Complicated acceleration problem?

Question

I want to cover a distance of D, and take exactly T time to do it. I want my initial velocity and final velocity to be zero. So, I'll be accelerating for a period <=.5T, maybe coasting for a while, then deccelerating equally and oppositely to my first acceleration. The easiest case to work with is when there's no coasting distance, and that's when the acceleration is a minimum. As my acceleration increases from trial to trial, there's more distance that I coast. I think. What I want is the distance that I coast expressed as a function of my acceleration, or vice versa. If that's possible in terms of T and D, great. If that turns out to be an unholy mess, it's fine if you let them equal one metre and one second, respectively. Thanks, folks!

Answer 1

During a V-T trial, a person has to travel a total distance (D) in total time taken (T). Another constraint of this journey is that initial and final velocities (at t=0 and t=T) are zero. He can vary acceleration (minimum) to a value where his distance of coasting has to be expressed as a function of “f” in such a way that : “his time of acceleration (t1)max = 0.5T (without coasting), and thereafter while coasting (t1) < 0.5 T (or T/2).

Case 1: No coasting:
For time t1, acceleration f, then an indeterminate cusp, peaking on abrupt boundary (f turning into – f, which you said may be coasting for a while, mechanically very hard to make f into -f? Just view it on the corresponding f-T curve), then for time t2 deceleration - f. both start and end with v=0,
s1 = (1/2) f t1^2, v1= f t1 and
s2 = v1 t2 - (1/2) f t2, 0 = f t1 – f t2 > t1 = t2
s2 = f t1 t2 - (1/2) f t2^2 = f t1^2 - (1/2) f t1^2 = (1/2) f t1^2
s2 = (1/2) f t1^2

Write equations for this case: D = s1+s2 = f t1^2 and T = t1 + t2 = 2t1
so D = f (T/2)^2 or

(f)min = 4D/T^2 ------ (1)
when t1 = (T/2) = 0.5 T time during which he accelerates to f

Case 1: With coasting:

In a particular trial, with coasting, let the person move from rest (initial velocity =0) and travel for time “t1”. Here he decides to coast along with a uniform velocity (let be “V” for another time “t2”. Then he decides to decelerate equally but oppositely ( - f ) for time “t3” before he came to rest (final velocity =0). This can easily we visualized on the V-T graph.
Write equations for case2:
During f: s1 = (1/2) f t1^2, and v1= f t1
During coasting s2 = v1 t2 and v1 = v1 or s2 = f t1 t2 ------ (2)
During - f: s3 = v1 t3 - (1/2) f t3^2 = f t1 t3 - (1/2) f t3^2 ---(3)
And 0 = v1 – f t3 = f t1 –f t3 >>>> t1 = t3 ---(4)
With >> (4), (3) becomes s3 = (1/2) f t1^2

Now D = s1+s2+s3 and T = t1+t2+t3 = 2 t1 + t2 ----- (5)

s2 = let distance coasted = Dc = D – s1 –s3 put values

Dc = s2= distance coasted = D - f t1^2 ----- (6)

From (2) we can put s2 in (6) , we get f t1 t2 = D - f t1^2
From (5) putting t2, it becomes f t1 (T – 2 t1) = D - f t1^2 or
f t1^2 – f T t1 + D = 0 ---- (7)

equation (7) is 2-degree in t1 (time of acceleration) which will give real or imaginary values of t2 or in turn that of Dc. It solution

t1 = [ (T/2) ± (T/2) * sqrt {1 – (4 D/fT^2)} ] as t1 is less than T/2 for coasting we retain the negative sign. For positive root of equation, t2 calculated from other side becomes negative – meaning that such a case would be non-existing. One can see from (1) that factor within square root is function of (f)min.

t1 = [ (T/2) - (T/2) * sqrt {1 – (4 D/fT^2)} ] ---- (8)

(6) becomes Dc = D – f [ (T/2) - (T/2) * sqrt {1 – (4 D/fT^2)} ]^2

Dc = D – (1/4) f T^2 [ 1 - sqrt {1 – (4 D/fT^2)} ]^2 Answer

Dc = D – (1/4) f T^2 [ 1 - sqrt {1 – (fmin/f)} ]^2 from (1)

You please give your opinion on the solution – can use my email.

Answer 2

It sounds as if you're assuming at acceleration will first be a constant A, then 0, then a constant -A. But you didn't actually say that.

Assuming that's true, let S be the time you spend accelerating. S will also be the time you spend decelerating. Let U will be the time you spend coasting. T = 2*S + U.

Let V = your coasting speed.

V = AS

D = 1/2 AS^2 + UV + 1/2 AS^2.

You take it from there (after checking my work).