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My vehicle needs to pace another vehicle |
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| May2-10, 12:54 PM | #1 |
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My vehicle needs to pace another vehicle
This is not really homework but it is work (fun) and I'm doing it at home. I’m a software guy and haven’t done a lot of physics in awhile. Actually, I’m building a dial display for a car rally computer. It’s basically a clock like display that reads out the odometer. I’m working on the stepper motor control loop for this. I’m trying to predict where the dial needs to move to in order to synch up with the car’s position. In order not to confuse the issue too much, I’ve boiled it down to the following problem. I think I have a solution that works except that the answer depends on the position where the two vehicles merge (t3). Is there a way solve this without having to know t3?
1. The problem statement, all variables and given/known data
2. Relevant equations (I'll try to attached my scanned in PDF but here goes) d2 = 1/2at22 + v0t2 v2 = at2 + v0 d3 = -1/2a(t3-t2)2 + v2(t3-t2) + d2 v3 = v1 = -a(t3-t2) + v2 d3 = v1t3 + d1 I know v0, v1, d1, and a. I want to solve for v2 without depending on t3. 3. The attempt at a solution (see my attached my scanned in PDF) v2 = Sqr((2av1t3 + 2ad1 + v12 + v02)/2) I plotted this in Excel and it seems to work. But it seems to me that I should be able to sove this without using t3 or d3. Thanks in advance. |
| May3-10, 12:16 AM | #2 |
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I figured it out.
For the part of my problem where my vehicle is accelerating to meet the other vehicle, this equation applies: t2 = (v2 - v0) / a For the second part where I'm trying to slow down to pace the other vehicle, this one applies: t3 - t2 = (v2 - v1) / a If I solve for t3 I get: t3 = (2v2 - v1 - v0) / a I my attached PDF work notes, the last equation was this: 2v22 = 2av1t3 + 2ad1 + v12 + v02 If I substitute in my t3 from above, I get this: 0 = 2v22 -4v1v2 + (2v0v1 + v12 - v02 - 2ad1) I used the quadratic formula to solve this for v2. |
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