Supposing I have a linear axis with linear acceleration/deceleration. I would like it to move to its target location and arrive after a specific amount of time.
How can I calculate the maximum velocity given the time, distance and acceleration?
It seems like a rather easy problem to solve at first sight. However I strugle with the fact that the time/distance needed for acceleration/deceleration changes with the max velocity.
On
Divide the time $t$ into $\Delta t + T + \Delta t$, where the $\Delta t$ is the acceleration \deceleration time, and $T$ the time at constant speed.
Space traveled is the area of the trapetium you draught $s= V_{max} \cdot (2T+ 2\Delta t )/2$.
$a=v_{max}/ \Delta t$
You have all the elements for the answer ...
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Assumption - the body accelerates and decelerates for the same amount of time.
Solution : (Graphical) The question of max velocity becomes the question of attaining the max height on the graph you posted while keeping the area under the graph and the slope constant since the distance and acceleration are fixed. Thus, logically it can be attained by reducing the 'zero slope' component to zero.
Solution : (Theoretical) Suppose the body accelerates and decelerates for time t and moves with constant velocity for t0. Total time T, dist D and acc. A (are fixed). Then T = 2t + t0 OR t0 = T - 2t
Now, D = (1/2)At^2 + At·(t0) + At·t - (1/2)At^2 OR D = At^2 + At(t0) OR D = ATt - 2A(t^2) [ eliminate t0 ] Now differentiate to maximize t since the max velocity is At. This gives t = T/2.
Therefore, max velocity = AT/2
On
Let the distance be $D$, time for acceleration be $t_1$ and that for deceleration be $t_2$.
So, $t_1 = \frac{v_{\max}}{a} \ , t_2 = -\frac{v_{\max}}{d} , (d <0)$
Now $$D = \text{ Area of the trapezium } = \frac 1 2 (t + t - (t_1+t_2))\times v_{\max} $$
$$D = \frac 12\bigg[2t - v_\max\bigg(\frac1 a - \frac 1 d\bigg)\bigg]v_\max$$
$$ \bigg(\frac1 a - \frac 1 d\bigg)v_\max^2 - 2tv_\max + 2D = 0$$
Substitute $D,a,d (d<0) $ ,solve the above quadratic equation and find the one that yields the maximum velocity(+ve) .
Let the maximum speed be $v_{max}$ and total time, distance, acceleration and deceleration be $t_{total}$, $s_{total}$, $a_{1}$ and $a_{2}$ respectively, with $a_{2} < 0$.
In order to find the maximum speed, you will need to consider each section of the journey individually to find an expression for $v_{max}$, as it is dependent on the acceleration and deceleration.
Considering the journey in three stages, labelled as follows:
where $s_{m}$ and $t_{m}$ represent the distance and duration for which $v = v_{max}$.
As the total distance and time are known, we can say $s_{total} = s_{1} + s_{m} + s_{2}$ and $t_{total} = t_{1} + t_{m} + t_{2}$.
For the middle section travelled at the constant speed $v_{max}$, we can say $v_{max}= \frac{s_{m}}{t_{m}}$.
As $a_{1}$ and $a_{2}$ are known, we will be able to find $t_{1}$, $t_{2}$, $s_{1}$ and $s_{2}$. So, $$v_{max}= \frac{ s_{total} - s_{1} - s_{2}}{t_{total} - t_{1} - t_{2}}$$.
$t_1$ and $t_2$ can be found using $v = u + at$:
For $t_1$, $u = 0$ so $t_1 = \frac{v_{max}}{a_1}$.
For $t_2$, $v = 0$ so $t_2 = \frac{-v_{max}}{a_2}$.
$s_1$ and $s_2$ can be found using $s=ut+\frac{1}{2}at^2$:
For $s_1$, $u=0$ so $$s_1 = \frac{1}{2} * \frac{v_m^2}{a_1}$$
For $s_2$, $u=v_{max}$ so $$s_2 = v_{max}t_2 + \frac{1}{2}a_2t_2^2$$
Substituting these expressions into the expression for $v_{max}$ will give a quadratic in $v_{max}$:
$$v_{max} = \frac{ s_{total} - (\frac{-v_max^2}{a_2} + \frac{v_max^2}{a_2})-\frac{v_max^2}{a_1} }{t_{total} -\frac{v_max}{a_1} + \frac{v_{max}}{a_2} }$$
which simplifies to
$$(\frac{1}{2a_2} - \frac{1}{2a_1})v_{max}^2 + v_{max}t_{total} - s_{total} = 0$$
EDIT The geometrical method used in Ak19's answer is much more efficient, however you may still be interested to see how an equivalent result can be derived from the basic kinematics formulae.