One Dimensional Kinematics

The First Step: Choosing Coordinates

Before beginning a problem in kinematics, you must set up your coordinate system. In one-dimensional kinematics, this is simply an x-axis and the direction of the motion is usually the positive-x direction. Though displacement, velocity, and acceleration are all vector quantities, in the one-dimensional case they can all be treated as scalar quantities with positive or negative values to indicate their direction.

The positive and negative values of these quantities are determined by the choice of how you align the coordinate system.

Velocity

Velocity represents the rate of change of displacement over a given amount of time. The displacement in one-dimension is generally represented in regards to a starting point of $x_1$ and $x_2$. The time that the object in question is at each point is denoted as $t_1$ and $t_2$ (always assuming that $t_2$ is later than $t_1$, since time only proceeds one way). The change in a quantity from one point to another is generally indicated with the Greek letter delta, $\Delta$.

Using these notations, it is possible to determine the average velocity (vav) in the following manner:

$v_{av} = \dfrac{(x_2 - x_1)}{(t2 - t1)} = \dfrac{\Delta x}{\Delta t}$. If you apply a limit as $\Delta t$ approaches 0, you obtain an instantaneous velocity at a specific point in the path. Such a limit in calculus is the derivative of x with respect to t, or $\dfrac{dx}{dt}$. Acceleration

Acceleration represents the rate of change in velocity over time. Using the terminology introduced earlier, we see that the average acceleration ($a_{av}$) is: $a_{av} = \dfrac{(v_2 - v_1)}{(t_2 - t_1)} = \dfrac{\Delta x}{\Delta t}$

Again, we can apply a limit as $\Delta t$ approaches 0 to obtain an instantaneous acceleration at a specific point in the path. The calculus representation is the derivative of v with respect to t, or $\dfrac{dv}{dt}$. Similarly, since v is the derivative of x, the instantaneous acceleration is the second derivative of x with respect to t, or d2x/dt2. Constant Acceleration

In several cases, such as the Earth's gravitational field, the acceleration may be constant - in other words the velocity changes at the same rate throughout the motion. Using our earlier work, set the time at 0 and the end time as t (picture starting a stopwatch at 0 and ending it at the time of interest). The velocity at time 0 is $v_0$ and at time t is v, yielding the following two equations:

$a = \dfrac{(v - v_0)}{(t - 0)};\\ v = v_0 + at$

Applying the earlier equations for $v_{av}$ for $x_0$ at time 0 and x at time t, and applying some manipulations (which I will not prove here), we get: $x = x_0 + v_0t + 0.5at_2 v_2 = v_02 + 2a(x - x_0)$

$x - x_0 = \dfrac{(v_0 + v)t}2$

The above equations of motion with constant acceleration can be used to solve any kinematic problem involving motion of a particle on a straight line with constant acceleration.