much farther will car A travel before it is overtaken by car B?

So to start off, I graphed the problem and came up with this:

http://album6.snapandshare.com/3936/45466/853951.jpg [Broken]
where series 1 is car A and series 2 is car B.

I also know that this has to do with position and time, so I am most likely going to use one of my kinematical equations that deals with position and time, so I chose the equation: $$x_{f}=x_{i}+v_{0}t+\frac{1}{2} a t^2$$ because I know that the accelleration is constant and is equal to 0 since the cars are not speeding up. I also know that the final positions have to be equal, but how do I equate the final positions of the two cars? Or is this even the right direction for me to solve the problem?

I mean, I could just graph the two lines, find the equations, and then find the intersection of those two lines, but I would like to learn how to do it using the kinematical equations.

Thanks

Last edited by a moderator: May 2, 2017
2. Oct 29, 2006

### BishopUser

You are on the right track. By finding the intersection point on that graph you have calculated how much time passes before the two cars come together. Now you must just realize that the problem is just asking how far does car A travel during this time.

The 2 lines that you graphed are the exact same thing as the kinematic equations btw. You probably graphed something like y = 80x vs y = 110x -45.

Now look at the kinematic equation

xf = xi + vit + 1/2at^2
you stated that acceleration is 0 so its just
xf = xi + vit

final position is just the independent variable (y). xi is the initial position (y-intercept). vi is the initial velocity in the problem (coefficient of x), and T is the variable (x).

Once you set up the 2 equations you just find the intersection point by setting the 2 equations equal to each other. then you can solve for the time

Last edited: Oct 29, 2006
3. Oct 29, 2006

### prace

So that's it? Just do it mathmatically and it is fine? Awesome, I solved for the intersection time to be .6 hours and then plugged that back into my equation for Car A and got a distance of 4.8 km. Does that sound good?

Thank you for your help Bishop!!

4. May 28, 2010

### bdblankster

I'm currently working on this problem and I agree with that I need to find equations for both cars and then set them equal to each other. However, I did not get .6 hrs for the time. I ended up with 1.5 hours.