Finding Right car Acceleration
Most recent answer: 05/24/2011
Q:
Hi,
I have a car:
- at a distance D away from destination
- current velocity V0
- time to go t : the time period I want the car to spend until reaching destination
I discretize time into seconds. At every every second, I want to find suitable acceleration (or deceleration)
I apply this equation
D = V0 t + (1/2)at^2
I came up with something strange:
If t is sufficiently small, then the car arrives exactly at t once. (case 1)
If I set t too high, the car arrives at the destination early (much less than t), go past destination for some time, then go backwards
to reach destination at given t. (case 2)
My question: is there any constraint for t to always yield case 1, preventing case 2 happening.
To be specific, I put an example in the spreadsheet here:
https://spreadsheets.google.com/spreadsheet/ccc?key=0AtCnYAzIH013dElQdlpRQ3djTGYzRHB1V1RWVjc3ZEE&hl=en_US&authkey=CIyymf8E
Table 2 is where the case 2 happened. Round-up error is ignored.
Thanks much
- Tung Le (age 28)
Randwick, NSW, Australia
- Tung Le (age 28)
Randwick, NSW, Australia
A:
I'm puzzled as to why you would discretize this equation. It seems that you're looking for an equation with fixed a. You can just directly solve:
a=2*(D-V0t)/t2. That gets you to D after time t. It's a unique solution that exists for any t > 0.
If t is short, D/V0 > t , so you have to speed up (a > 0) to arrive at D on time. We're assuming fixed a , so you keep speeding up and never return.
If t is long, we need a < 0. When you reach D at time t, your velocity is
VF = V0+at = 2D/t-V0.
For D/V0 < t < 2D/V0 , that gives VF > 0 when you reach D at time t. But with a < 0, after you go past D you turn back and recross it from the other side. For t > 2D/V0, we get VF < 0. That means that at the time we solved for, we're crossing D going backwards. It's got to be the second time we cross D.
So there are three ranges:
t < D/V0 , accelerate, cross D once.
D/V0 < t < 2D/V0, decelerate, cross D again backwards after t.
t > 2D/V0, decelerate, cross D once before t, then going backwards at t.
There are some special crossovers:
t = D/V0 , a=0, simple.
t = 2D/V0 just reach D with VF = 0, turn back without crossing.
Mike W.
a=2*(D-V0t)/t2. That gets you to D after time t. It's a unique solution that exists for any t > 0.
If t is short, D/V0 > t , so you have to speed up (a > 0) to arrive at D on time. We're assuming fixed a , so you keep speeding up and never return.
If t is long, we need a < 0. When you reach D at time t, your velocity is
VF = V0+at = 2D/t-V0.
For D/V0 < t < 2D/V0 , that gives VF > 0 when you reach D at time t. But with a < 0, after you go past D you turn back and recross it from the other side. For t > 2D/V0, we get VF < 0. That means that at the time we solved for, we're crossing D going backwards. It's got to be the second time we cross D.
So there are three ranges:
t < D/V0 , accelerate, cross D once.
D/V0 < t < 2D/V0, decelerate, cross D again backwards after t.
t > 2D/V0, decelerate, cross D once before t, then going backwards at t.
There are some special crossovers:
t = D/V0 , a=0, simple.
t = 2D/V0 just reach D with VF = 0, turn back without crossing.
Mike W.
(published on 05/24/2011)