# Kinematic Equations - Why can't I solve like this?

1. Sep 16, 2010

### DaleSwanson

Ok, pretty basic Physics-101 question. However, I don't understand why my method for solving isn't working.
1. The problem statement, all variables and given/known data
Two cars. Car one is at rest, while the car two is traveling at a constant 16.66 m/s (given as 60 km/hr). At the moment car two passes car one, car one begins accelerating at a constant 2 m/s2. Eventually car two overtakes car one. At that point what is the distance, time, and speed of car one?

Car one = s for sports car, car two = f for friends car.
Here are all the relevant variables:

car one or "s":
vs0 = 0 = initial velocity of car one
vs = ? = final velocity of car one
xs0 = 0 = starting distance of car one
xs = ? = final distance of car one
ts0 = 0 = starting time of car one
ts = ? = final time of car one
as = 2 m/s2 = acceleration of car one

car two or "f":
vf0 = 16.66 m/s = initial velocity of car two
vf = 16.66 m/s = final velocity of car two
xf0 = 0 = starting distance of car two
xf = ? = final distance of car two
tf0 = 0 = starting time of car two
tf = ? = final time of car two
af = 0 m/s2 = acceleration of car two

2. Relevant equations
kinematic equations

3. The attempt at a solution
So, just to get it out of the way the answers from the back of the book are: speed=33 m/s, distance=190 m, and time=11 s. I suspect these might be wrong. If something accelerates at 2 m/s2 for 11 seconds it will be going 22 m/s, not 33 m/s.

I figured that since we are trying to find the point at which the cars meet again then the time, and distance at the end must be equal.

ts = tf and xs = xf.

$$x_{s} = 0 + 0t_{s0} + \frac{1}{2} * 2 \frac{m}{s^{2}} * t^{2}_{s} = 1 \frac{m}{s^{2}} * t^{2}_{s}$$
$$x_{f} = 0 + 16.66 \frac{m}{s} * t_{f0} + \frac{1}{2} * 0 * t^{2}_{f} = 16.66 \frac{m}{s} * t_{f}$$
So I can just set xs equal to xf

$$1 \frac{m}{s^{2}} * t^{2}_{s} = 16.66 \frac{m}{s} * t_{f}$$

Since ts = tf, I should be able to change it to use all one or the other. Then solve for t. However, doing that gives t = 16.66 s.

I tried the same general idea except replacing xf with the other distance formula. That ended up giving me t = 8.33 s.

So I guess I have several questions:
1. Why doesn't my method work?
2. Are the provided answer correct?
3. How do you actually solve this?

Last edited by a moderator: May 4, 2017
2. Sep 16, 2010

### cepheid

Staff Emeritus
Seems right to me.

What other distance formula?

I think your first method does.

No, not as far as I can tell.

Since the car's positions are equal when one overtakes the other, you set the expressions for the positions vs. time of each car to be equal to each other.

3. Sep 16, 2010

### DaleSwanson

Sorry I was trying to cut down on vertical space so I just referenced the formulas in the link.
The distance formula I used above was:
x = x0 + v0t + .5at2
The second distance forumla I mentioned is:
x = x0 + .5(v0 + v)t

The second equation uses v, which is unknown for car one, but known for car two. Since the two are both solved for x, I tried using the first one with the car one variables in it. Then set it equal to the second with the car two variables plugged in. Earlier I got 8 m/s using that method. However, just now, I realize that is because I used 0 for vf0 instead of 16 m/s.

So I suppose the book is wrong. This is the second error in only a handful of problems we've done. I'm glad I didn't buy the book for \$150. If my method is correct then I have no problems figuring out the rest. Thanks for your help.