Question: I'm new to physics and I've been trying to solve a few high school olympiad questions. I've figured that I approach the questions by analyzing the forces acting on objects and trying to induce something from there, but it gets complicated, and I mean it (whole pages covered in equations). Then I've analytically checked how "they" solve those questions and I've realized their favourite method is to approach questions by considering conservation laws, and it dramatically decreases the amount of time and work required to solve a problem.

I've searched if there is a formal intellectual documentation on this matter but I failed to find one. I'm really looking forward to learn more on this particular situation.

Could you please enlighten me further? (any resource/comment appreciated)

Eg:
A and B have masses of 100kg each. For what value of d, will the mass A travel 3 meters then stop?

I don't know, maybe I was unable to write the correct force equations, but when I wrote them and tried to solve the question I miserably failed, due to high amount of equations, whereas the energy approach gets you to the solution in a few lines.

[![Problem](https://i.stack.imgur.com/RhVwB.jpg)](https://i.stack.imgur.com/RhVwB.jpg)
[![Solution](https://i.stack.imgur.com/N1RJh.jpg)](https://i.stack.imgur.com/N1RJh.jpg)

Answer: 
I really wish they would teach some very basic calculus as an introduction to physics.

Anyway, we you need to realize is that the displacement of an object is mathematically equivalent to the area under the velocity vs. time curve.

![fig1](https://i.stack.imgur.com/xcoEX.png)

So the kinematic relationship is just the *area under a trapezoid* if you remember your geometry.

$$ s = \frac{v\_f+v\_i}{2} (t\_f -t\_i ) $$

But why? well in physics you say on average during this time frame it moves with $v\_a = \frac{v\_f+v\_i}{2}$ speed, and so $s$ is the distance traveled under constant speed $v\_a$.

Geometrically, if you slice a trapezoid along its average height, and move the leftover triangle under the cut you can create a rectangle (of known height).

[![fig2](https://i.stack.imgur.com/D4zJ2.png)](https://i.stack.imgur.com/D4zJ2.png)

The final question on this topic, is why is displacement equals the area under the speed curve? The answer is that by definition speed is the slope of the displacement curve, and calculus shows us that area under the curve is the inverse operation of slope of a curve.
(This supplements the graphical answer by @JAlex, which implicitly assumes constant acceleration.)

In general, the *physics-textbook definition for average-velocity* is,
$$\Delta x \stackrel{textbook\ def}{=} v\_{avg} \Delta t.$$
For simplicity, set $t\_i=0$, and let $s=\Delta x$.

Then $$s=v\_{avg} t\_f.$$

When the **acceleration is constant**,

starting with the *real definition for "average velocity"*

(**the time-average of the velocity** *[not "half the sum of the initial and final velocities"]*),

we have

\begin{align}
v\_{avg} 
&\stackrel{def}{=}\frac{\int\_0^{t\_f} dt\ v}{\int\_0^{t\_f} dt }\quad
\stackrel{textbook\ def}{=} \frac{\Delta x}{\Delta t }\\\
&\stackrel{constant\ a}{=}\frac{\int\_0^{t\_f} dt\ (v\_i+at)}{\int\_0^{t\_f} dt }\\
&\stackrel{constant\ a}{=}\frac{v\_i t\_f+\frac{1}{2}at\_f^2}{t\_f }\\
&\stackrel{constant\ a}{=}v\_i +\frac{1}{2}at\_f\\
&\stackrel{constant\ a}{=}v\_i +\frac{1}{2}(v\_f-v\_i)\\
v\_{avg}&\stackrel{constant\ a}{=}\frac{1}{2}(v\_f+v\_i)\\
\end{align}
Thus,
$$s \stackrel{constant\ a}{=} \left(\frac{1}{2} (v\_f+v\_i)\right)t\_f$$
If the acceleration isn't constant, then this expression derived for $v\_{avg}$ does not hold.

So, this expression is useful for constant acceleration if you already know the initial and final velocity and the duration of the trip. (It could be derived from presumably more basic kinematics equations for constant acceleration:
$x\_f-x\_i=v\_i \Delta t +\frac{1}{2}a\_i (\Delta t)^2$
and
$v\_f-v\_i=a\_i \Delta t$.)