# What is the max speed of an object with startspeed/endspeed/acceleration/decelerations rates known?

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?

1. 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 $d$.
2. Assuming $a$ is positive and $d$ is negative (!), the maximum speed occurs after the object has finished accelerating at rate $a$.
3. 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*}

1. 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:

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

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

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

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

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

1. We multiply by $2ad$:

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

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

1. Collect terms involving $V_x$ :

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

1. 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:

$$V_x = \sqrt{\frac{aV_A^2 + aV_B^2 + 2a^2D}{2a}}$$

$$V_x = \sqrt{\frac{V_A^2 + V_B^2}{2} + aD}$$

• I need to make sure it works, I check and mark it – paycheck87 Dec 15 '17 at 9:53
• Ah, the final answers here do not have the VA added back in and I'm not sure how that changes things... – paycheck87 Dec 15 '17 at 14:30
• If you want to add V_A back in, you can solve in a manner analogous to the problem you asked earlier; see updated answer. – user326210 Dec 18 '17 at 4:06

Wolfram Alpha came through on this one, I just used the most basic kinematic and used (v - r + at / d) as the substitute for deceleration time:

apparently I don't have enough rep to actually link my wolfram image but that is the link to it.

t = time to max speed

a = acceleration

d = deceleration

e = total distance to travel

v = initial speed

r = end speed

Here is the code I used:

            var a = _acceleration;
var d = _decceleration;
var e = totalDist;
var v = _startSpeed;
var r = _endSpeed;
var term1 = (a + d) * (2 * e * a * d + a * r * r + d * v * v);
var term2 = Math.Sqrt(term1) + a * v + d * v;
var term3 = term2 / (a * a + a * d);

_timeToMaxSpeed = term3;