Reader Mike writes in with an interesting problem to work out. It runs thus:
You're in your car parked on the side of the road when you see your friend a distance d away, driving toward you at velocity v. You want to talk to him through your window, so what constant acceleration should you pick so that you're also traveling at v at the moment he reaches you?
The question as originally posed asked for the acceleration necessary to accomplish this in the minimum time, but in fact only one acceleration will cause him to catch up with you right as you both have the same speed. However long it takes is however long it takes. There's a reason for this. The graph of your position vs. time as an accelerating body is a parabola, and parabolas are completely specified by three points on the curve. In this case one is fixed by your starting position, one is fixed by your starting velocity (0, of course), and one is fixed by the requirement of matching speeds where your position intersects your friend's. There's nothing left to adjust to further optimize the problem.
Knowing that, let's look at the three possibilities. The x-axis is time, the y-axis is distance (with the zero set as your starting position). Keep in mind that in this kind of graph you can think of speed as being represented by the slope of the curve at that point. Units are arbitrary.
1) Too much acceleration: your friend never catches up.
2) Too little acceleration. your friend passes you up at speed, and then you pass him up at speed:
3) Just right. your friend catches up right as you both have the same speed:
So let's figure out what your acceleration needs to be. Give yourself the name "A" and your friend the name "B". The equations describing your positions are:
Here I'm using "d" to denote how far away your friend is when you begin the acceleration. The temptation is to set them equal to find the time of intersection and solve for t, but that would turn out to be the long way around. Let's differentiate those equations to find the velocities instead. We have:
Easy. Now let's impose the condition that the velocities have to be equal when the positions are the same. We don't know when that happens, but whatever the particular time happens to be, let's call it T. Since the velocities are equal, we can set the velocity equations equal to each other at that time. Solve that for T and we get:
Now at that time T we've said by definition that the two cars are at the same location. Therefore we can set those first two equations equal to each other, at that time T. So substitute our newly-found T into the equations. This gives:
Solve for the acceleration:
Which is a nice and simple final result. Accelerate at that velocity and when the two cars meet, they'll be moving at the same speed. Actually doing the calculation in time might be tricky, so in practice I think I'd just estimate and not be so picky about keeping a constant acceleration. Probably the experimentalist in me.
Interestingly, since always v_A^2 = 2ax_a this says, at the meeting position D, v^2 = 2aD and comparison with the acceleration found above shows that D = d.
Another way to do it is with a speed time diagram, using the fact that the areas of the curves are the distances traveled. One graph is horizontal line, and the other is a straight line through the origin. It's also clear from this diagram that the total distance traveled by the constant speed car is twice the distance traveled by the accelerating car.
OMB, you are right on the money. That is the high-school way of doing things with velocity/time graphs.
After the kids have had some calc, we usually go back and give it the "Built on Facts" treatment so they can live with non-constant accelerations.
Excellent job, by the way, in your use of graphs.
Another option would be to pull out directly in front of your friend's car, and let the collision accelerate your car and decelerate his car so that your velocities match.
To meet the criteria of the original problem, the rear window of your car still counts as a window, and there's a good chance the impact would shatter it to allow for unimpeded conversation.