Time for object to reach target Im' trying to work out out long an object would take to reach a target given that it decreases in velocity over distance fired. The graph of the object velocity over distance is shown by the following ballistic chart:

As can be shown by the graph it looks like a linear deceleration. This was the process I was going to use to determine the time it would take to reach the target:


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*Determine the acceleration. In this case negative because it's slowing down using the following formula: a = (v^2 − u^2 ) / 2s Where v means final velocity, u means starting velocity and s is the distance travelled between the starting and final velocity. I can calculate this off the graph.


From here i'm unsure what I need to do. Assuming I have the initial velocity, the distance to target, final velocity (0) how can i calculate the time? Do i need to use a calculus for this?
 A: As you already have the velocity given by the graph, you don't need to determine or deal with the acceleration since it won't help you directly to solve what you're asking to find.
First determine an appropriate equation for the velocity. As you've already done, let $s$ be the distance, so $\frac{ds}{dt}$ would be velocity at any given time. The graph, as you state, is very close to being a straight line, so let's assume it is. Using the fairly standard $y = mx + b$ formula, where $m$ is the slope and $b$ is the $y$-intercept, gives
$$\begin{equation}\begin{aligned}
\frac{ds}{dt} & \approx \left(\frac{1730 - 2650}{500}\right)s + 2650 \\
& = -1.84s + 2650
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
I eye balled your graph to get estimates of the values, so you may wish to check yourself more carefully to make any appropriate adjustments. Once you determine an appropriate equation, then this is a separable equation which can be solved as
$$\begin{equation}\begin{aligned}
\frac{ds}{-1.84s + 2650} & = dt \\
\left(\frac{1}{-1.84}\right)\left(\frac{ds}{s - \frac{2650}{1.84}}\right) & = dt \\
\left(\frac{1}{-1.84}\right)\ln\left|s - \frac{2650}{1.84}\right| & = t + C
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using the initial condition that $s = 2650$ when $t = 0$ gives
$$C \approx -3.86 \implies \left(\frac{1}{-1.84}\right)\ln\left|s - \frac{2650}{1.84}\right| \approx t - 3.86 \tag{3}\label{eq3A}$$
Note you are asking for basically $2$ potential things. You can get the time to get to a specific distance, including to the target, and you can also get the time it takes for the velocity to reach $0$. Note these $2$ times won't usually be equal to each other.
For the time to get to a particular distance $s_p$, you can just plug in $s_p$ into \eqref{eq3A}. On the other hand, for the time it takes for the velocity to reach $0$, first use $\frac{ds}{dt} = 0$ in \eqref{eq1A} to get $s_p = \frac{2650}{1.84}$, then plug this into \eqref{eq3A}.
However, you will note a problem here. As you can see, you then get on the LHS a factor of $\ln|0|$! This is because the graph can't remain linear as the velocity approaches $0$ since, as the velocity becomes closer and closer to $0$, the time it will take to travel even a short distance more becomes longer & longer. As such, the graph will need to curve down at some point, so the equations above won't hold in that range and, thus, will need to be adjusted accordingly. Also, this means that for the first part, if the velocity becomes relatively small at your final distance, you will likely need to adjust its final result as well.
A: First, an explanation of why your approach is wrong.
The kinematics equations (of which $v^2 = u^2 + 2as$ is one) only apply when acceleration $a$ is a constant. It is not constant here. To see why, consider what is known, that $\frac{dv}{ds}$ is a constant (let's call it $k$ where $k<0$). By chain rule, $\displaystyle \frac {dv}{ds} = \frac{\frac{dv}{dt}}{\frac{ds}{dt}}= \frac av$. So $a = kv$ and since $v$ is obviously not a constant, neither is $a$. So you can't use any of the standard kinematic equations.
Here's how to solve it:
Start with $\frac{dv}{ds} = k$
Write $ds = vdt$ and rearrange to give:
$\frac{dv}v = kdt$ (this is separation of variables)
Integrate both sides:
$\displaystyle \int_U^V \frac{dv}v = \int_0^T kdt$, where $U$ is initial velocity, $V$ is final velocity and $T$ is final time.
$\ln V - \ln U = kT$
$\ln \frac VU = kT$
$\frac VU = e^{kt}$
$V = Ue^{kT}$.
Since $k$ is negative, the object's velocity is decreasing over time.
You will note that $V$ will never quite be equal to zero in this theoretical model. In practical terms, objects will eventually come to rest, and that's because the model using the simple calculus treatment breaks down as the velocity decays enough.
For your purposes, you need to know the final velocity with which the object hits the target. If it's too low (too close to zero), the model will be inaccurate and you will grossly over-estimate $T$.
