I asked this question in the Physics section, and didn't get much help. I want to move a metre in exactly one second. I also want to start stationary and end that way too.
As you all know, distance=velocity*time for coasting, and while accelerating, distance=.5*acceleration*time^2
I would like to acclerate for a period, perhaps coast for a bit, and then deccelerate equally and oppositely, so my accelerations will be a, 0, then -a for the three stages of this movement. What my real question is, and I apologize if I've beat around the bush, is what is the relationship between acceleration and distance coasted. If you can express one as a function of the other, great, and it would be even better if you could solve in terms of T and D, intstead of using 1 second and 1 metre for those values.
2007-02-10 01:24:56 · 3 answers · asked by Mehoo 3 in Science & Mathematics ➔ Mathematics
0.5 * a * (t1)^2 + a * t1 * t2 + 0.5 * a * (t1)^2 = D
t1 * 2 + t2 = T from which you get t2 and put it in the first equation
from the first equation you get a as a function of D, T, t1
distance coasted is d = a * t1 * t2 = a * t1 * (T - 2 * t1) from which you get t1
you use t1 in a as a function of D, T, t1 and you get a as a function of D, T, d
2007-02-10 01:35:27 · answer #1 · answered by Ioana 2 · 0⤊ 0⤋
If you know how long you accelerate for before you coast, a x T gives you the speed you coast at and if you coast for time Tc, the distance you coast is simply a x T x Tc
2007-02-10 09:31:25 · answer #2 · answered by Gene 7 · 0⤊ 0⤋
I answered your question in other question just 12 hours back in physics section . I sent you an email also.
In my expression you can put the value and get the answer on distance and time.
2007-02-10 10:43:21 · answer #3 · answered by anil bakshi 7 · 0⤊ 0⤋