Thread: red car / green car physics problem

1. This is a real fun problem I think, but as hard as I try I can't seem to get it correct. This is my latest attempt.

Ch 2 Problem 40 a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x<sub>r</sub> = 0 and the green car is at x<sub>g</sub> = 260 m. If the red car has a constant velocity of 24.0 km/h, the cars pass each other at x = 45.5 m, and if it has a constant velocity of 41.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the acceleration of the green car?

My work:
Thanks to the known velocities and their associated times of the red car, we know that the green car has these positions and associated times: I get these values by dividing the velocity of the red car by the time it takes to get to that destination. Since the green car is at that same destination at that same time, we can apply these values to the green car in reverse, meaning that if we find the red car at a position of 40, we subtract 40 from 260 to get the distance and position of the green car from its origin.

Looking over my available forumlas to use, and assuming that the acceleration of the green car is constant since the problem is under titled "constant acceleration problems," two very useful formulas looks to be these ones: and I believe that is the average acceleration between two ponits, that is the position at time , is the initial position, is the initial velocity, and can be either or both the acceleration at a particular time and the average acceleration. If acceleration is constant, these two values should be the same.

Using the above values, I've worked out the average velocities of the green car from the initial position and and the initial position and  to     to    Knowing these two average velocites, I was able to set up two equations that both represented initial velocity and accelleration as variables. and by substitution:   What have I done wrong?

(I will edit my "change of" 's into mathamatical symbols once I find out how )  2.

3. Let the green cars initial speed and acceleration be u and a respectively.

In the first scenario, the time taken for the red car to travel 0.0455 km is hours. In that time, the green car travels km in the other direction. In the second scenario, the red car takes hours to cover 0.0766 km, in which time the green car travels km going the opposite way. So you have two simultaneous equations in u and a, which you can solve. Note however that your acceleration will be in so you must remember to convert it to :   4. I've got the same position and time as you, but it seems I am still having trouble. In order to eliminate the possibility of accidently entering in a 10 digit number incorrectly, I've deducted a nice little formula that should just spit the answer right out, but it doesn't seem to spit out a correct one.

First of all our relevant green car positions and times in km and hours this time:

[LaTeX ERROR: Compile failed]

Next is our nice equation. The following should be true under three conditions:
1) our initial position and time are both 0
2) acceleration is constant
3) beyond our initial position and time, we have two more pairs of position and time values We can next set up a nearly identical equation for our second pair of position and time Now we can set these two equations equal to eachother and solve for a, the only remaining variable.      This seems real nice. It seems like it is impossible to get an incorrect answer now, but when I plug in my two pairs of values, I get I am assuming that since every value entered into the equation was in km and hours, that this number is in km / h<sup>2</sup>. Using the conversion method Jane mentioned above, I should be dividing this number by 12960. This gives me the final answer of:

83.92978815 m / s<sup>2</sup>

however it seems this answer is still incorrect.  5. Originally Posted by Demen Tolden
I've got the same position and time as you, but it seems I am still having trouble. In order to eliminate the possibility of accidently entering in a 10 digit number incorrectly, I've deducted a nice little formula that should just spit the answer right out, but it doesn't seem to spit out a correct one.

First of all our relevant green car positions and times in km and hours this time:

[LaTeX ERROR: Compile failed]

Next is our nice equation. The following should be true under three conditions:
1) our initial position and time are both 0
2) acceleration is constant
3) beyond our initial position and time, we have two more pairs of position and time values We can next set up a nearly identical equation for our second pair of position and time Now we can set these two equations equal to eachother and solve for a, the only remaining variable.      This seems real nice. It seems like it is impossible to get an incorrect answer now, but when I plug in my two pairs of values, I get I am assuming that since every value entered into the equation was in km and hours, that this number is in km / h<sup>2</sup>. Using the conversion method Jane mentioned above, I should be dividing this number by 12960. This gives me the final answer of:

83.92978815 m / s<sup>2</sup>

however it seems this answer is still incorrect.
Wouldn't you need some kind of restraint in the cars ability to positively accelerate? To determine the maximum positive acceleration the car could achieve?

If the car had no top speed, then the car could conceivably positively accelerate from zero kilometers an hour to 216 kilometers a nanosecond. I wish these questions had more ties to everyday calculations.

I would bet the train system uses something like this. And at each check point the speed of the train is updated.

Sincerely,

William McCormick  6. Originally Posted by William McCormick
Wouldn't you need some kind of restraint in the cars ability to positively accelerate? To determine the maximum positive acceleration the car could achieve?
In this problem, acceleration is constant and does not fluctuate. There is no range of values for a, just a single answer. Originally Posted by William McCormick
If the car had no top speed, then the car could conceivably positively accelerate from zero kilometers an hour to 216 kilometers a nanosecond.
If the constant acceleration was that fast, our position and times would be very different. If you are thinking of the car starting with the velocity of 0, achieving a velocity of 216 km in a nanosecond, and then no longer increasing its velocity, this would be two changes in acceleration. I would rather not work out what the acceleration would be for the first change in velocity - its a large positive number, but after that the acceleration becomes 0. This is therefore not a constant acceleration. Originally Posted by William McCormick
I would bet the train system uses something like this. And at each check point the speed of the train is updated
Are you talking about a local train system or a mathematical model?  7. Originally Posted by Demen Tolden Originally Posted by William McCormick
Wouldn't you need some kind of restraint in the cars ability to positively accelerate? To determine the maximum positive acceleration the car could achieve?
In this problem, acceleration is constant and does not fluctuate. There is no range of values for a, just a single answer. Originally Posted by William McCormick
If the car had no top speed, then the car could conceivably positively accelerate from zero kilometers an hour to 216 kilometers a nanosecond.
If the constant acceleration was that fast, our position and times would be very different. If you are thinking of the car starting with the velocity of 0, achieving a velocity of 216 km in a nanosecond, and then no longer increasing its velocity, this would be two changes in acceleration. I would rather not work out what the acceleration would be for the first change in velocity - its a large positive number, but after that the acceleration becomes 0. This is therefore not a constant acceleration. Originally Posted by William McCormick
I would bet the train system uses something like this. And at each check point the speed of the train is updated
Are you talking about a local train system or a mathematical model?

I am talking about a local train system. I was working in the LIRR Manhattan Central control office. And I was looking at the system control board. They may have updated since then.

So today they may have a constant up to the second location for the trains now with all the high tech stuff.

But at the time I was there it looked like they just intermittently kept track of them. And manually verified where each one was at specific points.

Sincerely,

William McCormick  Posting Permissions
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