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Consider some movement along a path segment $s$ with constant acceleration/deceleration (see figure below). The initial speed is $v_0$ and the final speed is $v_1$. The constant acceleration is $a$ and the constant deceleration is $-a$.

speed vs time

So, to clarify, we know the values for $v_0$, $v_1$, $a$, and segment length $s_{t}$. We do not know the values for $v_x$, $t_x$, or $t_s$.

So for $t:[0,t_x]$ we have $v(t) = at + v_0$, and then at $t_x$ the acceleration switches to deceleration so we get $v_1$ at $t_s$ (the end of the path segment).

The ultimate goal is to find $t_s$, i.e., the time it takes to move along the entire segment.

One approach would be to find an expression for $v_x$ that lets me calculate

$$t_x = \frac{v_x - v_0}{a}$$

and

$$t_s = t_x + \frac{v_1 - v_x}{-a}.$$

But I am having trouble finding any expression that tells me the speed $v_x$ where the switch from acceleration to deceleration occurs.

I have tried solving the equation system

(1) $v(t) = at + v_0$

(2) $v(t) = -at + m$

with $m = -at_s + v_1$, which gives me

$$t_x = \frac{t_s}{2} + \frac{(v_1 - v_0)}{2a},$$

which substituted into (1) gives me

$$v_x = \frac{at_s + v_1 + v_0}{2}$$

but is dependent on $t_s$ which is the unknown I want to find.

I seem to be at a loss here and would appreciate any help whatsoever.

EDIT: Clarified what values are known.

UPDATE: I have realized that $v_x$ can be calculated from the fact that the segment length $s_t$ is known.

The equation

$$s(t) = \frac{a}{2}t^2 + v_0t$$

can be used in combination with the system of equations (1) and (2) above. If we start with (1) we get

$$t = \frac{v - v_0}{a},$$

which at $t_x$ becomes

$$t_x = \frac{v_x - v_0}{a}.$$

Substituting this into the equation for $s(t)$ we get the distance covered from $t=0$ to $t_x$ (referred to as $s_1$)

$$s_1 = \frac{a}{2}\left(\frac{v_x - v_0}{a}\right)^2 + v_0\left(\frac{v_x - v_0}{a}\right).$$

If we then look at (2) and consider $t_x=0$ we can write it as

$$v(t) = -at + v_x.$$

If we extract t from this equation we get

$$t = \frac{v_x - v}{a}.$$

This means that the time it takes to decelerate from $v_x$ to $v_1$ (referred to as $t_2$) is thus

$$t_2 = \frac{v_x - v_1}{a}.$$

Substituting this into the general equation for distance we get an expression for the deceleration distance, i.e., the distance covered from $t_x$ to $t_s$ (referred to as $s_2$)

$$s_2 = \frac{a}{2}\left(\frac{v_x - v_1}{a}\right)^2 + v_0\left(\frac{v_x - v_1}{a}\right).$$

Since the total segment length $s_t$ is known and it is the sum of the two parts

$$s_t = s_1 + s_2,$$

and $s_1$ and $s_2$ both only depend on $v_x$ we get an equation that has only one unknown, $v_x$, and can therefore be solved.

I have not tried solving it yet but unless I have done some fatal error, it should be straight forward.

UPDATE: The solution mentioned in the first update seems ok. If we set

$$s_t = s_1 + s_2$$

and solve for $v_x$ we get

$$v_x = \sqrt{as_t + \frac{v_0^2}{2} + \frac{v_1^2}{2}},$$

(which is the same expression as @narasimham mentioned in a comment to his answer).

This can be used in the approach mentioned at the top to get an analytic expression for $t_s$.

$$t_s = \frac{v_x - v_0}{a} + \frac{v_1 - v_x}{-a}$$ $$ = \frac{2v_x - v_0 - v_1}{a}$$ $$ = \frac{2\sqrt{as_t + \frac{v_0^2}{2} + \frac{v_1^2}{2}} - v_0 - v_1}{a}.$$

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    $\begingroup$ Unless you give either $t_x$ or $v_x$, you have an infinity of solutions. If you have the choice, I would recommend to decelerate from $v_0$ to $v_1$ directly... $\endgroup$ – Martigan Jun 15 '15 at 15:43
  • $\begingroup$ I clarified constraints (segment length) that should narrow it down to a single solution. I also came to a realization on how to do this. As mentioned in the update, I have not checked that it is correct yet but it feels straight forward and I will try first thing tomorrow. :) $\endgroup$ – mags Jun 15 '15 at 20:22
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There is not enough information to solve the problem. The expression you have gives the difference in the length of the intervals $[t_0,t_x]$ and $[t_x,t_s]$, but we have no constraint on the sum. You could have $t_x=t_0$, so the particle starts decelerating immediately and decelerates until it reaches $v_1$. You can have $t_x$ any higher value, then the period of deceleration will increase by $t_x$ and you will arrive at the same final $v_1$

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  • $\begingroup$ No, you don't, you have two similar equations, that is, one equation... with two variables... and that's right, for if the only parameters are $v_0$, $v_1$ and $\pm a$, you have an infinity of paths that works, the shortest going in a single direction, with a continuous movement. $\endgroup$ – Martigan Jun 15 '15 at 15:42
  • $\begingroup$ @Martigan: I believe $a$ is a given. If $a$ is to be solved for, you are correct that there is not enough information. $\endgroup$ – Ross Millikan Jun 15 '15 at 16:34
  • $\begingroup$ $a$ is a given all right. But even with that, there is not enough information. Your two equations are in fact the same and only... $\endgroup$ – Martigan Jun 15 '15 at 16:39
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    $\begingroup$ @Martigan: I see now you are correct. You are right that there is not enough information. I will replace this. $\endgroup$ – Ross Millikan Jun 15 '15 at 17:23
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No need labeling any time markers $ t_x, t_s $, it is unnecessary information.

Just find sum of individual times.

$$ t_s= \dfrac{(v_x-v_o) + (v_x-v_1)}{a}= \dfrac{2 v_x-v_o -v_1}{a}$$

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  • $\begingroup$ I am sorry if it was not clear from my question but $v_x$ Is not known, only $v_0$, $v_1$, $a$, and the segment length. $\endgroup$ – mags Jun 15 '15 at 19:25
  • $\begingroup$ Please re-write or edit to list what is given and what is required. $\endgroup$ – Narasimham Jun 15 '15 at 19:32
  • $\begingroup$ Edited to clarify what is known. $\endgroup$ – mags Jun 15 '15 at 19:44
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    $\begingroup$ Apart from above equation another distance equation is to be considered. $ 2 a\, s_t = 2 v_x^2 - v_1^2 -v_0^2 $ ... because time is same whether for acceleration or deceleration. Solving the two equations I got solution in terms of stated unknowns. Shall be back after the net connection improves. $\endgroup$ – Narasimham Jun 15 '15 at 21:30
  • $\begingroup$ Updated question with possible answer. Will confirm today. $\endgroup$ – mags Jun 16 '15 at 5:46

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