Displacement, velocity, and acceleration

Position, velocity, and acceleration are three different things; each is the rate of change of the one before it. A car at rest has $v = 0$ but can have large $a$ the instant the driver hits the gas; a car cruising at $30$ m/s has large $v$ but $a = 0$. The number for one tells you nothing about the number for the other.

Pick a positive direction (call it east) once and use it for the whole problem. Displacement is the change in position from start to end:

$$\Delta x = x_f - x_i.$$

The signs are not optional. If $x_i = -3$ m and $x_f = +5$ m, then $\Delta x = (+5) - (-3) = +8$ m. The displacement is $+8$ m east. Subtract a negative; do not skip the parentheses.

Displacement is not distance. Distance is total path length: always positive, accumulates over every segment regardless of direction.

The two agree when motion never reverses and part company the moment it does. A jogger who runs one full $400$ m lap and stops at the start has distance $400$ m and displacement $0$ m.

Average velocity is the displacement divided by the time interval over which it happened:

$$\bar v = \dfrac{\Delta x}{\Delta t} = \dfrac{x_f - x_i}{\Delta t}.$$

The numerator is displacement (signed); the denominator $\Delta t$ is always positive. So the sign of $\bar v$ comes entirely from $\Delta x$. It tells you which side of the start the object ended up on.

Worked example. A car drives $20$ m east in $4$ s, sits at rest for $2$ s, then drives $10$ m west in $4$ s. East is positive. Total time is $\Delta t = 4 + 2 + 4 = 10$ s. Net displacement is $\Delta x = (+20) + 0 + (-10) = +10$ m east.

$$\bar v = \dfrac{\Delta x}{\Delta t} = \dfrac{+10 \text{ m}}{10 \text{ s}} = +1 \text{ m/s}.$$

So $\bar v = +1$ m/s east. This is not the speed at any instant (the car was at $+5$, $0$, then about $-2.5$ m/s on the three legs). "Average" here means net displacement over total time.

Compare against average speed, which uses path length, not displacement:

$$\text{average speed} = \dfrac{\text{total distance}}{\Delta t} = \dfrac{20 + 0 + 10}{10} = 3 \text{ m/s}.$$

Average speed and $|\bar v|$ are different ($3$ vs $1$ m/s) because the reversal cancelled part of leg one's displacement but added to its distance. Whenever the path reverses, $|\bar v| < $ (average speed), and AP problems exploit that gap.

The same shape, one level up. Average acceleration is the change in velocity divided by the time interval:

$$\bar a = \dfrac{\Delta v}{\Delta t} = \dfrac{v_f - v_i}{\Delta t}.$$

Final minus initial, signs preserved. Acceleration is to velocity what velocity is to position.

Worked example. A car is travelling east at $v_i = +12$ m/s. The driver brakes uniformly and the car slows to a stop ($v_f = 0$) in $\Delta t = 4$ s. East is positive.

$$\Delta v = v_f - v_i = 0 - (+12) = -12 \text{ m/s}.$$ $$\bar a = \dfrac{\Delta v}{\Delta t} = \dfrac{-12 \text{ m/s}}{4 \text{ s}} = -3 \text{ m/s}^2.$$

So $\bar a = -3$ m/s$^2$, that is, $3$ m/s$^2$ west. The car moves east, the acceleration points west: that's what slowing down looks like in 1D.

Same way → speeding up. Opposite ways → slowing down. The sign of $\bar a$ is its direction, not the answer to "speeding or slowing." That depends on the signs of $\vec v$ and $\vec a$ together.

An object is accelerating whenever $\vec v$ is changing: magnitude, direction, or both. Three cases catch most students out, none of them "speeding up" in the everyday sense.

Reversing at the same speed. A ball going east at $+4$ m/s bounces off a wall and returns west at $-4$ m/s. Speed unchanged, but $\Delta v = (-4) - (+4) = -8$ m/s. Over a short contact time, $\bar a$ is large and points west.

$v = 0$ during constant deceleration. A cart at $+6$ m/s with $\bar a = -3$ m/s$^2$ passes through $v = 0$ at $t = 2$ s. At that instant the acceleration is still $-3$ m/s$^2$; whatever's decelerating the cart doesn't pause at the turnaround. Treating $v = 0$ as $a = 0$ is one of the most common errors on this topic.

Turning at constant speed. A car rounding a bend at a steady $30$ mph has constant speed but changing direction of $\vec v$. It is accelerating, with $\vec a$ pointing toward the inside of the curve (size computed in Topic 1.5).

Accelerating does not always mean getting faster, and $v = 0$ does not mean $a = 0$.

Three failure modes that collapse distinctions the test measures separately.