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

[DaleSwanson]

Ok, pretty basic Physics-101 question. However, I don't understand why my method for solving isn't working.

Homework Statement

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":
v_s0 = 0 = initial velocity of car one
v_s = ? = final velocity of car one
x_s0 = 0 = starting distance of car one
x_s = ? = final distance of car one
t_s0 = 0 = starting time of car one
t_s = ? = final time of car one
a_s = 2 m/s² = acceleration of car one

car two or "f":
v_f0 = 16.66 m/s = initial velocity of car two
v_f = 16.66 m/s = final velocity of car two
x_f0 = 0 = starting distance of car two
x_f = ? = final distance of car two
t_f0 = 0 = starting time of car two
t_f = ? = final time of car two
a_f = 0 m/s² = acceleration of car two

Homework Equations

kinematic equations
http://wiki.answers.com/Q/What_are_the_kinematic_equations"

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.

t_s = t_f and x_s = x_f.

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

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

Since t_s = t_f, 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 x_f 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?

 

[cepheid]

Since t_s = t_f, 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.

Seems right to me.

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

What other distance formula?

So I guess I have several questions:
1. Why doesn't my method work?

I think your first method does.

2. Are the provided answer correct?

No, not as far as I can tell.

3. How do you actually solve this?

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.

 

[DaleSwanson]

What other distance formula?

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 v_f0 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.