# My vehicle needs to pace another vehicle

1. May 2, 2010

### charlievictor

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

• I’m in a vehicle traveling at a constant speed of v0. I could be traveling forward (+) or reverse (-).
• Another vehicle is traveling at a constant speed of v1. It also could be traveling forward (+) or backward (-).
• The other vehicle is a distance of d1 from me at the start (with both vehicals moving). It could be ether in the positive or negative direction.
• I want to change my vehicle speed while undergoing a constant acceleration (a) and de-acceleration (-a) to arrive at the other vehicle and pace it at its speed of v1.
• I would like to know what my maximum speed (v2) I would reach during this maneuver.

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.

File size:
792.4 KB
Views:
69
File size:
593.3 KB
Views:
35
File size:
762.3 KB
Views:
60
2. May 3, 2010

### charlievictor

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

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.