RusselB wrote: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÷2

r) 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.

RusselB wrote: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+

v²

c )÷(

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.

Mr. Programmer

AOO 4.1.5 Build 9789 on MacOS 10.11.6. The locale for any menus or Calc formulas in my posts is English (USA).