Displacement, Velocity, and Acceleration: Study Guide (Physics 1 1.2)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra and graphs. If you are taking the calculus-based Physics C: Mechanics course, it has its own separate guide for its Topic 1.2.

Start with an object and an axis

To describe motion we first simplify the moving thing. In the object model, a car, a ball or a runner is treated as a single point. We ignore its size, shape and what is happening inside it. We only track where that point is.

Next we choose a coordinate system. In one dimension that means three choices:

1. Where is the origin (x = 0)?
2. Which direction is positive?
3. What unit do we measure in? (In this course: metres, m.)

Write the choice down every time, for example “+x to the right, origin at the start line”. The physics does not depend on this choice, but every sign in your answer does.

Position, x, tells you where the object is relative to the origin. A position of x = −3.0 m means 3.0 m on the negative side of the origin.

Displacement and distance

Displacement is the change in position:

Δx = x − x₀

Here x₀ is the initial position and x is the final position. Δx has a sign. Positive Δx means the net change was in the +x direction.

Distance travelled is the total length of path covered. It has no direction and is never negative. If you walk 5 m forward and 5 m back, your distance travelled is 10 m but your displacement is 0 m.

Displacement only cares about where you start and finish. Distance cares about every step on the way.

Average velocity and average speed

Averages use only the initial and final states over a time interval Δt = t − t₀.

Average velocity: v_avg = Δx / Δt

Average speed = distance travelled / Δt

Average velocity carries a sign and has units m/s. Average speed is never negative. The two are equal in size only when the object never turns around.

Average acceleration

Velocity can change. Average acceleration measures how fast velocity changes:

Average acceleration: a_avg = Δv_x / Δt = (v_x − v_x0) / Δt

The unit is (m/s) per s, written m/s². An object is accelerating whenever its velocity changes. In one dimension that means its speed changes, its direction reverses, or both.

Speeding up or slowing down?

The sign of a_x alone does not tell you whether an object is speeding up. Compare the signs of v_x and a_x:

v_x	a_x	What happens to the speed
+	+	speeding up (moving in +x, faster)
−	−	speeding up (moving in −x, faster)
+	−	slowing down
−	+	slowing down
any	0	constant velocity

Same signs → speeding up. Opposite signs → slowing down.

Instantaneous velocity without calculus

A speedometer shows velocity at an instant, not over an interval. How can an average become an instantaneous value? Make the interval very short.

Take a cart that starts from rest with +x to the right. Its position is x = (1.0 m/s²)t², so at t = 2.0 s it is at x = 4.0 m. Work out average velocities starting at t = 2.0 s:

Interval	Δx (m)	Δt (s)	v_avg = Δx / Δt (m/s)
2.0 s → 3.0 s	9.0 − 4.0 = 5.0	1.0	5.0
2.0 s → 2.5 s	6.25 − 4.0 = 2.25	0.5	4.5
2.0 s → 2.1 s	4.41 − 4.0 = 0.41	0.1	4.1
2.0 s → 2.01 s	4.0401 − 4.0 = 0.0401	0.01	4.01

As Δt shrinks, the average velocity settles on 4.0 m/s. That is the instantaneous velocity at t = 2.0 s.

On a position–time graph, each average velocity is the slope of a secant line joining two points on the curve. As the second point slides towards the first, the secant turns into the tangent line: the straight line that just touches the curve at that instant.

To measure a tangent slope on a real graph: draw the tangent carefully with a ruler, pick two points far apart on the tangent line (not on the curve), and divide rise by run. In Figure 1 those points are (1.0 s, 0 m) and (3.0 s, 8.0 m).

The same idea works for acceleration. Instantaneous acceleration is the average acceleration over a very short interval: the slope of the tangent on a velocity–time graph.

Reading a velocity–time graph

Two facts to read from Figure 2:

• The slope of a velocity–time graph is the acceleration: (−3.0 − 3.0) m/s ÷ 4.0 s = −1.5 m/s².
• The area between the line and the time axis is the displacement, with area below the axis counting as negative. Each triangle has area ½ × 2.0 s × 3.0 m/s = 3.0 m.

Notice that a_x is negative throughout. For the first 2 s the cart slows down; for the last 2 s it speeds up. The sign comparison in the table above explains both halves. Topic 1.3 develops these graph links in full.

Worked example 1: distance, displacement and two kinds of average

Question. Take +x as east, with the origin at a bus stop. A cyclist starts at x₀ = +20 m. She rides east to x = +140 m in 30 s, then rides west to x = +80 m in a further 20 s. Find (a) the distance travelled, (b) the displacement, (c) the average velocity and (d) the average speed for the whole 50 s.

1. Distance, leg 1: |140 − 20| = 120 m. Leg 2: |80 − 140| = 60 m. Total distance = 120 + 60 = 180 m.
2. Displacement: Δx = x − x₀ = 80 m − 20 m = +60 m (60 m east of where she started).
3. Average velocity: v_avg = Δx / Δt = +60 m ÷ 50 s = +1.2 m/s (east).
4. Average speed: 180 m ÷ 50 s = 3.6 m/s.

Check. The leg velocities are +120 m ÷ 30 s = +4.0 m/s and −60 m ÷ 20 s = −3.0 m/s. Their plain mean is +0.5 m/s, not +1.2 m/s, because she spent longer on the first leg. A time-weighted mean agrees: (4.0 × 30 + (−3.0) × 20) ÷ 50 = +1.2 m/s. Average speed is bigger than the size of the average velocity because she turned around.

Worked example 2: positive acceleration while slowing down

Question. Take +x as east. A car travels west and brakes. Its velocity changes from v_x0 = −18 m/s to v_x = −6.0 m/s in 4.0 s. Find its average acceleration and say whether it is speeding up or slowing down.

1. Change in velocity: Δv_x = v_x − v_x0 = (−6.0 m/s) − (−18 m/s) = +12 m/s.
2. Average acceleration: a_avg = Δv_x / Δt = +12 m/s ÷ 4.0 s = +3.0 m/s².
3. Compare signs: v_x is negative (moving west) and a_x is positive. Opposite signs, so the car is slowing down.

Interpretation. “Positive acceleration” here means the acceleration points east, opposite to the motion. Its speed falls from 18 m/s to 6.0 m/s.

Check. If you dropped the signs and wrote (6 − 18) ÷ 4 = −3.0 m/s², you would get the right size but the wrong direction. Always substitute signed velocities.

Limiting cases worth knowing

Testing extreme cases is a quick way to check your understanding.

• Very short Δt. The average value becomes the instantaneous value (the table and Figure 1).
• Back to the start. If x = x₀, then Δx = 0 and v_avg = 0, even though the object moved. Its average speed is not zero.
• Momentarily at rest is not the same as zero acceleration. Throw a ball straight up and take +y upward. With g = 9.8 m/s², its velocity changes from +14.7 m/s to 0 at the top after 1.5 s, and to −4.9 m/s after 2.0 s. Average acceleration over 0–2.0 s = (−4.9 − 14.7) m/s ÷ 2.0 s = −9.8 m/s². At the top v_y = 0, but its velocity is still changing, so a_y = −9.8 m/s², not 0.
• Zero acceleration does not mean at rest. A puck gliding at a steady +2.0 m/s has a_x = 0.

Preview: constant-acceleration equations

When a_x is constant, Topic 1.3 gives three equations that link x, v_x, a_x and t:

Equation	Missing quantity
v_x = v_x0 + a_x t	x
x = x₀ + v_x0 t + ½ a_x t²	v_x
v_x² = v_x0² + 2a_x(x − x₀)	t

They are valid only when acceleration is constant. This course describes changing acceleration in words and sketches, but does not ask you to calculate with it. The first equation is just the definition of average acceleration rearranged, because for constant acceleration the average and instantaneous values are equal.

Common misconceptions

• “Negative acceleration means slowing down.” No. Compare the signs of v_x and a_x (Worked example 2 and Figure 2).
• “Displacement and distance are the same thing.” Only when the object never reverses direction (Worked example 1).
• “Average velocity is the mean of the starting and ending velocities.” That only works for constant acceleration. Use Δx / Δt.
• “If v = 0, then a = 0.” A ball at the top of its flight has v_y = 0 and a_y = −9.8 m/s².
• “The slope of a curve is the slope between two points on it.” That is a secant (an average). The instantaneous value needs the tangent line.
• Reading the height of a position–time graph as velocity. Velocity is the slope, not the height. A high, flat line means far away and at rest.
• Forgetting to state the axis. Without “+x to the right” (or similar), a negative answer has no meaning.

Where this leads

These definitions are the language of all of kinematics. Topic 1.3 (Representing Motion) adds the full graph links and the constant-acceleration equations; Topics 1.4 and 1.5 extend them to moving reference frames and two dimensions. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.