How to calculate the final position of a particle under variable accelaration and its instantenous velocity?

Question

I'm a first-semester physics student who was recently on a train. On a screen, it said the instantaneous velocity of the train was 176 km / h. We had 4 min left until our destination. I wanted to calculate the distance based on these facts, so I tried using the kinematic equations. More specifically, this one: $X_f = X_i + \frac{1}{2} (V_{x_i} + V_{x_f}) t $, which worked, but I realized these kinematics equations are only valid for particles under constant acceleration. The train obviously speeds up and down depending on the road's curvature. My question is whether we can still model it using the average acceleration. We can think of it as beginning at $V_{x_i}$ (which is 176 km / h) and slowing down to 0 km / h over the course of 4 minutes. Does this model give us the same results? Why do we even need the term "particle under constant acceleration" then?

And the second part of my question. Would it be possible to calculate the instantaneous velocity of this train at any point t, only based on the given information? I wanted to take the derivative, but I don't have access to a function that gives us $x$ in terms of $t$. In all the questions I've solved so far, I was given a similar function. Something like $ x = t^3 $. Then it's easy to take its derivative, of course. Can we construct such a function on our own? How would we go about doing this?

This isn't my homework or anything. I was on a train and simply curious. Thank you!

Answer 1

The time shown on the screen(4 mins) is also instantaneous and based on the speed of the train. That's why ETA decreases when you reach closer to the destination. So, $d=V(inst) × t = 11.73 km$

No need of calculus here it seems. About your second part, the relation of $x$ with $t$ generally appears when there is a constant external force in play. Like when the train reaches uphill and is rolling down with engine off on a defined curvature (gravitational force). Or a object thrown at an angle with some initial velocity, or a charged particle traveling in a constant or varying (defined by nature of its source) Magnetic field, etc.

Answer 2

Even though others have provided the correct answer to the specific question, I want to go over some basics on how to estimate distance from a graph of speed.

On the top above is an example of the actual speed vs. time graph. The distance traveled is found by the area under the curve. With calculus notation this is $$\Delta x = \int \limits_0^{\Delta t} v(t)\,{\rm d}t$$

But often you can find an approximation of the above without doing calculus. In the example above the shape is broken down to two triangles and a rectangle, all of which have known area formulas based on geometry.

$$ \Delta x_{\rm est} = x_1 + x_2 + x_3$$

With $$\begin{aligned} x_1 & = \tfrac{1}{2} t_1 v_{\rm est} \\ x_2 & = (t_2-t_1) v_{\rm est} \\ x_3 & = \tfrac{1}{2} (t_3-t_2) v_{\rm est} \end{aligned}$$

You can visually see that an error was introduced by making the smooth rounded curves into sharp corners in the graph. The more segments used to describe the shape the less the error.