5.7: Kinematic analysis as a function of time t
We will show how the motion, velocities and accelerations of a system can be determined by solving equations of motion using time differentiation and integration.
5.7.1 Time differentiation
We have already shown how the velocity and acceleration vectors can be determined from the position vector by two time differentiation steps that are indicated by arrows:
\[\begin{align} \overrightarrow{\boldsymbol{r}}_{i} & \rightarrow \tag{5.27} \label{5.27}\\[4pt] \overrightarrow{\boldsymbol{v}}_{i} & =\dot{\overrightarrow{\boldsymbol{r}}}_{i} \rightarrow \tag{5.28} \label{5.28}\\[4pt] \overrightarrow{\boldsymbol{a}}_{i} & =\dot{\overrightarrow{\boldsymbol{v}}}_{i} \tag{5.29} \label{5.29}\end{align}\]
And the same can be done along the path coordinate \(s\), simplifying the problem to scalar differentiation:
\[\begin{align} s_{i}(t) & \rightarrow \tag{5.30} \label{5.30}\\[4pt] v_{s, i}(t) & =\dot{s}_{i}(t) \rightarrow \tag{5.31} \label{5.31}\\[4pt] a_{s, i}(t) & =\dot{v}_{s, i}(t)=\ddot{s}_{i}(t) \tag{5.32} \label{5.32}\end{align}\]
During each of these steps, an ordinary differential equation (ODE) is solved, which is of the form of Equation 2.12, \(\frac{\mathrm{d} f(x)}{\mathrm{d} x}=y(x)\), where \(x\) is replaced by \(t, f(x)\) is replaced by the known function \(s(t)\) or \(v_{s}(t)\) and \(y(x)\) is replaced by the unknown function \(v_{s}(t)\) or \(a_{s}(t)\) that we would like to determine. The details
of the procedure to solve these first order differential equations using either indefinite or definite integrals is explained in Sec. 2.4.
5.7.2 Time integration
To go in the opposite way, from acceleration to velocity and position, an equation of motion needs to be solved to obtain \(s_{i}(t)\) from \(a_{s, i}(t)\), which is of the form:
\[\ddot{s}_{i}=a_{s, i}(t) \tag{5.33} \label{5.33}\]
Because the function \(a_{s, i}(t)\) only depends on time, this differential equation can be solved by integrating twice over time. Here we show how that is done by integration along the path curve, and in Equation 5.57 it will be shown how it can be done in free space. Although these integrals can become difficult analytically, they can always be performed using numerical methods as will be discussed concisely in section 5.8.5.
If the path acceleration function \(a_{s, i}(t)\) is known, it is possible to determine the path velocity and path position by integration of Eqs. (5.30)-(5.32) as follows.
\[\begin{align} a_{s, i}(t) & \rightarrow \tag{5.34} \label{5.34}\\[4pt] v_{s, i}\left(t_{2}\right) & =v_{s, i}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} a_{s, i}(t) \mathrm{d} t \rightarrow \tag{5.35} \label{5.35}\\[4pt] s_{i}\left(t_{2}\right) & =s_{i}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} v_{s, i}(t) \mathrm{d} t \tag{5.36} \label{5.36}\end{align}\]
During each of the integration steps, an ODE is solved of the form of Equation 2.13. From equations like Equation 5.35 it is noted that time integration to obtain final velocity and position can only be performed if the position \(s_{i}\left(t_{1}\right)\) and velocity \(v_{s, i}\left(t_{1}\right)\) are known at a certain instant \(t_{1}\). That is why one needs to known the initial conditions to solve the equations of motion.
To illustrate the kinematic analysis in time, let’s consider the motion of a car along a track (Figure 5.7), with path curve \(\overrightarrow{\boldsymbol{r}}_{s}(s)=s \hat{\boldsymbol{\imath}}\) and path coordinate \(s_{A}=x_{A}\). It is given that the position of the car \(A\) on the track is:
\[s_{A}(t)=\alpha t^{3} \tag{5.37} \label{5.37}\]
Then we can obtain the velocity and acceleration by differentiation using Eqs. (5.30)-\((5.32)\):
\[\begin{align} v_{s, A}(t) & =\dot{s}_{A}=\alpha \frac{\mathrm{d}}{\mathrm{d} t} t^{3}=3 \alpha t^{2} \tag{5.38} \label{5.38}\\[4pt] a_{s, A}(t) & =\dot{v}_{s, A}=\alpha \frac{\mathrm{d}}{\mathrm{d} t} 3 t^{2}=6 \alpha t \tag{5.39} \label{5.39}\end{align}\]
So the velocity is the slope (=time-derivative) of the \(s_{A}(t)\) graph and the acceleration is the slope of the \(v_{s, A}(t)\) graph, as is shown by the arrows in Figure 5.8.
Suppose now that instead, the acceleration of the car \(a_{s, A}(t)=6 \alpha t\) is given and we want to determine the distance travelled after a certain time. It is given that the car starts at \(t_{1}=0\) from rest, such that initial velocity and position of the car are both zero \(\left(v_{s, A}\left(t_{1}\right)=s_{A}\left(t_{1}\right)=0\right)\). From the acceleration the functions that describe velocity and position at an arbitrary time \(t_{2}\) can be obtained using equations \((5.34)-(5.36)\) :
\[\begin{align} a_{s, A}(t) & =6 \alpha t \tag{5.40} \label{5.40}\\[4pt] v_{s, A}\left(t_{2}\right) & =v_{s, A}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} 6 \alpha t \mathrm{~d} t=0+\left[3 \alpha t^{2}\right]_{t_{1}}^{t_{2}}=3 \alpha t_{2}^{2} \tag{5.41} \label{5.41}\\[4pt] s_{A}\left(t_{2}\right) & =s_{A}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} 3 \alpha t^{2} \mathrm{~d} t=0+\left[\alpha t^{3}\right]_{t_{1}}^{t_{2}}=\alpha t_{2}^{3} \tag{5.42} \label{5.42}\end{align}\]
So the velocity change is the area (=time-integral) under the \(a_{s, A}(t)\) graph and the position change is the area (=time-integral) under the \(v_{s, A}(t)\) graph, as is shown by the red areas in Figure 5.8.
Comparing the first 3 equations of this example to the last 3 equations demonstrates that the time differentiation and integration results in the same equations for position, velocity and acceleration, as expected.