5.12 Kinematics (Displacement, Velocity, Acceleration)

1. Fundamental Concepts

Consider a straight line and a fixed point $O$ on it. A body is moving on the line (forwards or backwards).

The displacement $s$ is $\pm$ the distance $|OA|$ from the fixed point $O$. The displacement $s$ (in meters) is given as a function of time $t$.

For example, let $s = t^2 - 4t + 3$. This means that:

  • at time $t=0$, the displacement is $3$ m from the fixed point $O$
  • at time $t=1$, the displacement is $0$ m (it goes back to point $O$)
  • at time $t=2$, the displacement is $-1$ m (the body is before point $O$)
  • at time $t=3$, the displacement is $0$ m (at point $O$ again)
  • at time $t=4$, the displacement is $3$ m (it is moving forward)

By definition, the rates of change are given by the derivatives:

  • Velocity = rate of change of displacement: $v = \dfrac{ds}{dt}$
  • Acceleration = rate of change of velocity: $a = \dfrac{dv}{dt}$

Notice also that $a$ is the second derivative of $s$. That is, $a = \dfrac{d^2 s}{dt^2}$.

Displacement ($s$) $\implies$ Velocity ($v$) $\implies$ Acceleration ($a$)
(Derivative)

2. Evaluating Motion

EXAMPLE 1

Consider $s = t^3 - 12t + 15$ representing the motion of a particle along a straight line, where the displacement $s$ is given in m (meters), and the time $t$ is given in sec (seconds).

Solution:
We find the functions for velocity and acceleration by taking derivatives: $$ \begin{aligned} v &= \dfrac{ds}{dt} = 3t^2 - 12 \\ a &= \dfrac{dv}{dt} = 6t \end{aligned} $$ Notice that $v$ is measured in m/sec while $a$ is measured in m/sec$^2$.

For example:
  • at time $t=1$: $s = 4$ m, $v = -9$ m/sec, $a = 6$ m/sec$^2$
  • at time $t=3$: $s = 6$ m, $v = 15$ m/sec, $a = 18$ m/sec$^2$
Notice also that the body is stationary when the velocity is 0.
Hence, let us solve $v = 0$: $$ \begin{aligned} v = 0 &\iff 3t^2 - 12 = 0 \\ &\iff t^2 = 4 \\ &\iff t = 2 \text{ sec} \end{aligned} $$ (notice that time is always positive, thus we reject $t = -2$).
At that time ($t=2$), $s = -1$ m, $v = 0$ m/sec, and $a = 12$ m/sec$^2$.

NOTICE on Signs

  • If $s > 0$, the body is to the right of the fixed point $O$.
  • If $s < 0$, the body is to the left of the fixed point $O$.
  • If $v > 0$, the body moves to the right (forward).
  • If $v < 0$, the body moves to the left (backward).
  • If $v = 0$, the body is stationary (it stops, usually to reverse direction).

3. Speed and Acceleration Analysis

Speed is the absolute value of the velocity. Thus, Speed = $|v|$.

We say that the body accelerates (its speed increases) when velocity and acceleration have the same sign (both positive or both negative). In this case $a \cdot v > 0$.

We say that the body decelerates (its speed decreases) when velocity and acceleration have opposite signs. In this case $a \cdot v < 0$.