TITLE: What is the max speed of an object with startspeed/endspeed/acceleration/decelerations rates known?
QUESTION [0 upvotes]: My question is very similar to:
Determining the peak speed of an accelerating/decelerating body between two point 
but with a fairly significant difference, deceleration rate is not necessarily negative acceleration rate.
Using:
-Va for starting velocity
-Vb for ending velocity
-Vx for max velocity
-D for distance traveled
-a for acceleration
-d for deceleration 
how do I solve for Vx?

REPLY [0 votes]: Let $s$ be the length of time the object spends accelerating at rate $a$, and $t$ be the length of time the object spends accelerating at rate $d$.
Assuming $a$ is positive and $d$ is negative (!), the maximum speed occurs after the object has finished accelerating at rate $a$. 
The average velocity of an object over a certain time period is equal to the difference between the ending and starting velocities, divided by the length of the time period. From this, we find that 

$$\begin{align*}
a = \frac{V_x - V_A}{s} &\qquad&  d = \frac{V_B-V_x}{t}\\
s = \frac{V_x - V_A}{a} &\qquad&  t = \frac{V_B-V_x}{d}\\
as = (V_x-V_A) & \qquad & dt = V_B - V_x
\end{align*}$$

If an object starts with velocity $v$, then travels for time $\tau$ at constant acceleration $\alpha$, it travels a distance equal to $v\tau + \frac{1}{2}\alpha\tau^2$. From this we find that:

$$D = \left[V_As + \frac{1}{2}as^2\right] + \left[V_xt + \frac{1}{2}dt^2\right]$$

Using our formulas for $as = (V_x-V_A)$ and $dt = (V_B - V_x)$, we get:

$$D = s\left[V_A + \frac{1}{2}(V_x - V_A)\right] + t\left[V_x + \frac{1}{2}(V_B - V_x)\right]$$
$$D = s\left[\frac{1}{2}(V_x + V_A)\right] + t\left[\frac{1}{2}(V_B + V_x)\right]$$

Using our formulas for $s = \frac{V_x-V_A}{a}$ and $t=\frac{V_B-V_x}{d}$, we get:

$$D = \left[\frac{1}{a}(V_x - V_A)\right] \left[\frac{1}{2}(V_x + V_A)\right] + \left[\frac{1}{d}(V_B - V_x)\right]\left[\frac{1}{2}(V_B + V_x)\right]$$

We multiply by $2ad$: 

$$2adD = d\left(V_x - V_A\right)\left(V_x + V_A\right) + a\left(V_B - V_x\right)\left(V_B + V_x\right)$$
$$2adD = d\left(V_x^2 - V_A^2\right) + a\left(V_B^2 - V_x^2\right)$$

Collect terms involving $V_x$ :

$$(d-a)V_x^2   = dV_A^2 - aV_B^2 + 2adD $$

Solve for $V_x$, as required:

$$V_x = \sqrt{\frac{aV_B^2 - dV_A^2  - 2adD}{a-d}}$$

Note that in the special case that $a = -d$, we recover the solution to your earlier problem:
$$V_x = \sqrt{\frac{aV_A^2 + aV_B^2 + 2a^2D}{2a}}$$
$$V_x = \sqrt{\frac{V_A^2 + V_B^2}{2} + aD}$$
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