Fundamentals of Physics/Motion in One Dimension

The goal of one dimensional motion is to understand how acceleration ($a$), drives changes in speed ($v$) and position ($x$) along a single direction of motion (one-dimension). An "object" means anything that can move, like a ball, car, truck, or person. One-dimensional means the object will only move along a straight line, typically along the $x$-axis if it's moving left or right or $y$-axis if it's moving up or down. There are two equations you need for this,

$x=x_0+v_0 \Delta t+\frac{1}{2}a \Delta t^2$,

and the second is

$v=v_0+a \Delta t$.

If the object is at $x_0$ with speed $v_0$, then $x$ and $v$ will be the object's new position and speed at some time interval $\Delta t$ later. These two equations allow you to compute the new position and speed of an object ($x$ and $v$), based on its old position and speed ($x_0$ and $v_0$), given some acceleration $a$ that is acting on the object, over a time interval $\Delta t$. $\Delta t$ is sometimes called the "time step" and is a small interval of time that separates when the object has $x_0$ and $v_0$, and when it will have $x$ and $v$.

We said that $a$ drives changes in $v$ and $x$. Notice in these equations if $a=0$, then $v=v_0$, meaning that $v$ doesn't change between time steps; $v$ is constant if $a=0$. In order for $v$ to change, $a$ must be nonzero. In other words, an object's speed can change only if it has an acceleration. For the $x$-axis (left-right motion), we have that

$x=x_0+v_{0x}\Delta t+\frac{1}{2}a_x\Delta t^2$

and

$v_x=v_{0x}+a_x \Delta t$.

For the $y$-axis (up-down motion), we have that

$y=y_0+v_{0y}\Delta t+\frac{1}{2}a_y\Delta t^2$

and

$v_y=v_{0y}+a_y \Delta t$.

These equations are the same, just the notation is different, being very specific as to which axis it pertains.

A numerical example

As an example, Suppose you have a sphere at $x=5$ m with speed $v=1$ m/s and an acceleration of $a=0.5$ m/s$^2$. When the next frame comes up, say $\Delta t=0.1$ s later, where will the sphere be and what will its speed be? Use the equations to get that $x=5\mathrm{m}+(1\mathrm{m/s})(0.1\mathrm{s})+(0.5)(0.5\mathrm{m/s}^2)(0.1\mathrm{s})^2$ or $x=5.1025$ m and $v=1\mathrm{m/s}+(0.5\mathrm{m/s}^2)(0.1\mathrm{s})$ or $v=1.05$ m/s. Be sure you see how the equations allowed you to compute the new position and speed of the object over the time step $\Delta t$. You can iteratively use this new $x$ and $v$ as a new $x_0$ and $v_0$ (i.e. $x\rightarrow x_0$ and $v\rightarrow v_0$) for computing still another $x$ and $v$ another $\Delta t$ in the future. Can you find $x$ and $v$ after another $\Delta t$ has gone by? In the Figure 3.22 find the acceleration of the masses and the tension in the string.

Signs

Be very aware of signs. Think of a cartesian coordinate system with $+x$ to the right, $-x$ to the left, $+y$ up and $-y$ down (assume $\Delta t$ is always positive). Positive values of position mean the object is to the right ($x$) (or up, $y$) relative to the origin. Negative means the object is left ($x$) (or down, $y$) relative to the origin. Positive values of speed mean the object is moving toward the right ($v_x$) or up ($v_y$), negative means to the left ($v_x$) or down ($v_y$). The sign of $a$ alone doesn't immediately help to characterize the object's motion. If, however, $a$ and $v$ have the same sign, $v=v_0+a \Delta t$ will predict an increase in $v$ (that is if $v$ and $a$ have the same sign, an object will speed up). Likewise, an object will slow down if $v$ and $a$ have opposite signs.

A case where opposite signs of $v$ and $a$ persist means $v$ will get smaller and smaller, until eventually $v=0$ at which case the object will stop. If $a$ still persists, then $v$ will begin to increase in the same direction as $a$; now the object is speeding up, but in the opposite direction to its original motion. All told the object slowed down, stopped, then started speeding up in the opposite direction. All combinations of signs between $v$ and $a$ are possible. $v>0$ and $a<0$ is a slow-down and potential turn-around case, as is $v<0$ and $a>0$. $v>0$ and $a>0$ or $v<0$ and $a<0$ are speed up cases, but in opposite directions. Lastly, you should be able to draw arrows on an object, representing its $v$ and $a$ and that instant. The arrow should point in the direction of a given parameter and its length should be proportional to its strength. For example, if on an object the arrow for $v$ and the arrow for $a$ were opposite, you'd know the object was slowing down.