# Time Diagram

One considers a point moving on a curve. One locates the position of this point on a date T , by his curvilinear X-coordinate S (T) . In Kinematic, the graph is called time diagram movement.

One makes use of it usefully for the crossings of the trains and the evaluation of the correspondences or shuntings.

Moreover, it is essential to include/understand well the difference between the data of v= F (T) or v=g (S).

## Crossings

### The TGV

it is a traditional example: the TGV makes Paris-Marseilles in 3:05 min with 05 min of stop in Avignon;

and the TGV Marseilles-Lyon-Paris in 3:10 min with 10min of stop in Lyon.

One wonders where and when cross train T1 and it T2 train knowing that T1 leaves in 15:00 and T2 leaves to 16:00.

Answer : the diagram shows that the stops do not play any part in this problem: the trains thus pass each other with 17:00, in Valence, city such as Paris-Valenc e= 480 km, Valence-Marseilles = 240 km.

There exists multitude of of the same problems left in the collections preparation with the Certificate of Primary Studies.

### Juggling

A case a little more difficult is this one:

A juggler vertically launches the B1 ball which goes up to the height H.

Other hand, it launches the B2 ball of an identical movement, just at the moment when the B1 ball starts to go down again:

where do the balls cross?

The answer is: with (3/4) H because the crossing will take place with half of the time of descent of the B2 ball (it is enough to plot the 2 time diagrams, to ensure itself some).

The kinematics of the Jonglerie is a pretty exercise of permutation between the various movements of hands and balls.

### Trains of Foucault

This example is famous, because it makes it possible to see " to turn Terre" (cf Pendulum of Foucault), without looking at stars, but simply by looking at a kinematic phenomenon interns with the reference frame Ground. To simplify the explanation, we will suppose the experiment made with the south pole S: on two circular ways, centered on the axis of the poles, two scooters of of the same snows circulate ABSOLUTE angular velocity, $\ omega_o$ (compared to stars, consequently), but one towards the East and the other towards the West. They cross in a point which derives continuously towards the East, and which makes 15° per hour, i.e. a turn per day. To be convinced, remake the reasoning in the Arctic of it, with the north pole. In addition, the traces of the 2 scooters will not be the same ones, because, compared to the ground of the Earth which turns, their speed is not the same one:

$\ omega_1: = \ omega_o + \ Omega_T; \ omega_2: = \ omega_o - \ Omega_T$

## Distinction v (T) and v (S)

The problem is not the same one on the road, if one raises
• speed in each place v (S) or
• speed at each time v (T).

### Case v (T)

In this second case (easiest), one plots curve v (T) of t=0 with to. The surface under the curve gives space traversed S (to):

Example: if speed increases v linearly (T) = At, displacement will be S (to) = to*ato/2 (surface of the right-angled triangle); either S (t0) = 1/2 has to^2: the movement known as is uniformly accelerated.

### Case v (S)

In this case, one speaks about Diagramme of spaces: it should be considered that, during a small way ds, time dt put to traverse it is dt = ds/v (S). All thus should be summoned the dt to obtain the duration of the course (of s=0 with so). It is thus necessary to trace 1/v (S), and to take the surface under the curve, which this time will be a time.

### Traditional example: the return ticket

Pierre makes the way go from has to B (AB = 10km) at the speed of 12km/h. To the return, more tired, it runs to 8 km/h. Question: will it arrive before or after Jacques who runs the return ticket to 10km/h?

The answer is: Pierre arrives after Jacques; indeed it ran less than one hour to 12km/h and more than one hour to 8km/h, therefore its average speed is lower than 10km/h.

This qualitative reasoning is with being controlled well: the harmonic mean speed is increasingly smaller than the arithmetic mean (cf Moyenne).

### Example of Torricelli

v (S) = sqrt (2gs).

The surface under the curve 1/sqrt (2gs) is sqrt (2so/g), therefore run time to is such as S (to) = so = 1/2 G to^2.

Historically, the first to make this calculation is Evangelista Torricelli (of Motu 1641). It is of course still about the uniformly accelerated movement. Galileo had hesitated to use a curve which left infinite at the beginning. Torricelli, raises initially of Castelli, then of Cavalieri, did not have these scruples and exceeded the Master.

### Relativistic example of Bertozzi

this example is of level definitely higher. It is the modification in restricted relativity of the example of Torricelli . One will not be astonished that, at the beginning of the movement, the result is close to that of Torricelli, but that at the end of the movement, v remainder limited by the maximum speed C.

The experiment was carried out by Bertozzi and gave the results anticipated by the relativity restricted with a remarkable precision.

v (Z) is given by the equation, said kinetic theorem of energy of Einstein:

$\ frac \left\{1\right\} \left\{\ sqrt \left(1-v^2/c^2\right)\right\} mc^2 = mc^2 +mgz$

One draws v (Z), and it " do not remain any more but with faire" calculations: one advises to use the graphic or numerical method; but those which have more mathematical skill will find the following expressions:

• # C ² /g (sqrt-1)

• dz/dt = v = gt/sqrt

who satisfy the preceding equation whose solution was single: it is thus the good one (it is a good exercise to check it, if one knows the operation of derivation).

What to note? if gt/c <<1, v= WP and Z = 1/2gt ²: Galileo was right.

But if T becomes large in front of c/g, speed v saturates with the maximum speed C and

X = ct - C ² /g + O (C ² /gt): Einstein rightly: the particle cannot be further that ct.

Note: erudite People will have recognized that if the " is posed; rapidité" R such as v = C HT R, then, it is the speed which grows linearly R = WP with time " propre" T, which is in conformity with the principle of relativity galiléenne (it is enough to reason at every moment T in the tangent reference frame galiléen: this very elegant solution indeed is of Einstein itself)

It is noted that T = c/g HS gT/c and thus at the end of T = 3c/g, T = ~ c/2g exp gT/c, i.e. the clock of the particle slows down T~ c/g Ln enormously (2gt/c)

finally one can check that Z = C ² /g -1

Thus a little strange world of the fast particles goes: it is checked thousands of times in the accelerators; Einstein simply improved Galileo and Newton: restricted Relativity is a reality, banal for which travels at a speed close to C.

But it is not very probable that the pointsmen of trains have never to make use of these formulas!

## Movement of Torricelli

It is historically the first periodic case of movement, theoretically being able to constitute a CLOCK. But Torricelli did not consider the practical realization of it: only the mathematical phenomenon interested it.

It is about the case: v^2 (S) = Vo^2-2.a. |S|.

To take the case where at initial time, the mobile M is in s=0, with speed +V0: it will move towards the line until S = S1 = sqrt (Vo^2/2a). This course will have taken time T1 (precisely that calculates in the example of the preceding paragraph: V0/2a = sqrt (2S1/a).

But the mobile does not stop there, like analyzed it well Galileo. Acceleration remaining negative, the mobile sets out again in the other direction, with same speed at the same points: thus it is right the same movement but in opposite direction, and the mobile is found at the origin at time 2t1, with speed - Vo. It remakes then towards the left exactly what occurred on the right. On the whole, the movement is periodic of period T = 4t1, and is composed of two uniformly accelerated movements.

In experiments, Galileo operated on two tilted levels forming a V; for practical reasons, the corner is reamed, and it is to better take a heavy ball which rolls without slipping, with a low rolling friction. " can; tricher" , to compensate for light damping, while inclining the track out of V gives rhythm some, so that S1 remains the same one.

If a ball rebounded in an elastic way on a perfect racket, one would have exactly the same type of clock, on the condition of controlling the movement of the racket (cf Problem of Fermi-Ulam Chaos controlled).

This is a very simple example of movement in a Puits of potential

## Movement of Kepler according to Leibniz

It is a very famous case of movement in a Puits of potential.

Since 1689, Leibniz knew to include/understand the radial movement of a satellite by writing ITS equation of the kinetic energy (at the time, one said equation of the lifeblood):

It is thus about the movement in a Puits of potential U (R), if Eo is negative.

The limits of this well are called SP = R minimum = distance perigee and maximum SA = R = distance apogee, roots of the equation U (R) = Eo.

That is to say Lo ² /2m (1/r) ² - mgR ² (1/r) - Eo = 0, quadratic equation in 1/r:

Leibniz noticed immediately that half the sum 1/2 (1/SA +1/SP), called mean harmonic and equal to 1/p is independent of Eo (rule 1 of Leibniz), and that sum (SA +SP) = 2a was independent of Lo (rule 2 of Leibniz): cf Keplerian Movement.

The equation rewritten with 2a and p becomes:

U (R) /m - Eo/m = Lo ² /2m ² (1/r ² - 2/pr + 1/pa) = -1/2 (dr/dt) ².

L'" astuce" usual in this kind of problem is to consider the variable phi such as r= has - c.cos$\ phi$, phi varying of 0 to pi while passing from the perigee to the apogee. Then integration is much easier, and gives celebrates it equation of time of Kepler (see Keplerian Mouvement:

By expressing the reciprocal function, one obtains phi (T) and thus R (T). (cf Keplerian Movement). Here $\ omega$ represents the pulsation of the periodic movement in the well. One finds the third law of Kepler:

$\ omega^2 \ cdot a^3 = gR^2$, independent of the eccentricity of the described elliptic trajectory.

The equation giving the polar angle is via the integration of the second law of Kepler: $C = \ frac \left\{\ pi has B\right\} \left\{T\right\} = r^2 \ dowry \left\{\ theta\right\}$

## See too

 Random links: Cybercafé | Catopsilia | Jean-Pierre Hemmen | Bburago | Conservancy Central Park | Le_vers_l'avant