5.8: Kinematic analysis with the path coordinate s

In some cases the velocity \(v_{s, i}(s)\), acceleration \(a_{s, i}(s)\) or time \(t_{s}(s)\) as a function of path coordinate \(s\) is known, but the explicit time dependence is unknown. In those cases it can be more convenient to choose \(s\) as an independent differentiation or integration variable than \(t\).

Note well, the function \(v_{s, i}(s)\) is mathematically different from the function \(v_{s, i}(t)\), and that also holds for and \(a_{s, i}(s)\) and \(a_{s, i}(t)\). These functions are connected by the relations \(v_{s, i}(s(t))=v_{s, i}(t)\) and \(a_{s, i}(s(t))=a_{s, i}(t)\) that should hold at any time. The difference by these functions will be designated by clearly indicating between brackets the variable on which they depend \(\left(v_{s, i}(t)\right.\) or \(\left.v_{s, i}(s)\right)\).

5.8.1 Path differentiation

If we take the time derivative of the function \(v_{s, i}(s)\) we obtain using the chain rule:

\[a_{s, i}(s)=\frac{\mathrm{d} v_{s, i}(s)}{\mathrm{d} t}=\frac{\mathrm{d} v_{s, i}(s)}{\mathrm{d} s} \frac{\mathrm{d} s_{i}}{\mathrm{~d} t}=\frac{\mathrm{d} v_{s, i}(s)}{\mathrm{d} s} v_{s, i}(s) \tag{5.43} \label{5.43}\]

This equation can be used to obtain the acceleration function \(a_{s, i}(s)\) from \(v_{s, i}(s)\).

5.8.2 Path integration

When the acceleration \(a_{s, i}(s)\) is known as a function of the path coordinate \(s\) instead of as a function of time, the following differential equation needs to be solved to obtain the velocity \(v_{s, i}(s)\) :

\[a_{s, i}(s)=\frac{\mathrm{d} v_{s, i}}{\mathrm{~d} t}=\ddot{s}_{i} \tag{5.44} \label{5.44}\]

After multiplying the equation on both sides by \(\mathrm{d} s\), and using that \(\frac{\mathrm{d} v_{s, i}}{\mathrm{~d} t} \mathrm{~d} s=\) \(\mathrm{d} v \frac{\mathrm{d} s_{i}}{\mathrm{~d} t}=v_{s, i} \mathrm{~d} v_{s, i}\) the differential equation can be solved by integration:

\[\int_{s_{1}}^{s_{2}} a_{s, i} \mathrm{~d} s=\int_{v_{1}}^{v_{2}} v_{s, i} \mathrm{~d} v_{s, i}=\left[\frac{1}{2} v_{s, i}^{2}\right]_{v_{1}}^{v_{2}}=\frac{1}{2}\left(v_{2}^{2}-v_{1}^{2}\right) \tag{5.45} \label{5.45}\]

Where \(v_{1}\) and \(v_{2}\) are the velocity of the point mass at the positions \(s_{1}\) and \(s_{2}\), \(v_{1}=v_{s, i}\left(s_{1}\right)\) and \(v_{2}=v_{s, i}\left(s_{2}\right)\). This equation can be rewritten in a form that allows finding \(v_{s, i}\) from \(a_{s, i}\) :

\[v_{s, i}^{2}\left(s_{2}\right)=v_{s, i}^{2}\left(s_{1}\right)+2 \int_{s_{1}}^{s_{2}} a_{s, i}(s) \mathrm{d} s \tag{5.46} \label{5.46}\]

We note that the relation \(a_{s} \mathrm{~d} s=v_{s} \mathrm{~d} v_{s}\) that appears under the integrals in Equation 5.45 is often useful to simplify expressions for differentiation and integration with the path coordinate \(s\).

5.8.3 Velocity dependent acceleration

A final case that can become of relevance, is when acceleration \(a_{s, i}(v)\) depends on velocity. To obtain the velocity as a function of time \(v_{s, i}(t)\) one needs to solve this differential equation:

\[\frac{\mathrm{d} v_{s, i}}{\mathrm{~d} t}=a_{s, i}(v) \tag{5.47} \label{5.47}\]

The strategy for solving this kind of ODE is to rearrange all terms such that terms depending on \(t\) are on one side of the equal side and terms depending on \(v\) are on the other side. Here that is done by multiplying both sides by \(\mathrm{d} t\) and dividing them by \(a_{s, i}\) and then integrating:

\[\int_{v_{1}}^{v_{s}} \frac{\mathrm{d} v_{s, i}}{a_{s, i}(v)}=\int_{t_{1}}^{t_{s}} \mathrm{~d} t=t_{s}-t_{1} \tag{5.48} \label{5.48}\]

By evaluating this integral and using the information on the initial condition at \(t_{1}\), one obtains a function \(t_{s}\left(v_{s}\right)\), which can be inverted to give the velocity as a function of time \(v_{s}\left(t_{s}\right)\). Alternatively, if one is looking for the function \(v_{s}(s)\) it is possible to multiply both sides of the differential equation (5.47) by \(\mathrm{d} s\), divide by \(a_{s, i}\) and integrate to obtain:

\[\int_{v_{1}}^{v_{s}} \frac{\mathrm{d} v_{s, i}}{a_{s, i}(v)} v_{s, i}=\int_{s_{1}}^{s} \mathrm{~d} s^{\prime}=s-s_{1} \tag{5.49} \label{5.49}\]

Using the initial condition this gives \(s\left(v_{s}\right)\) which can be inverted to obtain \(v_{s}(s)\).

5.8.4 Overview integration and differentiation in kinematics

After having discussed several ways to differentiate and integrate the kinematics when motion is along a one-dimensional path curve, we provide here an overview table with all equations for reference. The equations for integrating and differentiating between the different kinematic variables \(a, v, s\) and \(t\) as a function of each other are summarised in Table 5.1 below. Please note that the cases with \(t\) as dependent variable are usually the most important. Moreover, realise that it is better to learn how to properly derive these equations yourself by integrating and differentiation than to memorise equations.

5.8.5 Numerical differentiation and integration

Often, expressions are too complicated to differentiate or integrate using analytical equations. However in those cases the integration can usually still be performed numerically.

• Advantages of numerical methods:
• Can solve almost all problems
• No need to perform (difficult) analytical integration or differentiation
• Disadvantages of numerical methods:

	\(a_{s}()\)	\(v_{s}()\)	\(s()\)	\(t_{s}()\)
\(t\)	\(a_{s}(t)=\frac{d v_{s}(t)}{d t}\)	\(v_{s}(t)=\frac{\mathrm{ds}(t)}{\mathrm{d} t}\)	\(s(t)=t_{s}^{t^{-}}\)
\(t\)		\(v_{s}(t)=v_{s}\left(t_{1}\right)+\int_{t_{1}}^{t} a_{s}\left(t^{\prime}\right) \mathrm{d} t^{\prime}\)	\(s(t)=s\left(t_{1}\right)+\int_{t_{1}}^{t} v_{s}\left(t^{\prime}\right) \mathrm{d} t^{\prime}\)
s	\(a_{s}(s)=v_{s}(s) \frac{\mathrm{d} v_{s}(s)}{\mathrm{d} s}\)	\(v_{s}(s)=\left(\frac{\mathrm{d} t_{s}(s)}{\mathrm{d} s}\right)^{-}\)		\(t_{s}(s)=s^{\langle-1\rangle}(t)\)
\(s\)		\(v_{s}^{2}(s)=v_{s}^{2}\left(s_{1}\right)+2 \int_{s_{1}}^{s} a_{s}\left(s^{\prime}\right) \mathrm{d} s^{\prime}\)		\(t_{s}(s)=t_{s}\left(s_{1}\right)+\int_{s_{1}}^{s} \frac{d_{s} s^{\prime}}{\left.v_{s}\right)}\)
\(v\)	\(a_{s}(v)=\left(\frac{d t_{s}(v)}{d v}\right)^{-}\)		\(s(v)=v_{s}^{\langle-1\rangle}(s)\)	\(t_{s}(v)=v_{s}^{\langle-1\rangle}(t)\)
\(v\)	\(a_{s}(v)=v\left(\frac{\mathrm{d} s(v)}{\mathrm{d} v}\right)\)		\(\left.s(v)=s\left(v_{1}\right)+\int_{v_{1}}^{v} \frac{v_{s}^{\prime}\left(v_{v}\right.}{v^{\prime}}\right)\)	\(\left.t_{s}(v)=t_{s}\left(v_{1}\right)+\int_{v_{1}}^{v} \frac{d v^{\prime}}{\left.a_{s}\right)^{\prime}}\right)\)

Table 5.1: Equations to relate different kinematic quantities along the path curve \(a, v, s\) and \(t\) as a function of each other. The top row indicates the function and dependent variable, while the left column indicates the independent variable. For example, the function \(s(t)\) can be obtained in 2 ways: either by time integrating the function \(v_{s}(t)\) or by taking the inverse function of the function \(t_{s}(s)\). Note that \(t_{s}^{\langle-1\rangle}(s)\) in this table indicates the inverse function of \(t_{s}(s)\), not \(1 / t_{s}(s)\).

• Less accurate and fast than computing an analytic solution
• Programming required
• Gives less insight because only one specific solution is found, not the dependence on variables

To numerically determine the time derivative of a known function \(s_{A}(t)\), you determine the position \(s_{A}\) at two closely spaced points in time \(s_{A}(t)\) and \(s_{A}(t+\Delta t)\), with a small time difference \(\Delta t\). The simplest numerical differentiation in formula form is then performed using this equation:

\[v_{s, A}(t) \approx \frac{s_{A}(t+\Delta t)-s_{A}(t)}{\Delta t} \tag{5.50} \label{5.50}\]

This equation can be applied for all values of \(t\) to obtain \(v_{s, A}(t)\). According to the definition of the derivative, this expression becomes exactly correct in the limit \(\Delta t \rightarrow 0\) which allows improving the numerical accuracy as much as needed by reducing the time step \(\Delta t\).

If instead the velocity \(v_{s, A}(t)\) is known, the position \(s_{A}\) can be determined by numerical time integration by just rewriting the previous equation to:

\[s_{A}(t+\Delta t)=s_{A}(t)+v_{s, A}(t) \Delta t \tag{5.51} \label{5.51}\]

This equation can be applied iteratively to obtain \(s_{A}(t+\Delta t)\) at a next time from its value \(s_{A}(t)\) at a previous time. For example the next step is:

\[s_{A}(t+2 \Delta t)=s_{A}(t+\Delta t)+v_{s, A}(t+\Delta t) \Delta t \tag{5.52} \label{5.52}\]

This integration procedure, which is called the Euler forward method, can be iterated to get the complete numerical integral for \(s_{A}\) for all times. Reducing
the time step \(\Delta t\) helps to increase the accuracy. The method can also be applied to integrate acceleration and velocity vectors, by performing scalar integration along each of the axes in a Cartesian system like in eq. (5.57). There are multiple methods for numerically solving integrals and obtaining derivatives. An important integration method is Euler’s forward method, which is the simplest but by far not the most efficient and accurate method. To be sure of a result, it can be a good strategy to solve a problem both analytically and numerically. Then the numerical solution can function as a plausibility check for the analytical solution or vice versa.