Ok, pretty basic Physics-101 question. However, I don't understand why my method for solving isn't working.(adsbygoogle = window.adsbygoogle || []).push({});

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/s^{2}. 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/s^{2}= 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/s^{2}= acceleration of car two

2. Relevant equations

kinematic equations

http://wiki.answers.com/Q/What_are_the_kinematic_equations" [Broken]

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/s^{2}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}.

[tex]

x_{s} = 0 + 0t_{s0} + \frac{1}{2} * 2 \frac{m}{s^{2}} * t^{2}_{s} = 1 \frac{m}{s^{2}} * t^{2}_{s}

[/tex]

[tex]

x_{f} = 0 + 16.66 \frac{m}{s} * t_{f0} + \frac{1}{2} * 0 * t^{2}_{f} = 16.66 \frac{m}{s} * t_{f}

[/tex]

So I can just set x_{s}equal to x_{f}

[tex]

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

[/tex]

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?

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

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