Kinematics: time for a given acceleration, deceleration and distance

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Homework Statement:

Find the minimum time it takes for a an object to start and end at rest for a given fixed acceleration, fixed deceleration and distance. Assume there is no maximum speed.

This can be modelled as a train travelling a given distance by accelerating until it hits the point where it needs to decelerate again. There is never a point of constant speed.

Relevant Equations:

Kinematic equations:
d = 0.5 * a * t^2
Note: Maths has always been kinda a weak point for me in school. Anyway.

From hereon I'm going to talk about this in the context of a train for simplicity, even though this is actually needed for an entirely different context - but I don't think this is relevant for the problem.

In terms of my own individual working, I've attempted to break down the problem into two different sections:
  1. Where the train accelerates to its maximum velocity,
  2. Then when it decelerates from this maximum velocity to rest.
This is hard because maximum velocity is of course not fixed and is entirely dependant on both acceleration and deceleration. I've worked on this by looking at a speed-time graph, using the the idea of triangular geometry (maintain the same area under the graph while squashing it in various ways). I'm sure this is a really primitive way of looking at it and would love to be shown a better way of calculating this. A friend has told me to look at integrals which I will get to later on in the post.

Anyway, my current equations are as follows:
distance = 0.5 * velocity * time (based off from the area of a triangle)​
acceleration = velocity / time (well known equation)​
therefore: distance = 0.5 * acceleration * time^2 (also well known equation but figured I'd create it myself for the sanity checking)​

Now I start breaking things down into the acceleration/deceleration sections:
Total Time (T) = ta + td​
Total Distance (D) = da + dd​
da = 0.5 * a * ta^2​
dd = 0.5 * d * td^2​
therefore: D = 0.5 * ((a * ta^2) + (d * td^2))​

And this is where I'm stumped with trying to get D in terms of (ta+td) which would then give me an equation for T.

So I've gone to my friend for help, who's told me about using integration. I'll just paste screenshots of what was given to me for this:
243222

as a representation of the problem to then plug into Wolfram Mathematica to solve for t:

243223

So of course we're never going to have negative time, so the positive answer has been used in my equations, however this is very much a black-box solution for me and therefore not really something I can work with and check it's correct, which is definitely an issue right now because I'm getting odd answers for T when putting in values of acceleration and deceleration. For instance, when I put in a higher value for deceleration, t goes up. Also I think if I swap acceleration and deceleration, t should remain the same, but they currently don't?

Can someone either verify for me that what's been given to me is definitely correct, or propose a different way for me to solve this, or just generally help/sanity check this for me?

Many thanks, and apologies for the wordy post! Edit: I also think I've put this under the wrong forum, but I'm unsure. Feel free to move this if that would be better!
 
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Answers and Replies

  • #2
DrClaude
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Problem Statement: Find the minimum time it takes for a an object to start and end at rest for a given fixed acceleration, fixed deceleration and distance. Assume there is no maximum speed.

This can be modelled as a train travelling a given distance by accelerating until it hits the point where it needs to decelerate again. There is never a point of constant speed.
The problem is ill-posed because there is no such minimum time.

Once you fix the distance, the acceleration, the deceleration, and the fact that the velocity must be zero at the end points and there is no coasting, there is no variable left. There will be a single moment/position where the deceleration can be applied such that the train come to a stop at the proper point.
 
  • #3
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The problem is ill-posed because there is no such minimum time.

Once you fix the distance, the acceleration, the deceleration, and the fact that the velocity must be zero at the end points and there is no coasting, there is no variable left. There will be a single moment/position where the deceleration can be applied such that the train come to a stop at the proper point.
Fair. The wording could do with changing. Basically, I've badly implied that the reason why we're not coasting at a fixed speed is because the traversal needs to be done within the fastest time possible.
 
  • #4
PeroK
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Problem Statement: Find the minimum time it takes for a an object to start and end at rest for a given fixed acceleration, fixed deceleration and distance. Assume there is no maximum speed.

This can be modelled as a train travelling a given distance by accelerating until it hits the point where it needs to decelerate again. There is never a point of constant speed.
Relevant Equations: Kinematic equations:
d = 0.5 * a * t^2

Note: Maths has always been kinda a weak point for me in school. Anyway.

From hereon I'm going to talk about this in the context of a train for simplicity, even though this is actually needed for an entirely different context - but I don't think this is relevant for the problem.

In terms of my own individual working, I've attempted to break down the problem into two different sections:
  1. Where the train accelerates to its maximum velocity,
  2. Then when it decelerates from this maximum velocity to rest.
This is hard because maximum velocity is of course not fixed and is entirely dependant on both acceleration and deceleration. I've worked on this by looking at a speed-time graph, using the the idea of triangular geometry (maintain the same area under the graph while squashing it in various ways). I'm sure this is a really primitive way of looking at it and would love to be shown a better way of calculating this. A friend has told me to look at integrals which I will get to later on in the post.

Anyway, my current equations are as follows:
distance = 0.5 * velocity * time (based off from the area of a triangle)​
acceleration = velocity / time (well known equation)​
therefore: distance = 0.5 * acceleration * time^2 (also well known equation but figured I'd create it myself for the sanity checking)​

Now I start breaking things down into the acceleration/deceleration sections:
Total Time (T) = ta + td​
Total Distance (D) = da + dd​
da = 0.5 * a * ta^2​
dd = 0.5 * d * td^2​
therefore: D = 0.5 * ((a * ta^2) + (d * td^2))​

And this is where I'm stumped with trying to get D in terms of (ta+td) which would then give me an equation for T.

So I've gone to my friend for help, who's told me about using integration. I'll just paste screenshots of what was given to me for this:
View attachment 243222
as a representation of the problem to then plug into Wolfram Mathematica to solve for t:

View attachment 243223
So of course we're never going to have negative time, so the positive answer has been used in my equations, however this is very much a black-box solution for me and therefore not really something I can work with and check it's correct, which is definitely an issue right now because I'm getting odd answers for T when putting in values of acceleration and deceleration. For instance, when I put in a higher value for deceleration, t goes up. Also I think if I swap acceleration and deceleration, t should remain the same, but they currently don't?

Can someone either verify for me that what's been given to me is definitely correct, or propose a different way for me to solve this, or just generally help/sanity check this for me?

Many thanks, and apologies for the wordy post! Edit: I also think I've put this under the wrong forum, but I'm unsure. Feel free to move this if that would be better!
The answer doesn't look quite right to me. For example, if you make ##a_1 = a_2## then your answer is:

##t = \sqrt{\frac{d}{a}}##

Can you solve the problem for the case where the accleration and deceleration have the same magnitude and confirm this is wrong?
 
  • #5
jbriggs444
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therefore: D = 0.5 * ((a * ta^2) + (d * td^2))
The reasoning to reach this equation is correct. And the equation itself is also correct. There is no point applying integrals to proceed further. The required integration already went into deriving this result -- ##d=\frac{1}{2}at^2## is the result of integrating acceleration twice to determine distance.

However, the above is one equation in two unknowns. (##t_a## and ##t_d##). You need another equation. The fact the that initial and final velocity are zero can give you another equation. Can you write that equation down?
 
  • #6
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The answer doesn't look quite right to me. For example, if you make ##a_1 = a_2## then your answer is:

##t = \sqrt{\frac{d}{a}}##

Can you solve the problem for the case where the accleration and deceleration have the same magnitude and confirm this is wrong?
It does indeed seem wrong. So this is what I've done to prove this wrong:

Using ##D = \frac{at^2}{ 2}## and:
## a_1 = a_2 ##, so therefore:​
##t_1 = t_2 ## and ##d_1 = d_2##​
##T = 2 * t_1##, and therefore ##t_1 = \frac{T}{2}##​
##D = 2 * d_1##​
##D = at_1^2##​
##= a (\frac{T}{2}^2)##​
##= \frac{aT^2}{4}##​
Rearrange for ##T##:​
##T = \sqrt{\frac{4D}{a}} \neq \sqrt{\frac{D}{a}}##​
Is this right?

jbriggs444: I'll work on an answer to your response and get back to you ASAP.

Thanks for the help, guys. :)
 
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  • #7
PeroK
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It does indeed seem wrong. So this is what I've done to prove this wrong:

Using ##D = \frac{at^2}{ 2}## and:
## a = d ##, so therefore:​
##t_a = t_d ## and ##d_a = d_d##​
##T = 2 * t_a##, and therefore ##t_a = \frac{T}{2}##​
##D = 2 * d_a##​
##D = at_a^2##​
##= a (\frac{T}{2}^2)##​
##= \frac{aT^2}{4}##​
Rearrange for ##T##:​
##T = \sqrt{\frac{4D}{a}} \neq \sqrt{\frac{D}{a}}##​
Is this right?

jbriggs444: I'll work on an answer to your response and get back to you ASAP.

Thanks for the help, guys. :)
Yes, that's the correct answer when ##a_1 = a_2##.

Note that the distance travelled is the area under a velocity vs time graph. If you draw a graph and use some geometry, you may find an easier way to get this answer; and, this may help for the case where ##a_1 \ne a_2##.

PS here's another interesting thing that you may be able to see from a graph. Don't worry if you don't understand this yet. From the graph I could see that the answer should be symmetric in ##a_1, a_2##. That means that if you swapped ##a_1## and ##a_2## round, you should get the same time.

E.g. if the acceleration was ##a_1 = 5m/s^2## and the deceleration was ##a_2 = 2m/s^2##, then that should take the same time as the other way round with ##a_1 = 2m/s^2## and ##a_2 = 5m/s^2##.

Now, the original answer your friend got was not symmetric is ##a_1, a_2##. If you swap them over you get a different answer. So, you can see that that answer must be wrong.
 
  • #8
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The reasoning to reach this equation is correct. And the equation itself is also correct. There is no point applying integrals to proceed further. The required integration already went into deriving this result -- ##d=\frac{1}{2}at^2## is the result of integrating acceleration twice to determine distance.

However, the above is one equation in two unknowns. (##t_a## and ##t_d##). You need another equation. The fact the that initial and final velocity are zero can give you another equation. Can you write that equation down?
So after more asking around, someone's presented to me the idea that maximum velocity would be the same for acceleration and deceleration which is pretty obvious, but I didn't think to include this in the equation. Is this what you're getting at?

So therefore using ##v = at##:
##a_1 t_1 = a_2 t_2##, so therefore ##t_2 = \frac{a_1 t_1}{a_2}## (eqn. 1)
Substituting (eqn. 1) into ##T = t_1 + t_2## and rearranging would give us:​
##T = t_1(1 + \frac{a_1}{a_2})## (eqn. 2)
Leaving us needing to find ##D = \frac{1}{2} ( a_1 t_1^2 + a_2 t_2 )## (eqn. 3) in terms of ##t_1(1 + \frac{a_1}{a_2})## to find ##T##:
Substituting (eqn. 1) into (eqn. 3):​
##D = \frac{1}{2} ( a_1 t_1^2 + a_2 \frac{a_1 t_1}{a_2})##​
Doing a bunch of rearranging:​
##D = \frac{a_1 t_1}{2} \cdot t_1 (1 + \frac{a_1}{a_2})##​
Substituting (eqn. 2) and rearranging for ##T##:​
##T = \frac{2D}{a_1 t_1}## (eqn. 4)
So to get rid of the final unknown ##t_1##, from (eqn. 2) we have:
##t_1 = \frac{T}{1+\frac{a_1}{a_2}}## (eqn. 5)
Substitute (eqn. 5) into (eqn. 4) and rearrange:
##T = \sqrt{\frac{2D (a_1 + a_2)}{a_1 a_2}}##​
So it looks like I have an equation for total time! I hope it's right. Will test it with a few variables.
 
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  • #9
PeroK
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Here's how I did it, using the idea that ##D## is the area under the velocity time graph, which is made up of two triangles. The height of the triangles is ##a_1t_1 = a_2t_2##; and its length is ##T##. Therefore:

##D = \frac12 T a_1t_1##

Now, using your equation (2), we have ##t_1 = \frac{a_2}{a_1 + a_2}T##, hence:

##D = \frac12 T^2 \frac{a_1 a_2}{a_1 + a_2}##

And, the result follows.
 
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