AP Physics 1 · Kinematics · free lesson

Displacement, Velocity, and Acceleration

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Describing where an object is, and how that changes

To describe motion you have to say where something is, and how its place changes with time. Real objects have a size, a shape and parts that move inside them. When you use the object model, you ignore all of that: the object is treated as a single point that carries its mass, its charge and its other extensive properties. A car on a highway, a thrown ball and a runner can each be modeled as a point, as long as the question is about where the whole object goes and not about how it spins or bends.

Along a straight line, the point's position $x$ is its signed distance from a chosen origin: positive on one side, negative on the other. A change in position is a displacement.

Definition.

Displacement is the change in an object's position: $\Delta x = x - x_0$ , the final position minus the initial position. It is a vector. Its sign gives the direction of the change, and its size is the straight-line distance between the two positions.

\begin{aligned} d_a &= x_1 - x_0 \\ &= 9\,\mathrm{m} - 2\,\mathrm{m} \\ &= 7.00\,\mathrm{m} \\ d_b &= x_2 - x_1 \\ &= \left(-3\,\mathrm{m}\right) - 9\,\mathrm{m} \\ &= -12.0\,\mathrm{m} \end{aligned}

A cart starts at the position $2\,\mathrm{m}$ on a track, rolls to $9\,\mathrm{m}$ , and then rolls back to $-3\,\mathrm{m}$ . The first displacement is $7.00\,\mathrm{m}$ and the second is $-12.0\,\mathrm{m}$ : the sign is the direction, and the second is negative because the position decreased. Displacement depends only on where the object starts and where it ends, not on the path between.

Note.

Displacement is not distance. The cart rolled a distance of $19.0\,\mathrm{m}$ along its path, but its displacement from the very start to the very end is $-5.00\,\mathrm{m}$ . Distance adds up every step; displacement compares only the ends.

Average velocity

Velocity says how fast the position changes and in which direction. Over a stretch of time, the simplest description uses only the state at the beginning and at the end of the interval.

Definition.

Average velocity is the displacement divided by the time interval in which it occurs: $v_{avg} = \dfrac{\Delta x}{\Delta t}$ , where $\Delta t = t - t_0$ . It is a vector, so it has the sign of the displacement. Its unit is m/s.

\begin{aligned} v_a &= \frac{dx}{dt} \\ &= \frac{\left(-120\,\mathrm{m}\right)}{15\,\mathrm{s}} \\ &= -8.00\,\mathrm{m/s} \end{aligned}

A runner whose displacement is $-120\,\mathrm{m}$ in $15\,\mathrm{s}$ has an average velocity of $-8.00\,\mathrm{m/s}$ , which is $8.00\,\mathrm{m/s}$ in the negative direction. Average velocity and average speed are different things. Average speed is the total distance divided by the time and is never negative. If the runner goes $120\,\mathrm{m}$ down the track and comes back to the start in $15\,\mathrm{s}$ , the average speed is not zero but the average velocity is, because the displacement is zero.

Average acceleration

Velocity itself can change. Acceleration says how fast it does.

Definition.

Average acceleration is the change in velocity divided by the time interval in which that change occurs: $a_{avg} = \dfrac{\Delta v}{\Delta t} = \dfrac{v - v_0}{t - t_0}$ . It is a vector, and its unit is m/s², meaning meters per second, each second.

\begin{aligned} dv &= v_1 - v_0 \\ &= 27\,\mathrm{m/s} - 12\,\mathrm{m/s} \\ &= 15.0\,\mathrm{m/s} \\ a_a &= \frac{dv}{T} \\ &= \frac{15\,\mathrm{m/s}}{5\,\mathrm{s}} \\ &= 3.00\,\mathrm{m/s^{2}} \end{aligned}

A car goes from $12\,\mathrm{m/s}$ to $27\,\mathrm{m/s}$ in $5\,\mathrm{s}$ . Its velocity changed by $15.0\,\mathrm{m/s}$ , so its average acceleration is $3.00\,\mathrm{m/s^{2}}$ : each second, its velocity grows by that much. Both averages use only the initial and the final state of the interval. They say nothing about what happened in between.

From average to instantaneous

An average hides detail. A car that covers 100 m in 10 s has an average velocity of 10 m/s, whether it creeps and then races or goes steadily. To describe the velocity at one instant, shrink the interval: the smaller $\Delta t$ is, the closer the average velocity over it comes to the velocity at the instant it surrounds.

A cart starts at $0\,\mathrm{m}$ with a velocity of $2\,\mathrm{m/s}$ and speeds up at $0.500\,\mathrm{m/s^{2}}$ . At t = $3\,\mathrm{s}$ it is at $8.25\,\mathrm{m}$ . Its average velocity over intervals that start at that moment and get shorter:

Time interval starting at 3 s	Average velocity
$2\,\mathrm{s}$	$4.00\,\mathrm{m/s}$
$1\,\mathrm{s}$	$3.75\,\mathrm{m/s}$
$0.1\,\mathrm{s}$	$3.53\,\mathrm{m/s}$
$0.01\,\mathrm{s}$	$3.50\,\mathrm{m/s}$

The averages close in on $3.50\,\mathrm{m/s}$ , the instantaneous velocity at 3 s. The same goes for acceleration: the average acceleration over a very small interval is very close to the instantaneous acceleration. On a position-time graph the average velocity is the slope of the line joining two points, and the shorter the interval, the closer that line comes to the tangent at the first point. Move the slider and watch the red line.

Rule.

Instantaneous velocity is what the average velocity approaches as the time interval shrinks toward zero. Instantaneous acceleration is the same for the average acceleration. In everyday words, "velocity" and "acceleration" at a moment mean these instantaneous values.

What it means to accelerate

An object is accelerating if the magnitude of its velocity, the direction of its velocity, or both are changing. Along a line, acceleration tells you how the velocity component changes, and the signs of the velocity and the acceleration together tell you what the speed does.

Velocity and acceleration	Direction of the acceleration compared to the motion	What happens to the speed
same sign	same direction	speeds up
opposite signs	opposite direction	slows down
acceleration is zero	no acceleration	constant velocity, speed unchanged

A positive acceleration does not always mean "speeding up", and a negative one does not always mean "slowing down". A car moving in the negative direction with a negative acceleration speeds up. The animation shows a cart that starts moving right at a speed of 6 m/s with a constant acceleration pointing left. The green arrow is the velocity and the red arrow is the acceleration. The cart slows, stops for an instant, and then speeds up to the left, and the acceleration arrow never changes.

For the first 3 seconds the velocity is positive and the acceleration is negative, so the cart slows down. At 3 seconds the velocity is zero, but the acceleration is not: the velocity is passing through zero on its way from positive to negative. After that both are negative, and the cart speeds up to the left.

Constant speed, changing velocity

Because velocity has a direction, an object can accelerate with no change in speed. A car going around a bend at a steady 20 m/s has a velocity that points a different way at every moment, so its velocity is changing, and so it accelerates. The picture shows a car going counterclockwise around a circular track at constant speed, with its velocity at two instants. Subtract the first from the second ( $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ ) and the change is not zero: it points toward the inside of the curve, and so does the acceleration.

Both arrows on the left have the same length, because the speed is $20\,\mathrm{m/s}$ at both points (each is $4.00$ squares long, one square for $5\,\mathrm{m/s}$ ). On the right they are drawn from a common point, and the green arrow from the head of $\vec{v}_1$ to the head of $\vec{v}_2$ is $\Delta\vec{v}$ , with a length of $5.66$ squares, which stands for $28.3\,\mathrm{m/s}$ . The speed did not change at all, yet the velocity did.

Warning.

Three traps. A large velocity does not mean a large acceleration: a plane cruising at a steady 250 m/s has zero acceleration. A velocity of zero does not mean zero acceleration: a ball at the top of its flight has zero velocity for an instant and is still accelerating. And "accelerating" does not only mean "speeding up": slowing down and turning are accelerations too.

Worked examples

Five examples of the kinds the AP exam asks on this topic: displacement and average velocity, average acceleration, deciding whether an object speeds up or slows down, estimating an instantaneous velocity, and justifying a claim about acceleration.

Example.

Example 1. Displacement, distance and average velocity

A delivery robot in a long hallway starts at the position $4\,\mathrm{m}$ , rolls to $34\,\mathrm{m}$ in $25\,\mathrm{s}$ , and then rolls back to $10\,\mathrm{m}$ , arriving $40\,\mathrm{s}$ after it started. Find the robot's displacement, the distance it rolled, its average velocity and its average speed for the whole trip.

Solution

Displacement is the final position minus the initial position. Distance adds the length of each leg. Both averages divide by the same time.

\begin{aligned} d_x &= x_2 - x_0 \\ &= 10\,\mathrm{m} - 4\,\mathrm{m} \\ &= 6.00\,\mathrm{m} \\ D &= \left|x_1 - x_0\right| + \left|x_2 - x_1\right| \\ &= \left|34\,\mathrm{m} - 4\,\mathrm{m}\right| + \left|10\,\mathrm{m} - 34\,\mathrm{m}\right| \\ &= 54.0\,\mathrm{m} \\ v_a &= \frac{d_x}{t_2} \\ &= \frac{6\,\mathrm{m}}{40\,\mathrm{s}} \\ &= 0.150\,\mathrm{m/s} \\ s_a &= \frac{D}{t_2} \\ &= \frac{54\,\mathrm{m}}{40\,\mathrm{s}} \\ &= 1.35\,\mathrm{m/s} \end{aligned}

The graph shows position against time. The blue line is the robot's motion; the red line joins the first and last points, and its slope is the average velocity.

Result.

The displacement is $6.00\,\mathrm{m}$ and the distance rolled is $54.0\,\mathrm{m}$ . The average velocity is $0.150\,\mathrm{m/s}$ , in the positive direction, and the average speed is $1.35\,\mathrm{m/s}$ . The average speed is larger because the robot turned back: distance counts the whole path, displacement only the ends.

Example.

Example 2. Average acceleration of a braking car

A car traveling east at $25\,\mathrm{m/s}$ brakes and slows to $7\,\mathrm{m/s}$ in $4.50\,\mathrm{s}$ . Take east as positive. Find its average acceleration and say which way it points.

Solution

Average acceleration is the change in velocity divided by the time interval.

\begin{aligned} dv &= v_1 - v_0 \\ &= 7\,\mathrm{m/s} - 25\,\mathrm{m/s} \\ &= -18.0\,\mathrm{m/s} \\ a_a &= \frac{dv}{T} \\ &= \frac{\left(-18\,\mathrm{m/s}\right)}{4.5\,\mathrm{s}} \\ &= -4.00\,\mathrm{m/s^{2}} \end{aligned}

Result.

The average acceleration is $-4.00\,\mathrm{m/s^{2}}$ . It is negative, so it points west, opposite to the car's velocity, and that is why the car slows down. Each second the velocity decreases by $4.00\,\mathrm{m/s}$ . The car is not "decelerating" in some separate sense: it has an acceleration, and its direction is opposite to the motion.

Example.

Example 3. Speeding up or slowing down?

Three carts move along a line, with right as positive. Cart 1 has a velocity of $8\,\mathrm{m/s}$ and an acceleration of $3\,\mathrm{m/s^{2}}$ . Cart 2 has $8\,\mathrm{m/s}$ and $-3\,\mathrm{m/s^{2}}$ . Cart 3 has $-8\,\mathrm{m/s}$ and $-3\,\mathrm{m/s^{2}}$ . Find each cart's velocity and speed $2\,\mathrm{s}$ later, and say whether it is speeding up or slowing down.

Solution

Under a constant acceleration, the velocity after a time is $v = v_0 + a t$ .

\begin{aligned} v_1 &= u_1 + a_1t \\ &= 8\,\mathrm{m/s} + 3\,\mathrm{m/s^{2}} \times 2\,\mathrm{s} \\ &= 14.0\,\mathrm{m/s} \\ v_2 &= u_2 + a_2t \\ &= 8\,\mathrm{m/s} + \left(-3\,\mathrm{m/s^{2}}\right) \times 2\,\mathrm{s} \\ &= 2.00\,\mathrm{m/s} \\ v_3 &= u_3 + a_3t \\ &= \left(-8\,\mathrm{m/s}\right) + \left(-3\,\mathrm{m/s^{2}}\right) \times 2\,\mathrm{s} \\ &= -14.0\,\mathrm{m/s} \end{aligned}

Cart	Velocity and acceleration	Speed before, after	Result
1	same sign	$8.00\,\mathrm{m/s}$ , $14.0\,\mathrm{m/s}$	speeds up
2	opposite signs	$8.00\,\mathrm{m/s}$ , $2.00\,\mathrm{m/s}$	slows down
3	same sign	$8.00\,\mathrm{m/s}$ , $14.0\,\mathrm{m/s}$	speeds up

Result.

Cart 1 ends at $14.0\,\mathrm{m/s}$ and cart 3 at $-14.0\,\mathrm{m/s}$ : both speed up, although one acceleration is positive and the other negative. Cart 2 ends at $2.00\,\mathrm{m/s}$ and slows down. What matters is whether the velocity and the acceleration have the same sign, not the sign of the acceleration alone.

Example.

Example 4. Estimating an instantaneous velocity

Photogates record the position of a cart on a track to the nearest 0.0001 m. At t = $3\,\mathrm{s}$ the cart is at $12\,\mathrm{m}$ . A time $1\,\mathrm{s}$ later it is at $20\,\mathrm{m}$ , $0.1\,\mathrm{s}$ later at $12.71\,\mathrm{m}$ , and $0.01\,\mathrm{s}$ later at $12.0701\,\mathrm{m}$ . Estimate the cart's instantaneous velocity at $3\,\mathrm{s}$ .

Solution

Find the average velocity over each interval that starts at $3\,\mathrm{s}$ . The instantaneous velocity is what these averages approach as the interval shrinks.

\begin{aligned} w_1 &= \frac{x_b - x_a}{h_1} \\ &= \frac{20\,\mathrm{m} - 12\,\mathrm{m}}{1\,\mathrm{s}} \\ &= 8.00\,\mathrm{m/s} \\ w_2 &= \frac{x_c - x_a}{h_2} \\ &= \frac{12.71\,\mathrm{m} - 12\,\mathrm{m}}{0.1\,\mathrm{s}} \\ &= 7.10\,\mathrm{m/s} \\ w_3 &= \frac{x_d - x_a}{h_3} \\ &= \frac{12.0701\,\mathrm{m} - 12\,\mathrm{m}}{0.01\,\mathrm{s}} \\ &= 7.01\,\mathrm{m/s} \end{aligned}

Result.

The averages are $8.00\,\mathrm{m/s}$ , $7.10\,\mathrm{m/s}$ and $7.01\,\mathrm{m/s}$ , and they are getting closer to a limit. The shortest interval gives the best estimate: the instantaneous velocity at $3\,\mathrm{s}$ is about $7.01\,\mathrm{m/s}$ , in the positive direction. An average over a long interval describes the whole interval, not the instant at its start.

Example.

Example 5. Justifying a claim about acceleration

A ball hits the floor moving down at $7.50\,\mathrm{m/s}$ and leaves it moving up at $7.50\,\mathrm{m/s}$ , with the floor contact lasting $0.0120\,\mathrm{s}$ . A student claims: "The ball's speed is the same before and after the bounce, so its acceleration during the bounce is zero." Is the claim correct? Justify your answer. Take up as positive.

Solution

Acceleration is the change in velocity divided by the time, and velocity is a vector. The speeds are equal, but the velocities are not: the first is $-7.50\,\mathrm{m/s}$ and the second is $7.50\,\mathrm{m/s}$ .

\begin{aligned} dv &= v_1 - v_0 \\ &= 7.5\,\mathrm{m/s} - \left(-7.5\,\mathrm{m/s}\right) \\ &= 15.0\,\mathrm{m/s} \\ a_a &= \frac{dv}{T} \\ &= \frac{15\,\mathrm{m/s}}{0.012\,\mathrm{s}} \\ &= 1250\,\mathrm{m/s^{2}} \end{aligned}

Result.

The claim is wrong. The speed does not change, but the direction of the velocity reverses, and a change in direction is a change in velocity. The average acceleration is $1250\,\mathrm{m/s^{2}}$ , pointing up, which is the direction of the change in velocity. It is large because the change happens in a very short time.

Practice

Answer AP-style questions on displacement, velocity, and acceleration. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise displacement, velocity, and acceleration with new numbers