[Solved] Travel time

[RusselB]

I'm trying to work out a formula that would calculate the amount of time required to travel distance A given a constant acceleration from 0 to B and deceleration from B to 0.
Eg: How long would it take to travel 1 km with a maximum speed of 50 km/hr?
I just realized that the amount of time spent travelling at the maximum speed can (will?) have a significant affect to the total time, so, for the example go with just hitting maximum speed before starting to decelerate..though if an additional variable C can be worked in, where C is the amount of time spent at maximum speed, that'd be great.

[MrProgrammer]

I'm trying to work out a formula that would calculate the amount of time required to travel distance A given a constant acceleration from 0 to B and deceleration from B to 0.
Eg: How long would it take to travel 1 km with a maximum speed of 50 km/hr?

This is a physics problem, and has nothing to do with Calc. I'd say the topic belongs in General Discussion.

By symmetry, the time for the first ½ kilometer will equal the time for the last ½ kilometer. It is simpler to consider just the first of these where the acceleration is constant and positive. You need information about the equations of motion, and specifically about Constant translational acceleration in a straight line. During the first ½ kilometer, we have r₀=0, r=½, v₀=0, v=50, want to determine t, and do not know a (except that it is a constant). From equation [4] a will be 2500 (v²÷2r) and then from equation [1] t will be v÷a = .02 hour or 1.2 minutes. The time for 1 kilometer will be double that, 2.4 minutes.

Edit: Oops: my earlier numbers for a and t were wrong because I forgot that r=½ here.

I'll presume you can determine the Calc formulas now.

I just realized that the amount of time spent traveling at the maximum speed can (will?) have a significant affect to the total time, so, for the example go with just hitting maximum speed before starting to decelerate..though if an additional variable C can be worked in, where C is the amount of time spent at maximum speed, that'd be great.

The problem is no longer well-defined. Let's call c the fraction of the time spent at maximum acceleration. c can vary from 0 (your initial scenario) to almost 1 (speed climbs very quickly form 0 to to v, continues at v, and decreases very quickly from v to 0). We have three periods: accelerating, constant speed, and decelerating; since this considers both acceleration and deceleration, r=1 for your example. The acceleration a during the first period will be (v²+v²c )÷(r-rc). As c approaches 1, the acceleration increases without bound since r-rc approaches 0. But vehicles have a maximum acceleration, thus c has an upper limit. The minimum acceleration needed is v²÷r, otherwise you cannot obtain speed v in distance r. For an acceleration a between this minimum and the maximum the vehicle allows, the time is (r÷v)+(v÷a).

Edit: Restated acceleration formula adding r; restated time formula replacing c by a. Added example below -- 2019-09-26 18:00 UTC

Example: The minimum acceleration is v²÷r or 2500. Suppose your vehicle can achieve 10000. Accelerating to 50, takes v÷a = .005 hour, and the distance covered is that time multiplied by the average accelerating speed of 25 = .125 kilometer. Decelerating covers the same distance. ¾ kilometer remains, with a constant speed of 50 and thus a time of .015 hour. Adding, .005+.015+.005 = .025 hour or 1.5 minutes, the same value as (r÷v)+(v÷a). Suppose your vehicle can only achieve 2000 (below the minimum). Accelerating to 50, takes v÷a = .025 hour, and the distance covered is that time multiplied by the average accelerating speed of 25 = .625 kilometer. But we have just half of the 1km distance for the acceleration phase.

[RusselB]

Thank you for the information.
I believe I now have the correct basis for creating the actual formulas.