Displacement, Velocity, and Acceleration

Some highways use average-speed cameras: they photograph your car at two points $2.0\ \text{km}$ apart and divide the distance by the time between the photos. A radar gun reads your speed at a single instant. You can pass the camera check at an average of $100\ \text{km/h}$ and still be caught by radar at $130\ \text{km/h}$, because an average over an interval says nothing about what happened inside it. Mechanics needs both kinds of number, and calculus is what connects them.

Treating the object as a point

Kinematics describes motion without asking what causes it, and it starts by simplifying the thing that moves. In the object model, a car, a ball or a planet is treated as a single point. Its size, its shape and anything going on inside it (spinning wheels, sloshing fuel, a flexing tennis racket) are ignored. The point still carries the object's mass and charge, because later units need them. The model works whenever you only care where the object as a whole is and how that changes. It stops working when rotation or deformation matters, which is why Unit 5 replaces it with a rigid body that can turn.

The point's location is its position, measured from an origin you choose: $x$ in one dimension, or the position vector $\vec{r}$ from topic 1.1 in two or three.

Displacement and the two averages

Displacement is the change in position, final minus initial:

It depends only on where the object starts and finishes. A delivery robot that drives $30\ \text{m}$ down a hallway and $10\ \text{m}$ back has covered a distance of $40\ \text{m}$ but has a displacement of $+20\ \text{m}$.

An average compares two moments, the start and the end of a time interval, and ignores everything between them:

Average speed is a different quantity: total distance divided by elapsed time. The robot above, taking $20\ \text{s}$ for both legs, has an average speed of $\frac{40\ \text{m}}{20\ \text{s}} = 2.0\ \text{m/s}$ but an average velocity of $\frac{+20\ \text{m}}{20\ \text{s}} = +1.0\ \text{m/s}$. The two agree only when the object never reverses or turns.

Velocity changes, and so an object accelerates, whenever its speed changes or its direction changes. A tennis ball reaches a wall at $8.0\ \text{m/s}$ and leaves at $6.0\ \text{m/s}$, in contact for $0.020\ \text{s}$. Take toward the wall as positive. The ball's speed dropped by only $2.0\ \text{m/s}$, but its velocity went from $+8.0$ to $-6.0\ \text{m/s}$:

that is, $700\ \text{m/s}^2$ directed away from the wall. Reversing direction takes far more acceleration than the change in speed suggests. The same logic means a car rounding a curve at a steady $50\ \text{km/h}$ is accelerating, since its direction keeps changing.

Instantaneous values are derivatives

The radar gun's number is the limit of an average. Take a cart whose position is $x(t) = (2.0\ \text{m/s}^3)\,t^3$ and ask for its velocity at $t = 1.0\ \text{s}$. Average over the interval from $1.0\ \text{s}$ to $1.0\ \text{s} + \Delta t$:

As $\Delta t$ shrinks, the extra terms vanish and the average closes in on $6.0\ \text{m/s}$, the instantaneous velocity at $t = 1.0\ \text{s}$. On a graph of $x$ against $t$, each average is the slope of a secant line through two points of the curve, and the limit is the slope of the tangent line.

That limit is the definition of a derivative, and it gives the precise statements:

Here $\frac{d}{dt}\left[(2.0\ \text{m/s}^3)\,t^3\right] = (6.0\ \text{m/s}^3)\,t^2$, which is $6.0\ \text{m/s}$ at $t = 1.0\ \text{s}$, matching the table.

Because the average over a very short interval is already close to the instantaneous value, a lab can measure instantaneous velocity without calculus. A photogate times how long a narrow flag on a cart blocks its light beam. A few milliseconds is too short for the velocity to change much, so the flag's width divided by that time is a good estimate of the velocity as the cart passes the gate.

The signs of $v_x$ and $a_x$ together tell you whether the object is speeding up. If they have the same sign, the acceleration points along the motion and the speed grows. If they have opposite signs, the speed shrinks. So "negative acceleration" alone does not mean slowing down: an object moving in the $-x$ direction with $a_x < 0$ is speeding up.

Undoing the derivative: integrals and initial conditions

Going the other way, from acceleration to velocity to position, takes integration. Since $a_x = \frac{dv_x}{dt}$, integrating from time $0$ to time $t$ gives

The integral alone gives only the change: the acceleration tells you how velocity changes, not what it started at. The initial values $v_{x0}$ and $x_0$ have to come from the problem, and they are where the constant of integration goes. Two carts with the same $a_x(t)$ but different starting velocities follow different paths.

The integral of velocity is displacement, which counts motion in the $-x$ direction as negative. Distance traveled is the integral of speed, $\int |v_x|\,dt$; in practice you find where $v_x = 0$, integrate each piece, and add the sizes. When the acceleration is constant, these integrals produce the three kinematic equations, which topic 1.3 develops along with reading slopes and areas off graphs.

Worked examples

Where students lose points

On the exam

Questions in this topic give one of $x(t)$, $v_x(t)$ or $a_x(t)$ as a function and ask for the others, a turnaround time, or an average over an interval. "Derive" means starting from $v_x = \frac{dx}{dt}$ or $a_x = \frac{dv_x}{dt}$ and showing the differentiation or the integration with its limits or constant. A calculator is allowed on both sections of the exam. "Momentarily at rest" or "changes direction" always points to $v_x = 0$ with a sign change, never to $a_x = 0$.