Rectilinear Motion

This section builds up rectilinear (straight-line) motion quantity by quantity, using vectors, and ends with the equations of motion for uniform acceleration, free fall, and one-dimensional relative velocity.

1. Displacement. If an object's position vector is $\vec{x}_1$ at time $t_1$ and $\vec{x}_2$ at time $t_2$, its displacement over that interval is $\vec{s} = \vec{x}_2 - \vec{x}_1$ — the vector difference of the two position vectors, pointing along the line of motion. Displacement has dimensions of length $[L^1 M^0 T^0]$; e.g. an object moving 1 m along the +x direction has a displacement of magnitude 1 m directed along +x, and moving the same 1 m along +y instead gives the same magnitude but a different direction.

2. Path length (distance travelled). This is the actual, always-positive length of the trail traced by the object, regardless of any reversals in direction. It is a scalar (no direction), also with dimensions of length. If an object moves from x = 2 m to x = 5 m, both the distance travelled and the displacement are 3 m. But if it then comes back to x = 4 m, the path length grows to 3 + 1 = 4 m even though the displacement (measured from the very first position, x = 2 m) has shrunk to 2 m. If it continues on to x = 1 m, the path length becomes 4 + 3 = 7 m while the displacement's magnitude is now only 1 m, and its direction has flipped to the -x side. This example is the standard illustration that path length can only grow, while displacement depends only on the start and end points.

3. Average velocity. Defined as displacement divided by the time interval over which it occurred: $\vec{v}_{av} = \dfrac{\vec{x}_2 - \vec{x}_1}{t_2 - t_1}$. Since displacement is a vector, average velocity is a vector too, with dimensions $[L^1 M^0 T^{-1}]$. For example, an object at x = +2 m at t = 0 and x = +4 m at t = 1 min has $\vec{v}_{av} = 2\,\hat{i}$ m/min.

4. Average speed. Defined as total path length divided by the same time interval: it is a scalar with the same dimensions as velocity. When the motion is entirely in one direction, path length equals the magnitude of displacement, so average speed equals the magnitude of average velocity; but the moment the object reverses direction, path length exceeds the displacement's magnitude, so average speed becomes larger than the magnitude of average velocity.

5. Instantaneous velocity. The velocity at one particular instant, obtained by shrinking the averaging interval $\Delta t$ around that instant down to zero: $\vec{v} = \lim_{\Delta t \to 0} \dfrac{\Delta \vec{x}}{\Delta t} = \dfrac{dx}{dt}$, the ordinary derivative of position with respect to time.

6. Instantaneous speed. Similarly the limiting value of average speed as $\Delta t \to 0$; in this limit the path length traversed becomes equal to the magnitude of the (infinitesimal) displacement, so instantaneous speed is always exactly equal to the magnitude of the instantaneous velocity — unlike their average counterparts, which need not match.

Always remember: for uniform rectilinear motion (constant velocity), average and instantaneous velocity are equal, and average and instantaneous speed are equal to the magnitude of that velocity. For nonuniform rectilinear motion, all of these generally differ from each other.

Graphical study of motion. Plotting position x against time t (an x-t graph) makes velocity visible as slope: a horizontal line means the object is at rest (zero velocity); a straight line with a positive slope means uniform velocity along +x; a straight line with a negative slope means uniform velocity along -x; a zig-zag line of constant slope-magnitude but alternating sign means oscillatory motion at constant speed; and a curved (nonlinear) line means nonuniform velocity, where the instantaneous velocity at any instant $t_0$ is the slope of the tangent to the curve at $t_0$ (the limit of chord-slopes over ever-smaller intervals around $t_0$).

7. Acceleration. The rate of change of velocity with time, a vector with dimensions $[L^1 M^0 T^{-2}]$. Average acceleration between times $t_1$ and $t_2$ (velocities $\vec{v}_1$, $\vec{v}_2$) is $\vec{a}_{av} = \dfrac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1}$, and instantaneous acceleration is its zero-interval limit, $\vec{a} = \dfrac{d\vec{v}}{dt}$ — the slope of the tangent to a velocity-versus-time (v-t) graph at that instant. On a v-t graph, the area under the curve between two instants equals the displacement in that interval (since displacement is the time-integral of velocity, $\Delta x = \int_{t_1}^{t_2} v\,dt$); this holds whether the acceleration is zero, uniformly positive, uniformly negative, or nonuniform.

Equations of motion for uniform acceleration (derived graphically from a v-t line rising from $u$ at $t=0$ to $v$ at time $t$): the slope of the line gives the first equation, $v = u + at$. The area under the line (a trapezium, split into a rectangle plus a triangle) gives the displacement and the second equation, $s = ut + \tfrac{1}{2}at^2$. Eliminating $t$ between these two, using the average velocity $v_{av} = \tfrac{u+v}{2}$, gives the third equation, $v^2 = u^2 + 2as$. Vector notation was not needed for these because the motion is confined to one line; only their magnitudes/signs vary. Always remember: for uniform acceleration the v-t graph is linear, its slope is the acceleration, and the area under it between two times gives the displacement in that interval; for nonuniform acceleration the v-t graph is nonlinear, but the area-under-curve rule for displacement still holds, while the acceleration at any instant is the slope of the tangent there (and the origin of the velocity axis must be taken as zero whenever using the area-under-the-curve method). …