Kinematics

Chapter 1 of Dynamics and Dynamical Systems

Kinematics is the analysis of motion without consideration of their causes. In the case of the runners in the picture, the motion of their arms and legs is oscillating, whereas the movement of the head is more nearly uniform and therefore easier to describe: it would be enough to know the horizontal displacement of the head versus time. To describe the motion of the legs, in addition to considering the horizontal displacement, it would also be necessary to consider the variation of some angle versus time.

1.1. Motion of rigid bodies
An object is in motion if its position changes at different instants; if the position remains constant, the object is at rest. To measure the position of the object, a reference frame must be used, namely, other objects used as reference. If the position of a body changes, relative to the reference frame, the body is in motion relative to that frame. Thus, motion is a relative concept since an object may be at rest relative to a given reference frame, but in motion relative to another reference frame.



The simplest type of motion of a rigid body, a translation without rotation, occurs when all points on the body follow similar trajectories (see figure). In that case, to describe the motion of the whole body it will be enough to describe the motion of just one point in it.

In the case of rotation around a fixed axis, all points in the axis remain at rest while the other points move. In the second part of the figure, the hammer rotates around an axis perpendicular to the page. In this type of motion the trajectories of different points are no longer identical but all are circular arcs with the same angle, differing only in the value of the radius. To describe this motion, it will be enough to know the value of the rotation angle at every instant.

Any other motion of the body, on the plane of the figure, will always be the superposition of a translation and a rotation around a fixed axis (third part of the figure). In that case, different parts of the body follow different trajectories. A convenient way to study that type of motion consists in studying the motion of a fixed point on the body and the rotation around an axis through that point; the rotation angle will the same independently of the point chosen as reference; changing to a different reference point would change the value of the radii of other points, relative to the axis, but would not change the rate of change of the angle.

There is a more general type of rotation, around a single fixed point, where there is only one point that remains at rest. In that case the trajectories of each point in the body would not be a circular arc but any curve on the surface of sphere centered at the point at rest. The most convenient way to study this type of motion is to determine the variation of three angles. The most general type of motion of a rigid body consists of the superposition of translation and rotation around a point. In that case it will be necessary to describe the motion of a fixed point on the body and the variation of three angles.



1.2. Motion and degrees of freedom
The degrees of freedom of a system are the variables needed to measure its exact position. For example, to determine the position of a fly in a "rectangular" room, one could measure its distance to the floor and two walls perpendicular to the room. Those three distances are the three coordinates (Cartesian or rectangular) of the fly, which are usually designated by the letters $x$, $y$ and $z$.

Thus, the motion of a point is a motion with three degrees of freedom. The trajectory of the body is a curve in space, which can be described by giving the expressions for the 3 Cartesian coordinates as functions of time. Since the most general motion of a rigid body is the superposition of the motion of a point and a rotation around that point, this movement has six degrees of freedom: three coordinates describing the motion of the point, plus three angles which describe the rotation. Other more simple types of motion have less degrees of freedom; rotation about an axis has only one degree of freedom, the translational has three degrees of freedom and a translation plus rotation about a fixed axis has four degrees of freedom.

This chapter is restricted to the motion of a point. This study will serve as the basis for the study of the more general motion of rigid bodies later on.

When a point is restricted to follow a predetermined path, the motion of that point will have a single degree of freedom. For example, the displacement of cart on the rails of a roller coaster, disregarding the inclination of the cart, is a motion with one degree of freedom. If the position of the cart in an initial instant is known, to determine the position at any later moment it will be enough to know the displacement along the rails from the initial time until that moment.



The translational motion of a car on a highway can also be considered a motion with one degree of freedom (see the illustration). If the car breaks down and the driver has to call a towing truck, it will suffice to tell the truck driver the kilometer of the highway where the car stopped. Thus, the motion of the cars on a motorway is characterized by a single degree of freedom, the displacement along the road.

Note that the distance traveled on the road is not measured in a straight line, but along a curve in space, however, as the detailed shape of the curve is already established, only one variable is needed to describe the position at every moment. If we were to build a system for automatic driving, we would have to introduce another variable, for example, the distance to the side of the road and the motion would then have two degrees of freedom. These two degrees of freedom only describe the trajectory of a point of the car, if we are interested in a complete description of the position of the car, considering as if it were perfectly rigid, it would take three more degrees of freedom.

If a point is restricted to move over a surface, only two coordinates are necessary to determine its position, so its motion has two degrees of freedom.



If a biologist were following the motion of a fox on a territory, he would only need to register its longitude and latitude, for example with a GPS device, at various instants. Two coordinates, rather than three, are enough when the topographic map of the region is known, as in the figure on the side, making it possible to locate a point from its longitude and latitude; the altitude of the points on the surface of the terrain has a predetermined value in accordance with the topography of the terrain. The third coordinate, the height above the ground surface, would only become important if it had significant variations compared with the variations of latitude and longitude.

Consequently, the motion of the fox is a motion with two degrees of freedom, because two coordinates are sufficient to determine its position. The latitude and longitude on the surface of the ground are not really distances but angles with their vertexes at the center of the Earth; nevertheless, they constitute two degrees of freedom that can have different values ​​at different times.

Returning to the example of the flight of a fly, which was considered as a single point with 3 coordinates $x$, $y$ and $z$, the fly can also change its orientation. To determine the orientation of a straight line across the fly's body, two angles can be used and another angle would be needed to indicate the rotation of the fly relative to that line. That increases the number of degrees of freedom to six. But the fly may also stretch or bend its body and open or close its wings, so from a physical point of view it has many more degrees of freedom. If the fly is modeled with three rigid bodies: the two wings and the block consisting of head, thorax and abdomen, to describe the motion of the first rigid body — head, thorax and abdomen — six degrees of freedom are needed. Each wing adds other three degrees of freedom — the angles of rotation around a fixed point where the wing is connected to the thorax — giving a total of 12 degrees of freedom.

1.3. Speed
In the rest of this chapter only motions with one degree of freedom will be considered; the total distance traveled along the trajectory, from an initial instant, will be represented by the variable $s$. It is important not to confuse $s$ with the displacement. If, for example, at some instant the position is the same as at the initial time, the displacement from the initial moment will be zero because the position has not changed; however, the total distance traveled will not necessarily be zero.

The variable $s$ is always positive and can not decrease over time. For example, for a point that moves along the $x$ axis, imagine that the starting position at $t_0$ was $x$ = 2 and that the point moved in the negative direction until it stopped in $x$ = -1 at the instant $t_1$, starting then to move in the positive direction of the $x$ axis reaching the position $x$ = 4 at $t_2$. The value of the total distance traveled $s$ in the 3 instants would be $s_0$ = 0 in $t_0$, $s_1$ = 3 in $t_1$ and $s_2$ = 8 in $t_2$. It was assumed that the motion started at $t_0$ and therefore, the distance traveled started at zero, but $s_0$ could have any other initial value. As already mentioned, a motion with a single degree of freedom does not have to be along a straight path, as in the previous example, but it could be along a curvilinear path.

The average speed, in a time interval between $t_i$ and $t_j$, is defined the as the total distance traveled in that interval, per unit time:

\begin{equation} \bar{v}_{ij} = \frac{s_j - s_i}{t_j - t_i} \end{equation}

where $t_j > t_i$. Since the total distance traveled during any time interval will always be positive or zero, $\bar{v}_{ij}$ will also be either positive or zero. The units of speed are those of distance over time; for example, meters per second, m/s, kilometers per hour, km/h, etc.

Example 1.1
A driver recorded the total distance traveled on a road at various instants, obtaining the values ​​in the following table:

Determine the average speed in each half-hour interval and plot the graphs of distance traveled and average speed.

Solution: Being $t_1$, $t_2$, ..., $t_5$, the 5 instants indicated in the table, the average speeds at the four intervals are:


 * $$ \bar{v}_{12} = \frac{60 - 0}{0.5 - 0} = \frac{60}{0.5} = 120\;\mathrm{\frac{km}{h}}$$


 * $$\bar{v}_{23} = \frac{90 - 60}{1 - 0.5} = \frac{30}{0.5} = 60\;\mathrm{\frac{km}{h}}$$


 * $$\bar{v}_{34} = \frac{10}{0.5} = 20\;\mathrm{\frac{km}{h}}$$


 * $$\bar{v}_{45} = \frac{40}{0.5} = 80\;\mathrm{\frac{km}{h}}$$

In the last two intervals $\bar{v} = \Delta s/\Delta t$ was written directly, where $\Delta$ represents the change in the corresponding variable in the interval considered.

To plot the graph of distance versus time, one can use the program Maxima (see appendix A). It is convenient to start by storing the values ​​of time and distance in a list and then use the function plot2d:

(%i1) s_t: 0,0], [0.5,60], [1,90], [1.5,100], [2,140$ (%i2) plot2d([discrete,s_t],[style,linespoints],[xlabel,"t (h)"],[ylabel,"s (km)"])$

The graph is as follows:

To plot the graph of the average speed versus time, one has to decide in which instant on each interval the corresponding average speed should be represented. Should it be placed at the beginning or end of the interval? We will simply represent each average speed at the midpoint of the corresponding interval, as shown in the following figure.

(%i3) v_t: 0.25,120], [0.75,60], [1.25,20], [1.75,80$ (%i4) plot2d([discrete,v_t],[x,0,2],[y,0,150],[style,linespoints],               [xlabel,"t (h)"],[ylabel,"v (km/h)"])$



The value of the average speed in an interval does not give precise information about the motion in that interval. For example, in the previous example, during the interval from $t$ = 1 to $t$ = 1.5 the average speed had its lowest value of 20 km/h. This does not imply that during this half hour interval the driver was always driving at that speed. He might have traveled faster up to a point where he stopped for a while before continuing the journey.

To determine more precisely the kind of motion, it is necessary to know the variations $\Delta\,s$ of the distance traveled at smaller time intervals $\Delta\,t$. The more accurate information will be obtained in the limit when $\Delta\,t$ approaches zero.

In order to obtain the speed at an instant $t$, one must determine the average speed in the interval from $t$ until a later instant $t+\Delta\,t$, and then calculate its limit when $\Delta\,t$ approaches zero:

\begin{equation} v(t) = \lim_{\Delta t\rightarrow 0} \dfrac{\Delta\,s}{\Delta\,t} \end{equation}

This limit is the derivative of $s$ with respect to $t$, which is represented by:

\begin{equation} \boxed{ v = \dfrac{\mathrm{d}\,s}{\mathrm{d}\,t} }\label{eq:vel} \end{equation}

An alternative notation for the derivative with respect to time is $v = \dot{s}$, where a dot above a variable indicates differentiation with respect to $t$.

In a car, the speed at each instant is given with good approximation by the speedometer. It may not give the exact value of the instantaneous speed, since it has a minimum response time. In a speedometer of good quality, with very low response time, or in situations where the speed does not undergo abrupt changes, it is reasonable to assume that the speedometer indicates the exact instant speed.

1.4. Acceleration
The acceleration along the trajectory is defined as the increase in speed per unit time. One may begin by defining an average value for a time interval, as was made in the case of the average speed; the average acceleration along the trajectory, $\bar{a}_{ij}$, in the time interval a time interval $\Delta\,t = t_j - t_i$ (greater than zero), is defined by:

\begin{equation} \bar{a}_{ij} = \dfrac{v_j - v_i}{t_j - t_i} = \dfrac{\Delta v}{\Delta t} \end{equation}

The units of acceleration are units of distance over time squared. For instance, meters per square second, m/s2.

The acceleration along the trajectory, in an instant $t$, is defined as the average acceleration in an infinitesimal time interval following that instant.

\begin{equation} a_\mathrm{t}(t) = \lim_{\Delta\,t\rightarrow 0} \dfrac{\Delta v}{\Delta\,t} \end{equation}

Using the notation that a dot over the variable represents the derivative of this variable with respect to time:

\begin{equation} \boxed{a_\mathrm{t} = \dot{v} = \ddot{s}} \label{eq:acel} \end{equation}

where the two points indicate the second derivative with respect to time.

Note that the distance traveled, $s(t)$, is a function of time, always positive and increasing or constant. Thus, the first derivative, $\dot{s} = v$, will always be positive but the second derivative, $\ddot{s} = a_\mathrm{t}$, may have any sign. An acceleration along the trajectory with negative value implies speed decrease with time (the object is slowing down) and if the value is positive, the speed will increase (the object is speeding up). A null acceleration along the trajectory implies constant speed.

The use of the term "acceleration along the trajectory", instead of just acceleration, is because as will be discussed in another chapter, the acceleration has another component perpendicular to the trajectory, which is not related to the change in speed but with the curvature of the trajectory.

Example 1.2
A ship is initially stopped on a channel; in the instant $t=0$ the engine is turned on for 5 minutes and then turned-off again, letting the boat to slow down to a halt by the resistance of the water. In SI units, the expression for the speed in terms of time $t$ is


 * $$v = \left\{ \begin{array}{ll} 12\,\left(1-\mathrm{e}^{-0.06\,t}\right), & 0\leq t \leq 300\\ 12\,\left(1-\mathrm{e}^{-18}\right)\,\mathrm{e}^{-0.06(t-300)}, & t \geq 300 \end{array} \right.$$

Find the expressions of the acceleration along the trajectory and the distance traveled, as functions of time. Plot the graphs of the speed, acceleration and distance traveled as functions of time. Calculate the total distance that the ship travels during the time that the engine was turned on and until it stops.

Solution: First, note that the expression given for the speed is continuous, as it should be, since the speed can not change abruptly at any moment. The acceleration along the trajectory is calculated by differentiating the expression of the speed. To do the calculations using Maxima, one can start by saving the two expressions for the speed in two different variables

(%i5) v1: 12*(1-exp(-0.06*t))$ (%i6) v2: 12*(1-exp(-18))*exp(-0.06*(t-300))$

The differentiation is done using the command diff

(%i7) a1: diff (v1, t); - 0.06 t (%o7)                    0.72 %e (%i8) a2: diff (v2, t); - 18   - 0.06 (t - 300) (%o8)       - 0.72 (1 - %e    ) %e

Thus, the expression for the acceleration along the trajectory is


 * $$a = \left\{ \begin{array}{ll} 0.72\,\mathrm{e}^{-0.06\,t}, & 0\leq t < 300 \\ -0.72\,\left(1-\mathrm{e}^{-18}\right)\,\mathrm{e}^{-0.06(t-300)}, & t \geq 300 \end{array} \right.$$

Note that the acceleration does not need to be continuous; in this case there is a discontinuity in $t=300$ s, where the acceleration changed from a positive value to the value -0.72 m/s2, because the engine is switched off suddenly at that instant. To obtain the expression for the distance traveled, it is necessary to integrate the expression for the speed from the initial instant.


 * $$ s(t) = \int_0^t v(t)\,\mathrm{d}t = \left\{\begin{array}{ll} \displaystyle\int_0^t 12\,\left(1-\mathrm{e}^{-0.06\,t}\right)\,\mathrm{d}t, & 0\leq t \leq 300\\ s(300) + \displaystyle\int_{300}^t 12\,\left(1-\mathrm{e}^{-18}\right)\,\mathrm{e}^{-0.06(t-300)}\,\mathrm{d}t, & t \geq 300 \end{array} \right.$$

In Maxima, these two integrals are calculated in the following way

(%i9) s1: integrate (v1, t, 0, t); 3 t       3 t                      - ---        --- 50        50                    %e      (3 t %e    + 50)   50 (%o9)         12 ( - --) 3              3  (%i10) s2: float (subst (t=300, s1)) + integrate (v2, t, 300, t); 3 t                                  18 - --- 50                 - 18   50   50 %e (%o10) 12 (1 - %e   ) (-- - -) + 3400.000003045996 3         3

These results can be written in a more compact form


 * $$s = \left\{ \begin{array}{ll} 4\,\left(3\,t + 50\,\mathrm{e}^{-0.06\,t} - 50\right), & 0\leq t \leq 300 \\ 3400 + 200\,\left(1-\mathrm{e}^{-18}\right)\,\left(1 - \mathrm{e}^{(18-0.06\,t)}\right), & t \geq 300 \end{array} \right.$$

To plot the graphs shown in the figure, the following commands were used:

(%i11) plot2d(if t<300 then v1 else v2,[t,0,400],[ylabel,"v"],           [y,0,14])$ (%i12) plot2d(if t<300 then a1 else a2,[t,0,400],[ylabel,"a"])$ (%i13) plot2d(if t<300 then s1 else s2,[t,0,400],[ylabel,"s"])$

The plots above provide useful information that might not be evident in the algebraic expressions. The speed graph shows that the ship's speed quickly reaches a limit of approximately 12 m/s during the first minute, and continues with almost constant speed until the instant when the motor is switched off; from that instant, the speed decreases rapidly, and in $t=360$ s (6 minutes) it is practically zero. The exponential nature of the expression for the speed implies that, in theory, the speed would never be exactly zero. In practice, this expression for the speed can no longer be valid when the obtained value is very small; for example, in $t= 400$ s the speed obtained with this expression is

(%i14) float (subst (t=400, v2)); (%o14)                .02974502566698016

about 3 centimeters per second. There are other phenomena such as wind currents in the water and waves on the water surface, which may produce speed fluctuations greater than this value. The expression given for the velocity must have been obtain from a model that can only be valid when the values ​​exceed the effects of all these fluctuations.

In the graph of acceleration, the discontinuity in $t=300$ s appears as a continuous straight line, due to the fact that Maxima's program plot2d does not detect the discontinuity at that point and treats both parts of the graph as a single continuous function. The graph for the distance traveled shows a linear increase in almost the entire interval of the first 5 minutes and a quick stop after those first few minutes. The distance traveled while the engine was turned on is $s(300)$, Which has been already calculated in (%i10), and it is approximately 3.4 km


 * $$s(300) = 4\,\left(850 + 50\,\mathrm{e}^{-18}\right) \approx 3400$$

according to the theoretical model, the ship would take an infinite time to stop; in practice, it takes only a little more than 6 minutes, as already discussed. To calculate the total distance traveled until it stops, using the theoretical model, it will necessary to calculate the limit when $t$ goes to infinity; that is done by replacing by zero the exponential function that appears in the expression for $s(t)$. In Maxima, this limit can be calculated as:

(%i15) float (limit (s2, t, inf)); (%o15)                      3600.0

Hence, the total distance traveled is 3.6 km.

1.5. Kinematic equations
The differential equations \eqref{eq:vel} and \eqref{eq:acel} obtained in the previous two sections are the kinematic equations that relate the three kinematic variables $s$, $v$, $a_t$ and the time $t$. Given a mathematical relation for any of those three variables as a function of time, The expressions for the other two variables may be obtained by solving the kinematic equations, as in exemplo 1.2.

In the case when it is given an expression for the speed in terms of the distance traveled $s$, the derivative of the speed with respect to time shall be calculated using the chain rule for implicit functions:

\begin{equation} a_\mathrm{t} = \dfrac{\mathrm{d}\,v}{\mathrm{d}\,t} = \dfrac{\mathrm{d}\,v}{\mathrm{d}\,s}\dfrac{\mathrm{d}\,s}{\mathrm{d}\,t} = \dfrac{\mathrm{d}\,v}{\mathrm{d}\,s}\,\dot{s} = v\,\dfrac{\mathrm{d}\,v}{\mathrm{d}\,s} \end{equation}

This is another kinematic equation. To summarize, there are four kinematic equations:

\begin{equation} \boxed{ v=\dot{s} \qquad a_\mathrm{t}=\dot{v} \qquad a_\mathrm{t}=\ddot{s} \qquad a_\mathrm{t}=v\,\dfrac{\mathrm{d}\,v}{\mathrm{d}\,s} }\label{eq:eq_mov} \end{equation}

and four variables: $t$, $s$, $v$ and $a_\mathrm{t}$. Each of these equations relates three of the four kinematic variables. In order to solve any of these first-order differential equations using traditional analytical methods, it is necessary to know a relation among the three variables in the equation in order to eliminate one of them, leaving an ordinary differential equation with only two variables.

For example, the equation $v=\dot{s}$ relates the three variables $v$, $s$ and $t$ (the dot implies that $t$ is involved in the equation). To solve this equation it would be necessary to know an expression that relates two or three of the variables $v$, $s$ and $t$, in order to eliminate one of them in the equation $v=\dot{s}$.

1.5.1 Motion along an axis
In some cases it is more convenient to work with the position instead of the distance traveled. To measure the position along the path, an origin and a positive direction along the path must be chosen. The position is indicated by a coordinate $x$ which may be positive, negative or zero. This coordinate can be measured along a rectilinear axis ($x$ axis) which does not match the trajectory of the point and, if so, $x$ indicates the position of the projection of the point on the $x$ axis. But it is also possible to use $x$ to represent the position measured along the path of the point and in this case, the $x$ axis may be a curve instead of a straight line.

The derivative of the $x$ coordinate with respect to time gives the $x$ component of the velocity, $v_x$, which can also have any sign and the derivative of $v_x$ with respect to time gives the $x$ component of the acceleration, $a_x$. The sign of $a_x$ will not be enough to conclude whether the point is speeding off or slowing down, since it will be necessary to take into account the sign of $v_x$ too.

In terms of the components along the axis, the kinematic equations have the same form as equations \eqref{eq:eq_mov}

\begin{equation} \boxed{ v_x=\dot{x} \qquad a_x=\dot{v}_x \qquad a_x=\ddot{x} \qquad a_x=v_x\,\dfrac{\mathrm{d}\,v_x}{\mathrm{d}\,x} }\label{eq:eq_mov_x} \end{equation}

The relation among of components of the velocity and acceleration and the speed and acceleration along the trajectory is:

\begin{equation} v=|v_x| \qquad a_\mathrm{t} = a_x\quad(\mbox{se }v_x>0) \qquad a_\mathrm{t} = -a_x\quad(\mbox{se }v_x<0) \end{equation}

1.5.2 Acceleration of gravity
Near the Earth's surface, all objects falling freely have the same acceleration, called acceleration of gravity and represented by the letter $g$. The value of $g$ has slightly different values at different geographical locations, but it is always very close to 9.8 m/s2. Air resistance produces another acceleration opposite to the velocity, but when this resistance is small, it is assumed that the value of the acceleration is constant and equal to $g$.

The acceleration along the trajectory due to gravity may be positive, negative or zero, as it can increase or decrease the speed of the object, and may have a value different from $g$ if the path is not vertical. But if the axis $y$ is set vertically and upward pointing, the $y$ component of the acceleration (acceleration for the vertical projection of the motion) will always have the constant value $a_y$ = -9.8 m/s2 (or +9.8 if the positive direction of the $y$ axis is downwards).

Example 1.3
A stone is thrown up from a bridge which is 5 m above a river; the vertical component of the speed with which the stone is released is equal to 9 m/s. The stone eventually sinks into the river. Calculate the speed with which the stone hits the surface of the river and the maximum height attained by it, measured from the surface of the river (assume that air resistance can be neglected).

Solution: Choosing the axis vertically, pointing up and with origin at the surface of the river, the starting position is $y_0$ = 5 and the $y$ component the acceleration is $a_y$ = -9.8 (SI units). Hence, we have an expression for the acceleration versus time: $a_y$ é is a constant function with value -9.8.

This example could be solved by integrating the expression for the acceleration to find the expression for speed versus time and integrating again to find the height versus time; the time when the height is zero could then be calculated (moment of the impact with the river surface) and substituted into the expression for the speed to get the value of the speed at that instant. However, there is a more direct way to obtain the impact velocity, using a more general method (\ keywordi separation of variables {}) to be useful in other cases it is not possible to directly integrate the expression for the acceleration.

The initial step of the method consists in replacing the expression for the acceleration (in this case $a_y$ =-9.8) In any of the kinematic equations \eqref{eq:eq_mov_x} (using $y$ instead of $x$), to obtain an equation with only two variables. Since the problem asks for $v$ from the given initial height $y_0$ and an expression for $a_y$ is known, $a_y$ =-9.8 should be replaced in the differential equation that relates the variables $a_y$, $v_y$ and $y$:


 * $$-9.8 = v_y\,\dfrac{\mathrm{d}\,v_y}{\mathrm{d}\,y}$$

which is an ordinary differential equation with variables, $y$ and $v_y$.

Next, the derivative in the above equation is regarded as a quotient between $\mathrm{d}\,v_y$ and $\mathrm{d}\,y$ and the terms that depend on $y$ are grouped on one side of the equation while the terms that depend on $v_y$ are grouped on the other side.


 * $$-9.8\,\mathrm{d}\, y = v_y\,\mathrm{d}\, v_y$$

It is said that the variables were separated on both sides of the equation. Once separated the variables, the two sides of the equation are integrated and the limits for the integrals can be given: the height varies from to $y_0$ = 5 to $y$ = 0 (limits of integration for $\mathrm{d}\,y$); on the right side, the $y$ component of the velocity varies from 9 until a value we want to calculate, $v_y$ and therefore, it is written in the limit of the integral as an undetermined variable:


 * $$-\int_{5}^{0} 9.8\,\mathrm{d}\,y = \int_{9}^{v_y} u\,\mathrm{d}\,u$$

wherein the variable of integration on the right side has been replaced by $u$ to avoid confusion with the upper limit of the integral.

The integrals are easy to calculate, but can also be obtained using Maxima (integrate(-9.8,y,5,0), integrate(u,u,9,vy)). The result obtained is:


 * $$9.8\times 5 = \dfrac{v_y^2}{2} - \dfrac{81}{2} \qquad \Longrightarrow \qquad v_y = -\sqrt{98 + 81}$$

(The second solution of the equation, $+\sqrt{98+81}$, corresponds to the speed with which the stone should have left the water surface, to pass through the bridge with a component of the velocity equal to 9 m/s upwards).

Therefore, the vertical component of the speed with which the stone enters the river is $v_y$ = -13.38 m/s. To determine the maximum height, notice that at the point where the stone completes its rise and begins to fall, the vertical component of its velocity should be zero. Thus, the same integrals calculated above can be repeated, but leaving the final height undetermined, $y$, while the component of the final velocity is set to 0:


 * $$-\int_{5}^{y} 9.8\,\mathrm{d}\, y = \int_{9}^{0} v_y\,\mathrm{d}\, v_y$$

the result is:


 * $$9.8(5-y) = - \frac{81}{2}\qquad \Longrightarrow \qquad y=9.13$$

The maximum height is 9.13 m. It is noteworthy that while the stone rises and falls, it may also be moving away from the bridge horizontally; the calculations were made for the projection of motion in the vertical plane and not for the motion along the path of the stone.

The previous example could have been solved using equations that are only valid motion for constant acceleration, in particular the equation $v_y^2=v_0^2 - 2\,g(y-y_0)$, but it is not worth memorizing and using this equation, which is valid only if the acceleration is constant and can be obtained easily integrating $-g\,\mathrm{d}\,y = v_y\,\mathrm{d}\,v_y$. It is advisable to start always from the kinematic equations, with the values ​​known and using the method of separation of variables.

In some differential equations it is impossible to separate the variables; in such cases there are other resolution techniques, but there are no general analytical methods for any equation. The approach used in this book will be to use numerical methods for obtaining approximate solutions when the method of separation of variables can not be used.

Example 1.4
Cinemática