This problem is fairly common in the game of cricket. The distance a cricket ball travels is calculated which is basically from point A to point B. However, this isn't entirely accurate as the some object stands in the way of the traveling ball.
What I'd like to calculate is potential distance the ball could have travelled had there not been any objects. With hawk-eye technology the cameras are able to track the trajectory of the ball.
My question is if we know the speed (s), time (t in secs) and change in speed or deceleration (d), can we calculate the potential distance the ball could have traveled?
How do we incorporate d in the equation dist = s*t
If you are assuming that the acceleration is constant and if you only want to track the horizontal distance, you can use the popular kinematic equation
$$ \Delta x = v_0t + \frac{1}{2}at^2 $$
or using your naming convention
$$ \text{dist} = st+\frac{1}{2}dt^2 \text{ .} $$
You could go a step farther by saying that the drag is proportional to the speed (or even squared speed) of the ball giving a differential equation for the motion.
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$\textbf{EDIT:}$
For this edit, I will adopt a more standard notation as follows.
If we assume that drag is not constant, the above kinematic equations no longer hold. We know that air resistance is dependent in some way on the speed of the ball $$\text{speed} = \| \vec{v} \|$$ and will oppose the direction of motion. Putting these notions together, we will explore two different methods of modeling the drag.
The "easy" way to compute drag is to assume that the flow is approximately non-turbulent and use
$$\vec{F}_D = -b\|\vec{v}\| \hat{v}$$
as a "linear" drag force. Observe that the above equation reduces as
$$\vec{F}_D = -b\|\vec{v}\| \hat{v} = -b\|\vec{v}\| \frac{\vec{v}}{\|\vec{v}\|} = -b \vec{v}$$
thus we can conclude that the $x$-component of the force is independent of the $y$-component of the force. Newton's Laws give us two equations
$$\begin{align} F_x=ma_x &= -bv_x\\ F_y=ma_y &= -bv_y - mg \text{ .}\end{align}$$
Since we know that velocity is the derivative of position and that acceleration is the derivative of velocity, we can write
$$ \begin{align} \dfrac{\text{d}^2x}{\text{d}t^2} &= -\frac{b}{m}\frac{\text{d}x}{\text{d}t}\\ \\ \dfrac{\text{d}^2y}{\text{d}t^2} &= -\frac{b}{m}\frac{\text{d}y}{\text{d}t} - g \text{ .}\end{align}$$
With $\beta = b/m$, this is a differential equation for the system
$$ \begin{align} x(t) &=x_0 + \frac{v_{0x}}{\beta}\big( 1-e^{-\beta t} \big) \\ \\ y(t) &=y_0 + \frac{1}{\beta}\big( v_{0y}+\frac{g}{\beta} \big)\big( 1-e^{-\beta t} \big) - \frac{g}{\beta}t\text{ .}\end{align} $$
Solve the above system for the time the ball impacts the ground and substitute that time into $x(t) - x_0$ for the $x$-distance traveled. Please verify the above kinematic equations for yourself! I could have easily made a sign error (or worse) somewhere.
The second model I will address is the more realistic model where we assume a turbulent flow and
$$ \vec{F}_D = -b\|\vec{v}\|^2 \: \hat{v} \text{ .} $$ Now we have that
$$ \vec{F}_D = -b\|\vec{v}\|^2 \: \hat{v} = -b\|\vec{v}\|^2 \: \frac{\vec{v}}{\|\vec{v}\|} = -b\|\vec{v}\| \: \vec{v}$$ therefore Newton gives us that
$$\begin{align} F_x=ma_x &= -bv_x\sqrt{v_{x}^2 + v_{y}^2}\\ F_y=ma_y &= -bv_y\sqrt{v_{x}^2 + v_{y}^2} - mg \text{ .}\end{align}$$
This is of course the coupled, non-linear system
$$ \begin{align} m\dfrac{\text{d}^2x}{\text{d}t^2} &= -b\dfrac{\text{d}x}{\text{d}t}\sqrt{\bigg[\dfrac{\text{d}x}{\text{d}t}\bigg]^2 + \bigg[\dfrac{\text{d}y}{\text{d}t}\bigg]^2}\\ m\dfrac{\text{d}^2y}{\text{d}t^2} &= -b\dfrac{\text{d}y}{\text{d}t}\sqrt{\bigg[\dfrac{\text{d}x}{\text{d}t}\bigg]^2 + \bigg[\dfrac{\text{d}y}{\text{d}t}\bigg]^2} - mg \text{ .}\end{align} $$
I'll let you bang your head against that one for a while.