Acceleraton and Deceleration Problem



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Topic: Science > Physics
User: "Kylie"
Date: 24 Feb 2006 03:58:55 AM
Object: Acceleraton and Deceleration Problem
Hi everyone,
How would I go about solving this problem?
A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.
Maximum speed allowed is 25 km/h.
What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.
Any help is much appreciated.
Kylie
.

User: "OG"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 05:54:52 PM
"Kylie" <nojunkmail@please.com> wrote in message
news:43fed8fd$0$17409$afc38c87@news.optusnet.com.au...

Hi everyone,

How would I go about solving this problem?


A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely
(FinalVelocity = 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.


Any help is much appreciated.

Draw a velocity / time graph.
There will be a triangular sloped area at the start of the graph as the
speed increases and a triangular sloped area at the end of the graph. In
between will be a flat zone which represents the max allowed speed of 25
km/s
The shape of the graph is a trapezium, where the sloped areas are the
acceleration / deceleration zones.
*The area under the graph is the distance travelled.*
If you are asking for the least time to travel 25km, I think you'll see that
the sloped areas at start and end have to be as short (i.e. steep gradient)
as possible (very high acceleration / deceleration).
An analytic solution to the problem is then simply a matter of adding areas.
.
User: "OG"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 05:56:18 PM
"OG" <owen@gwynnefamily.org.uk> wrote in message
news:469kmdFa7uu2U1@individual.net...
Dmn
Missed PD's excellent post!
.


User: "srp"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 11:31:08 AM
Kylie a écrit :

Hi everyone,

How would I go about solving this problem?

A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.

Any help is much appreciated.

Max distance 5000 m
Max velocity 25 km/h = 6.95 m/s
Moving horizontally (orthogonal to gravitation)
You need to accelerate for the first half of the distance and
to decelerate for the second half
so, acceleration
a= 2v^2/d = ((6.95)^2)/2500 = 0.019321 m/s^2
Deceleration is the same.
Your vehicle will be there in the shortest time possible, given
a max velocity of 25 km/h.
André Michaud
.
User: "Mike"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 01:00:40 PM
srp wrote:

Kylie a =E9crit :

Hi everyone,

How would I go about solving this problem?

A vehicle at stationary state (InitialVelocity =3D 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelo=

city

=3D 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.

Any help is much appreciated.


Max distance 5000 m
Max velocity 25 km/h =3D 6.95 m/s
Moving horizontally (orthogonal to gravitation)

You need to accelerate for the first half of the distance and
to decelerate for the second half

so, acceleration

a=3D 2v^2/d =3D ((6.95)^2)/2500 =3D 0.019321 m/s^2

Deceleration is the same.

Your vehicle will be there in the shortest time possible, given
a max velocity of 25 km/h.

Andr=E9 Michaud

If we are talking about a car things are not that simple. Shortest time
means utilizing maximum car power. This means that as the velocity
increases the ability to maintain the same acceleration decreases.
Remember that power is tforce times velocity. Thus, the only possible
velocity profile (v versus t) for minimum time is a parabolic one. That
is, as your speed increases, the slope of the profile must decrease.
This is an optimization problem. Find a function (in this case a
parabola) that minimizes an objective function given constraints:
min v(t) wrt a(t) given that v0=3D0, vf =3D 0, s0 =3D 0, sf =3D d and P(t)
Unless the power function P(t) of the car is known the problem cannot
be solved.
The triangular velocity profile you suggested minimizes time in the
case of constant acceleration (bang), like in the case of a rocket. In
practical applications of motion control using servo motors, a
trapezoidal profile is used to minimize time with maximum power until
max speed is reached, then constant speed followed by decelleration.
Mike
.
User: "CWatters"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 07:14:07 PM
"Mike" <eleatis@yahoo.gr> wrote in message
news:1140807640.611590.170030@e56g2000cwe.googlegroups.com...

If we are talking about a car things are not that simple.

Make the problem harder why don't you.
.
User: "Mike"

Title: Re: Acceleraton and Deceleration Problem 25 Feb 2006 04:12:48 AM
CWatters wrote:

"Mike" <eleatis@yahoo.gr> wrote in message
news:1140807640.611590.170030@e56g2000cwe.googlegroups.com...

If we are talking about a car things are not that simple.


Make the problem harder why don't you.

Idealizations and simplifications is a tool only for those who have a
deep understanding of physics. Simplifying problems that refer to real
situations in student problems has an adverse effect and leads to a
distortion of reality which is hard to change later. Yet, this is what
happens all the time. One result is the great number of cranks around.
These people supperimpose simplifications and idealizations on reality
to come up with bizzare thought experiments. Look for example at the
"Final Theory" book advertized in these groups. I don't think the
person is responsible for his distorted views. IT is his educations and
the fact that his teachers presented the simplified level he was taught
as a deep level of understanding without mentioning the caveats and
pitfalls. Now he disputes the dot product in the Work function and
thinks this is a conspiracy:)
Science and Technology has advanced tremendously and we cannot effort
to still give students distorted views of reality. At least
alternatives must be mentioned as early as in Elementary school. Hiow
many of you got the impression from your teachers that Newton believed
in mutual attraction and his gravity law was the end of the story?
Mike
.


User: "srp"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 03:18:41 PM
Mike a écrit :

srp wrote:

Kylie a écrit :

Hi everyone,

How would I go about solving this problem?

A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.

Any help is much appreciated.


Max distance 5000 m
Max velocity 25 km/h = 6.95 m/s
Moving horizontally (orthogonal to gravitation)

You need to accelerate for the first half of the distance and
to decelerate for the second half

so, acceleration

a= 2v^2/d = ((6.95)^2)/2500 = 0.019321 m/s^2

Deceleration is the same.

Your vehicle will be there in the shortest time possible, given
a max velocity of 25 km/h.

André Michaud



If we are talking about a car things are not that simple. Shortest time
means utilizing maximum car power. This means that as the velocity
increases the ability to maintain the same acceleration decreases.
Remember that power is tforce times velocity. Thus, the only possible
velocity profile (v versus t) for minimum time is a parabolic one. That
is, as your speed increases, the slope of the profile must decrease.

This is an optimization problem. Find a function (in this case a
parabola) that minimizes an objective function given constraints:

min v(t) wrt a(t) given that v0=0, vf = 0, s0 = 0, sf = d and P(t)

Unless the power function P(t) of the car is known the problem cannot
be solved.

The triangular velocity profile you suggested minimizes time in the
case of constant acceleration (bang), like in the case of a rocket. In
practical applications of motion control using servo motors, a
trapezoidal profile is used to minimize time with maximum power until
max speed is reached, then constant speed followed by decelleration.

Mike

I see. I agree
André Michaud
.



User: "Nicholas Sherlock"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 04:10:07 AM
Kylie wrote:

A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.

It seems like there should be another bound on this problem. Surely the
vehicle would reach point_2 in the shortest time by accelerating at an
infinite rate to 25 km/hr, and doing the same to stop at the end of the
journey.
Cheers,
Nicholas Sherlock
.

User: "PD"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 12:49:04 PM
Kylie wrote:

Hi everyone,

How would I go about solving this problem?


A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.


Any help is much appreciated.

Kylie

Here's one way to look at it.
Make a plot of velocity vs time for the car. An acceleration will be an
upwardly sloping line on this plot. A period at constant speed will be
a horizontal line on this plot. You know the highest horizontal line
you can draw on this plot is at 25 km/h. A deceleration will be a
downwardly sloping line on this plot. So, you're going to draw a
trapezoid starting at t=0 and v=0 that slopes up, goes horizontal at
some point, then slopes down again at some later point.
The area in this trapezoid is the distance traveled, and you know the
area is 5 km.
So the question is, what shape trapezoid do I make so that the area
stays the same but the base of the trapezoid is as narrow as possible
(shortest time between start and end)?
Obviously, the taller you make the trapezoid, the narrower horizontally
it can be.
But it can only be as tall as 25 km/h, because that's the maximum
speed.
I can make the trapezoid even narrow and keep the same area if I make
the sloping sides as steep as possible. If I know how steep the sloping
sides can be (maximum acceleration, maximum deceleration), then I've
got my answer. If I'm not given maximum accelerations, then the
narrowest trapezoid I can make is actually a rectangle of height 25
km/h and width 1/5 h. But that would be ramming the car from behind to
get it up to 25 km/h immediately, and stopping it with a brick wall at
the end.
PD
.
User: "tadchem"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 03:41:19 PM
Nicely put, PD.
Tom Davidson
Richmond, VA
.


User: "arvee"

Title: Re: Acceleraton and Deceleration Problem 25 Feb 2006 06:16:06 AM
Kylie wrote:

Hi everyone,

How would I go about solving this problem?


A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

There must also be a bound on acceleration and deceleration, since
otherwise the solution would be pretty obvious. Anyway, it is a
standard optimal control problem, but with a state-space constraint. In
this case, you have a two-dimensional state space with x1 =
x-coordinate and x2 = velocity. The equations of motion are dx1/dt = x2
and dx2/dt = u, where u is the acceleration. You want to minimize T
such that x1(T) = 5 and x2(T) = 0. There are constraints -U <= u < U,
where U = acceleration/deceleration bound (could make these different,
too) and x2(t) <= 25 for all t. You could use the Pontryagin maximum
principle, modified to handle state-space constraints. However, in this
case the answer is quite obvious (and you could use the maximum
principle to prove it): just accelerate as fast as possible to the 25
kph limit, hold at that velocity for a time, then very carefully time
your maximum deceleration so as to come to rest at 5 km distance. Of
course, if you are not allowed to accelerate quickly enough, you might
never reach 25 kph; in this case, you would accelerate maximally for
the first half trip then decelerate maximally for the second half
(assuming max acceleration = max deceleration).
R.G. Vickson


What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.


Any help is much appreciated.

Kylie

.
User: "CWatters"

Title: Re: Acceleraton and Deceleration Problem 25 Feb 2006 11:26:26 AM
"arvee" <C6L1V@shaw.ca> wrote in message
news:1140869766.057544.121530@e56g2000cwe.googlegroups.com...

Kylie wrote:

Hi everyone,

How would I go about solving this problem?


A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely

(FinalVelocity

= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.


There must also be a bound on acceleration and deceleration, since
otherwise the solution would be pretty obvious.

The OP already said there was a max acceleration specified in the problem.
.


User: "Jasen Betts"

Title: Re: Acceleraton and Deceleration Problem 25 Feb 2006 02:32:47 AM
On 2006-02-24, Kylie <nojunkmail@please.com> wrote:

Hi everyone,

How would I go about solving this problem?


A vehicle at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2 (5 Km apart). It MUST stop completely (FinalVelocity
= 0) after travelling 5 km and reaching to point_2.

Maximum speed allowed is 25 km/h.

What would be required acceleration (to start moving from point_1) and
required deceleration (to stop moving at point_2) in order that vehicle
reach to point_2 in the shortest time.

infinite acceleration, vehicle immediately starts at 25 Km/h
then at the other end infinite decelleration. (vehicle immediately stops)
Don't use the vehicle for carrying goods or people.
Bye.
Jasen
.

User: "tadchem"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 04:24:47 AM
Scenario:
You accelerate at 1 g (9.8 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 1 g to rest, stopping exactly where required.
You maintain maximum allowable speed most of the way.
You accelerate at 10 g (98 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 10 g to rest, stopping exactly where required.
You maintain maximum allowable speed even longer.
If you could accelerate *instantaneously* (infinite m/s/s) to 25 km/h,
maintain that and stop instantly when you got to the other end (ouch!)
you would maintain *maximum speed* over the distance for maximum time,
and thus achieve minimum transit time of 12 minutes (= 5 km / 25 km/h =
0.2 hour).
See where this is going?
The required acceleration for minimum time is infinite, and would be so
*whatever* the maximum speed is.
Tom Davidson
Richmond, VA
.
User: "Kylie"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 04:33:08 AM
But I can't accelerate *instantaneously* (infinite)!
The vehicle can reach to a pre-defined maximum acceleration which is also
know.
Thanks,
Kylie
"tadchem" <tadchem@comcast.net> wrote in message
news:1140776687.200876.5450@i40g2000cwc.googlegroups.com...

Scenario:

You accelerate at 1 g (9.8 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 1 g to rest, stopping exactly where required.
You maintain maximum allowable speed most of the way.

You accelerate at 10 g (98 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 10 g to rest, stopping exactly where required.
You maintain maximum allowable speed even longer.

If you could accelerate *instantaneously* (infinite m/s/s) to 25 km/h,
maintain that and stop instantly when you got to the other end (ouch!)
you would maintain *maximum speed* over the distance for maximum time,
and thus achieve minimum transit time of 12 minutes (= 5 km / 25 km/h =
0.2 hour).

See where this is going?

The required acceleration for minimum time is infinite, and would be so
*whatever* the maximum speed is.

Tom Davidson
Richmond, VA

.
User: "CWatters"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 06:13:41 AM
"Kylie" <nojunkmail@please.com> wrote in message
news:43fee103$0$17406$afc38c87@news.optusnet.com.au...

But I can't accelerate *instantaneously* (infinite)!
The vehicle can reach to a pre-defined maximum acceleration which is also
know.

In that case it should accelerate at that maximium pre-defined rate until it
reaches 25 km/h. Stay at that rate for awhile and decelerate at the max
pre-defined rate.
Look up the standard equations of motion and write down versions for each
phase.
With luck you will have as many equations as unknowns.
Then substitute
d1+d2+d3 = 5km
and whatever value you have for max acceleration and distance.
If the max rate of acceleration is the same as the max rate of deceleration
then the problem simplifies as d1=d3, t1=t3 etc
Check that d2 and t2 aren't <0 ....In which case the vehicle never reaches
max speed of 25km/h before it has to start slowing down in time....but I
hope teacher isn't that mean :-)
.
User: "Mike Yarwood"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 08:29:27 AM
"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:VLCLf.276674$vn3.8835172@phobos.telenet-ops.be...


"Kylie" <nojunkmail@please.com> wrote in message
news:43fee103$0$17406$afc38c87@news.optusnet.com.au...

But I can't accelerate *instantaneously* (infinite)!
The vehicle can reach to a pre-defined maximum acceleration which is also
know.


In that case it should accelerate at that maximium pre-defined rate until
it
reaches 25 km/h. Stay at that rate for awhile and decelerate at the max
pre-defined rate.

Look up the standard equations of motion and write down versions for each
phase.

With luck you will have as many equations as unknowns.

Then substitute

d1+d2+d3 = 5km

and whatever value you have for max acceleration and distance.

If the max rate of acceleration is the same as the max rate of
deceleration
then the problem simplifies as d1=d3, t1=t3 etc

Check that d2 and t2 aren't <0 ....In which case the vehicle never reaches
max speed of 25km/h before it has to start slowing down in time....but I
hope teacher isn't that mean :-)

Hi! What Colin said but said a dfifferent way ( just for the fun of it) :
work out how far you need to go at maximum acceleration to reach 25/3.6
metres/second.
work out what distance you need to decelerate from 25km/hr to zero using
your maximum deceleration, add these two distances to check they are < 5000
metres. If they are < 5km then the remaining distance is covered at 25km/hr
so you get the time for the constant velocity part, now go back and find out
how long it took in the deceleration and acceleration phases - total all
your times.
If they are > 5km then you've need to solve for a*s1=d*s2 where a and d are
your acceleration and deceleration maxima and so that s1+s2=5km , then,
knowing a,d,s1 and s2 you can calculate the time.
This is a completely different problem from the one that you first posted
though.
Best of luck - Mike
.


User: "Tony"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 06:51:03 AM
Kylie, you're going to have to make some guesses to slove this problem.
If the vehicle is driven by wheels, and you assume a coefficient of
friction (1 is an interesting choice and might be close to reality for
tires on a road) you can assume a 1 g acceleration and decceleration
rate. If it's jet or rocket powered you'll need to make some other kind
of assumptions.
.


User: ""

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 08:12:19 PM
Don1 wrote:

tadchem wrote:

Scenario:

You accelerate at 1 g (9.8 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 1 g to rest, stopping exactly where required.
You maintain maximum allowable speed most of the way.

You accelerate at 10 g (98 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 10 g to rest, stopping exactly where required.
You maintain maximum allowable speed even longer.

If you could accelerate *instantaneously* (infinite m/s/s) to 25 km/h,
maintain that and stop instantly when you got to the other end (ouch!)
you would maintain *maximum speed* over the distance for maximum time,
and thus achieve minimum transit time of 12 minutes (= 5 km / 25 km/h =
0.2 hour).

See where this is going?

The required acceleration for minimum time is infinite, and would be so
*whatever* the maximum speed is.

Tom Davidson
Richmond, VA


Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.

If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); not a=vi-vt/t; which is called
"instantaneous acceleration"; of which it is not.

Then you could have taught each other how to solve these sorts of
things.

.


User: "Kylie"

Title: Re: Acceleraton and Deceleration Problem 24 Feb 2006 06:38:50 PM
Thanks everyone for wonderful responses!
Now I want to design a new problem:
This is actually a Robotic problem:
A robot at stationary state (InitialVelocity = 0) starts moving from point_1
toward point_2.
It MUST reach to FinalVelocity of V at point_2.
We know ONLY the maximum velocity that robot can reach and the distance
between point_1 and point_2.
We have methods that can SET the following parameters of the robot:
AcceleratorPosition (i.e. Gas Pedal)
BrakePosition (i.e. Brake Pedal)
Of course the range of AcceleratorPosition and BrakePosition is known as
well.
At any time we can also GET the following information back from robot:
DistanceToPoint_2
Velocity
Acceleration
I was thinking of having a loop in the software, so that start increasing
AcceleratorPosition and every time GetVelocity until it reaches to the
maximum velocity and then stop increasing AcceleratorPosition so that
velocity remain constant.
In every run in the loop we also check to see what is the distance to
point_2. If this distance become less than "S", then SettAcceleratorPosition
to zero and start increasing BrakePosition until it reaches to point_2 with
the desired velocity of V.
Now the problem is how do I calculate "S"? Too many unknown! isn't it?
Any other idea?
Thanks in advance.
Kylie
.
User: "Mike Yarwood"

Title: Re: Acceleraton and Deceleration Problem 25 Feb 2006 03:13:45 AM
"Kylie" <nojunkmail@please.com> wrote in message
news:43ffa73a$0$15125$afc38c87@news.optusnet.com.au...

Thanks everyone for wonderful responses!
Now I want to design a new problem:

This is actually a Robotic problem:

A robot at stationary state (InitialVelocity = 0) starts moving from
point_1 toward point_2.
It MUST reach to FinalVelocity of V at point_2.
We know ONLY the maximum velocity that robot can reach and the distance
between point_1 and point_2.
We have methods that can SET the following parameters of the robot:

AcceleratorPosition (i.e. Gas Pedal)
BrakePosition (i.e. Brake Pedal)

Of course the range of AcceleratorPosition and BrakePosition is known as
well.

At any time we can also GET the following information back from robot:

DistanceToPoint_2
Velocity
Acceleration

I was thinking of having a loop in the software, so that start increasing
AcceleratorPosition and every time GetVelocity until it reaches to the
maximum velocity and then stop increasing AcceleratorPosition so that
velocity remain constant.
In every run in the loop we also check to see what is the distance to
point_2. If this distance become less than "S", then
SettAcceleratorPosition to zero and start increasing BrakePosition until
it reaches to point_2 with the desired velocity of V.

Now the problem is how do I calculate "S"? Too many unknown! isn't it?

Any other idea?

Maybe , if distance_to_point_2 < final_velocity^2/2/maximum_acceleration
then *backup_a_bit*
else *accelerate_at_maximum_for(final_velocity/maximum_acceleration)seconds*
..
where *backup_a_bit* is a routine which increases distance to point 2 and
leaves the vehicle stationary at the end and *accelerate_at_maximum_for(
t )seconds* is a routine that does just that, leaving the vehicle travelling
at maximum velocity at the end of time t and either at or heading towards
point 2.
Best of Luck - Mike
.



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