Representing motion

Topic 1.1 set up the language of vectors. Topic 1.2 turned instantaneous velocity and acceleration into derivatives. Topic 1.3 is where the algebra-form motion equations live, the ones that solve nearly every traditional kinematic problem. The constraint is constant acceleration. Within that constraint, finding a position, a velocity, or a time is a matter of picking the right equation and plugging in.

Three equations cover the case of constant $a_x$ in one dimension:

$v_x = v_{x0} + a_x t, \qquad x = x_0 + v_{x0}t + \tfrac{1}{2}a_x t^2, \qquad v_x^2 = v_{x0}^2 + 2 a_x (x - x_0).$

When $a_x$ varies in time, the algebra-form rules break and the calculus-form integrals take over: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t)\,dt$ and $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t)\,dt$. Same physics, two different tools. Choosing the right one is half the answer.

The diagnostic targets five failure modes that show up at the boundary between these tools: confusing a position with a displacement, mistaking path length for the net change in position, reading a $v$-vs-$t$ graph as if it were a picture of the trajectory, assuming $v = 0$ at an extremum means $a = 0$ there too, and dropping the sign on an integral with reversed bounds.

Three quantities sound similar and aren't, and every Topic 1.3 problem starts by sorting them out.

• Position $\vec{r}$ (or $x$ in 1D): where the object is, relative to a chosen origin. Vector quantity. A single point.
• Displacement $\Delta\vec{r}$ (or $\Delta x$ in 1D): the change in position from one moment to another, $\Delta x = x_f - x_i$. Vector quantity. An arrow from start to end.
• Distance $d$: total path length traversed. Scalar quantity. Always non-negative. Depends on the whole route, not just the endpoints.

Distance and displacement agree on a one-way trip and diverge the moment the motion reverses. A walker who goes $8$ m east then $5$ m back west covers $13$ m of path length but ends only $3$ m east of the start. Distance: $13$ m. Displacement: $\Delta x = +3$ m (taking east as $+\hat{\text{ı}}$). For a full loop ending where it began, the distance is the lap length and the displacement is zero.

Two traps live here. The first reads the path length as if it were the displacement, missing that the legs of a reversing trip cancel. The second reads a position off a graph as if it were a displacement, missing that displacement is the change between two positions, not either one of them.

On an $x$-vs-$t$ graph, the value of $x$ at one instant is a position. The displacement on the interval $[t_1, t_2]$ is the vertical difference between the two endpoints of the graph: $\Delta x = x(t_2) - x(t_1)$. Quoting either endpoint as the displacement is the most common version of this trap.

Position is where you are. Displacement is how far you've moved from the start, with direction. Distance is total path length, no direction. Distance and displacement agree only when the motion never reverses; position and displacement are equal only when the starting position is zero.

For motion in one dimension along $\hat{\text{ı}}$ with constant $a_x$, three equations cover every question you can ask about position, velocity, and time. Pick the one that has what you know and what you want, then solve:

$v_x = v_{x0} + a_x t \qquad (\text{links } v, t)$

$x = x_0 + v_{x0}t + \tfrac{1}{2}a_x t^2 \qquad (\text{links } x, t)$

$v_x^2 = v_{x0}^2 + 2 a_x (x - x_0) \qquad (\text{links } x, v;\ t \text{ does not appear})$

Three things to watch for:

• Each equation is written for motion along $\hat{\text{ı}}$. For motion along $\hat{\text{ȷ}}$ or $\hat{k}$, replace $x$ with $y$ or $z$ and adjust the sign of $a$ to match.
• Signs come from the chosen positive direction. Once $+\hat{\text{ı}}$ is set, $v_{x0}$, $a_x$, and $\Delta x$ all carry signs relative to that choice.
• The third equation has no $t$ in it. Reach for it when time doesn't appear in either what you have or what you want.

Worked example. A cart starts at rest at $x_0 = 0$ and accelerates uniformly at $a_x = 4$ m/s$^2$ along $+\hat{\text{ı}}$. Find its position at $t = 3$ s.

The position equation links $x$ and $t$: $x = x_0 + v_{x0}t + \tfrac{1}{2}a_x t^2 = 0 + 0 + \tfrac{1}{2}(4)(3)^2 = 18$ m. Both $x_0$ and $v_{x0}$ drop out because they are zero, but the $\tfrac{1}{2}$ does not drop out. $at^2$ without the half is the wrong answer.

One thing this list doesn't include: a kinematic equation that gives the displacement when the acceleration varies in time. There isn't one. For variable $a_x$, the three equations above don't apply; the calculus-form integrals in §6 take over.

For constant acceleration in 1D, three equations cover the surface: velocity-time, position-time, and the time-free third equation. Pick the one whose variables match what you know and what you want. The pieces are signed; the $\tfrac{1}{2}$ in front of $a_x t^2$ is part of the equation, not optional.

Near the surface of Earth, every object in free flight (no air drag, no contact forces, no propulsion) has the same vertical acceleration. Taking up as $+\hat{\text{ȷ}}$:

$\vec{a} = -g\hat{\text{ȷ}}, \qquad g \approx 10 \text{ m/s}^2.$

This is the same $g \approx 10$ used in Physics 1, and it is the value to use for near-Earth-surface problems — unless a different gravitational field is specified or has to be computed from $g = GM/r^2$ (Topic 2.6). Free fall is a special case of constant-acceleration motion, so the three equations of §3 apply directly with $a_y = -g$.

Worked example. A stone is dropped from rest and falls $45$ m to the ground. How long does the fall take?

Take down as $-\hat{\text{ȷ}}$ so $a_y = -g = -10$ m/s$^2$; let $y_0 = 0$ at the release point and $y_f = -45$ m at the ground. Use the position equation: $y_f = y_0 + v_{y0}t + \tfrac{1}{2}a_y t^2 = 0 + 0 - 5t^2$. Solving $-45 = -5t^2$ gives $t^2 = 9$, so $t = 3$ s. Equivalently, with down as positive: $45 = \tfrac{1}{2}(10)t^2$, same $t = 3$ s. Either convention works; what matters is that the signs of $a$ and $\Delta y$ are consistent with the chosen positive direction.

The most common trap in free fall is at the apex. A ball is thrown straight up. It rises, slows, momentarily reaches $v_y = 0$ at the top, and falls back. At that single instant where $v_y = 0$:

• The velocity is zero, $v_y = 0$. This is what "apex" means.
• The acceleration is not zero. It is still $a_y = -g = -10$ m/s$^2$, the same value it has during the rise and during the fall.

The instant of zero velocity is one moment in a continuous motion, not a pause. The acceleration is what brings $v_y$ to zero and then makes the ball fall back. Treating $v_y = 0$ as a state of rest with $a = 0$ is one of the most common Physics C errors. The correct picture: $a_y = -g$ throughout, while $v_y$ passes continuously from $+v_{y0}$ through zero down to $-v_{y0}$ at the height of release.

The same logic applies to any extremum of position. If $x(t)$ reaches a maximum at some instant, then $v_x = dx/dt = 0$ there, but $a_x = dv_x/dt$ is whatever the second derivative says at that moment. For $x(t) = -t^2 + 6t$, the maximum is at $t = 3$ s where $v_x = 0$, but $a_x = -2$ m/s$^2$ at every $t$, including the maximum. The first derivative being zero does not imply the second derivative is zero.

Free fall is constant-acceleration motion with $a = -g \approx -10$ m/s$^2$ taking up as positive. At an apex, the velocity is zero by definition; the acceleration is not. The acceleration is what makes the apex happen.

Three graphs come up everywhere: position-time, velocity-time, acceleration-time. Each shows the same motion a different way. The most important rule: the graph is not a picture of the trajectory. The vertical axis is a kinematic quantity, not a height; the horizontal axis is time, not distance.

The four operations that move between graphs:

• Slope of $x(t)$ at an instant $=$ velocity at that instant: $v_x = dx/dt$.
• Slope of $v(t)$ at an instant $=$ acceleration at that instant: $a_x = dv/dt$.
• Area under $v(t)$ on $[t_1, t_2]$ $=$ displacement: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t)\,dt$.
• Area under $a(t)$ on $[t_1, t_2]$ $=$ change in velocity: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t)\,dt$.

Three failure modes show up here. The first is the graph-as-picture trap: reading an $x$-vs-$t$ parabola as if the object were tracing a parabolic path through the air. The graph plots a coordinate against time; the trajectory is along whatever physical axis $x$ measures. A downward-opening $x(t)$ parabola is a 1D motion that rises, reaches a maximum position, and reverses, not a flight through 2D space.

The second is the slope-area swap: reading a slope at a point when the question asked for the area on an interval, or vice versa. Slope at an instant is a derivative; area on an interval is an integral. Two different operations on two different inputs.

The third is endpoint-as-displacement: reading the value of $x$ at one instant as the displacement when the displacement is the difference between two values. The graph shows positions; the displacement is the change.

Position, velocity, and acceleration each have their own graph. Slopes carry derivatives between adjacent graphs; areas carry integrals back the other way. The graph is a coordinate against time, never a picture of the trajectory.

When $a_x$ depends on time, the three constant-acceleration equations of §3 do not apply. The calculus-form integrals take over:

$\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t)\,dt, \qquad \Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t)\,dt.$

Two rules to watch. First, integration is the reverse of differentiation: if $v_x(t)$ is given, $\Delta x$ on $[t_1, t_2]$ is the area under $v_x(t)$ on that interval, and the definite integral computes it without any leftover constant. Second, the order of the bounds matters: $\displaystyle\int_{t_1}^{t_2}$ and $\displaystyle\int_{t_2}^{t_1}$ differ by a sign:

$\displaystyle\int_{t_1}^{t_2} f(t)\,dt = -\displaystyle\int_{t_2}^{t_1} f(t)\,dt.$

Reversing the bounds flips the sign of the integral. The unsigned area is the same; the integral itself records the orientation of the bounds. Computing $\displaystyle\int_3^1 v\,dt$ for $v(t) = 4t$ m/s and reporting the magnitude $16$ m without the minus sign drops half the answer. The correct value is $-16$ m, with the negative recording that the integration is going from $t = 3$ s back to $t = 1$ s.

Worked example. A particle has velocity $v_x(t) = 2t$ m/s. Find the displacement on $[0, 4]$ s.

$\Delta x = \displaystyle\int_0^4 2t\,dt = [t^2]_0^4 = 16 - 0 = 16$ m. The same integral with bounds reversed: $\displaystyle\int_4^0 2t\,dt = -16$ m.

This is a real distinction, not a notational quirk. The negative integral records the direction of integration: running from $t_2$ back to $t_1$ undoes the change you'd accumulate going from $t_1$ to $t_2$. Reporting the unsigned area as the answer drops half the information.

When the goal is the full function $v_x(t)$ (not just the change), use the indefinite integral and apply the initial condition: $v_x(t) = v_{x0} + \displaystyle\int_0^t a_x(\tau)\,d\tau$, and similarly $x(t) = x_0 + \displaystyle\int_0^t v_x(\tau)\,d\tau$. The initial values $v_{x0}$ and $x_0$ are not optional; they pick out which antiderivative is the right one for the motion.

For variable acceleration, integration replaces the three algebra-form equations. Bounds in reverse order flip the sign of the integral; the unsigned area is not always the right answer. When the goal is a function rather than a change, the initial conditions fix the constants of integration.

Each of the five below mistakes one kinematic object for a look-alike: path length for displacement, a position for a change, a graph for a path, a zero velocity for zero acceleration, a signed integral for a bare area.