Displacement, velocity, and acceleration

Topic 1.1 set up the language of vectors. This topic puts that language in motion. Three rates govern everything that follows: position changes at a rate called velocity; velocity changes at a rate called acceleration; and the changes themselves accumulate over time to give displacement.

In Physics C, those rates are derivatives. The instantaneous velocity is $v(t) = \dfrac{dx}{dt}$, the slope of the position-time graph at one instant. The instantaneous acceleration is $a(t) = \dfrac{dv}{dt}$, the slope of the velocity-time graph at one instant. The same operations run in reverse as integrals: $\Delta x = \int v\,dt$ is the area under $v(t)$, and $\Delta v = \int a\,dt$ is the area under $a(t)$.

The shift from Physics 1: "average over a small interval" becomes "value at an instant," and "area under a curve" becomes a definite integral with limits. The physics is the same; what's new is the calculus. The diagnostic targets the six failure modes that show up exactly where the calculus enters: treating $dx/dt$ as a fraction, computing an average when an instantaneous value was asked, and dropping the constant of integration on the way back from $a(t)$ to $v(t)$ to $x(t)$.

Three quantities sound similar and aren't.

• Position $\vec{r}$ (or $x$ in 1D): where the object is, relative to a chosen origin. Vector quantity. A point.
• Displacement $\Delta\vec{r}$ (or $\Delta x$ in 1D): change in position from one moment to another. Vector quantity. An arrow from start to end.
• Distance $d$: total path length traversed. Scalar quantity. Always non-negative, and 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 where they started. Distance: $13$ m. Displacement: $\Delta x = +3$ m (taking east as $+\hat{\text{ı}}$).

In 1D, displacement is the signed difference of positions: $\Delta x = x_f - x_i$. The sign comes entirely from $x_f - x_i$; it records direction along the chosen axis. In 2D, displacement is a vector with two signed components: $\Delta\vec{r} = (x_f - x_i)\hat{\text{ı}} + (y_f - y_i)\hat{\text{ȷ}}$.

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 the magnitude of displacement agree only when the path has no reversal or turn; position and displacement agree only when you start at the origin.

Average quantities compare endpoints to elapsed time.

$\overline{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_f - x_i}{\Delta t}, \qquad \overline{a} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_f - v_i}{\Delta t}.$

Both numerators are signed (displacement, change in velocity). Both denominators are positive elapsed times. The sign of the result reports direction along the chosen axis. The first equation is the basic definition of average velocity; the second is the same equation one level up: acceleration is to velocity what velocity is to position.

Worked example. A cart moves east at $\vec{v}_i = +12\hat{\text{ı}}$ m/s. The driver brakes uniformly and the cart slows to rest in $4$ s. Taking east as $+\hat{\text{ı}}$, what is the average acceleration?

Apply the definition. $\Delta v = v_f - v_i = 0 - (+12) = -12$ m/s. Divide by $\Delta t = 4$ s: $\overline{a} = \dfrac{-12 \text{ m/s}}{4 \text{ s}} = -3$ m/s$^2$. So $\vec{\overline{a}} = -3\hat{\text{ı}}$ m/s$^2$. The cart moves east, the acceleration points west: that's what slowing down in 1D looks like.

The sign of $\overline{a}$ is direction, not the answer to "speeding or slowing." That depends on the signs of $\vec{v}$ and $\vec{a}$ together: same direction means speeding up; opposite directions means slowing down. A few cases that catch students out:

• $v = 0$ at an instant doesn't mean $a = 0$. At the apex of a thrown ball, $v = 0$ but $a$ is still $-10\hat{\text{ȷ}}$ m/s$^2$. The acceleration is exactly what brought $v$ to zero and what makes the ball fall back.
• Reversing at the same speed still requires acceleration. A ball going $+4$ m/s that bounces off a wall and returns at $-4$ m/s has $\Delta v = -8$ m/s; over a short contact time, $\overline{a}$ is large and points west.
• Turning at constant speed is still acceleration. A car rounding a bend at $30$ mph has constant speed but changing direction of $\vec{v}$, so $\vec{a} \neq 0$ (size set by Topic 1.5).

Average velocity is displacement over time. Average acceleration is change in velocity over time. The sign reports direction. "Accelerating" doesn't mean "speeding up"; it means $\vec{v}$ is changing, in magnitude or direction or both.

Average velocity tells you the net rate over an interval. Instantaneous velocity tells you the rate at one specific moment. Shrink the interval around that moment: $\Delta t \to 0$. The ratio $\Delta x/\Delta t$ approaches a limit. That limit is the derivative:

$v(t) = \dfrac{dx}{dt} = \lim_{\Delta t \to 0} \dfrac{x(t + \Delta t) - x(t)}{\Delta t}.$

Geometrically, $v(t)$ is the slope of the position-time graph at the point with horizontal coordinate $t$. It is not a fraction of two small numbers. It is a single object: the slope of a tangent line. Treating $dx/dt$ as a literal fraction — for instance writing $\Delta x = v\,\Delta t$ when $v$ changes over the interval — is the central calculus trap of Topic 1.2. Differential notation can still appear in valid manipulations, such as the separation of variables you will meet in Topic 2.9; what fails is forcing the finite-ratio shortcut onto a changing $v$.

Worked example. A particle's position is $x(t) = 2t^3 - 5t + 1$ m. Find its instantaneous velocity at $t = 2$ s and its instantaneous acceleration at the same instant.

Differentiate term by term. The power rule gives $\dfrac{d}{dt}(t^n) = n t^{n-1}$, and constants disappear:

$v(t) = \dfrac{dx}{dt} = 6t^2 - 5 \quad \text{m/s}.$

Evaluate at $t = 2$: $v(2) = 6(4) - 5 = 24 - 5 = 19$ m/s. Differentiate $v(t)$ again to get $a(t)$:

$a(t) = \dfrac{dv}{dt} = 12t \quad \text{m/s}^2.$

At $t = 2$: $a(2) = 24$ m/s$^2$. The instantaneous values are $v(2) = 19$ m/s and $a(2) = 24$ m/s$^2$. If the same question had asked for averages, you'd compute $\Delta x/\Delta t$ over an interval ending at $t = 2$, and the two answers would differ. The derivative is a function: it gives one value at each $t$, no interval required.

Notation. "Differentiate $x$ with respect to $t$" is written $\dfrac{dx}{dt}$, $\dot{x}$, or $x'$. We'll use the first form. It is not a fraction. The Leibniz notation reads like "small change in $x$ over small change in $t$," but the operation is the limit, not literal division. Writing $dx = v\,dt$ is shorthand for setting up an integral, not algebra: it says the small change in $x$ over a tiny slice of the interval equals $v$ times that slice, and the total displacement is the integral of those slices.

Instantaneous velocity is the derivative $dx/dt$, the slope of $x(t)$ at one instant. Instantaneous acceleration is $dv/dt$, the slope of $v(t)$ at one instant. The derivative is a single operation, not a fraction; treat the notation as a name for "tangent slope," not a quotient.

The derivative goes from $x(t)$ down to $v(t)$ down to $a(t)$. Integration goes back up. Given $a(t)$, integrate once to recover $v(t)$. Given $v(t)$, integrate again to recover $x(t)$.

$v(t) = \int a(t)\,dt + C_1, \qquad x(t) = \int v(t)\,dt + C_2.$

The constant of integration $C$ matters. An antiderivative is defined only up to an additive constant: $\frac{d}{dt}(t^2) = 2t$ and $\frac{d}{dt}(t^2 + 7) = 2t$ are the same. The integral $\int 2t\,dt = t^2 + C$ doesn't specify which member of that family. The initial condition picks it.

If $v(0)$ is given, plug $t = 0$ into the antiderivative and solve for $C_1$ to match. Same for $x(0)$ and $C_2$. The shortcut is the definite integral:

$\Delta v = \int_{t_1}^{t_2} a(t)\,dt, \qquad \Delta x = \int_{t_1}^{t_2} v(t)\,dt.$

The definite integral handles the constant automatically: the endpoints subtract, the constant cancels. To get the function $v(t)$ itself (not just the change), use the indefinite form with $+C$ and apply the initial condition.

Worked example. A particle has acceleration $a(t) = 4t$ m/s$^2$ and initial velocity $v(0) = 3$ m/s. Find $v(t)$.

Integrate. The antiderivative of $4t$ is $2t^2$, so $v(t) = 2t^2 + C$. Use $v(0) = 3$ to fix $C$: $2(0)^2 + C = 3$, so $C = 3$. Therefore:

$v(t) = 2t^2 + 3 \text{ m/s}.$

Check: $v(0) = 3$ ✓. The constant of integration is not optional. Dropping it would give $v(t) = 2t^2$, which has $v(0) = 0$ instead of $3$. Two functions, $2t^2$ and $2t^2 + 3$, have the same derivative $4t$; only the initial condition tells you which to use. Same story when you integrate from $v(t)$ to $x(t)$: another constant, fixed by $x(0)$.

Integration reverses differentiation. Each integration brings a constant; the initial condition fixes it. Going from $a(t)$ to $x(t)$ is two integrations and two initial conditions. Dropping either constant silently puts the wrong initial value into the answer.

Six misconception codes, grouped into three pitfalls — each marks a spot where a derivative or an integral gets mistaken for a different operation.