B. Kinematics · Kinematics: Motion in One and Two Dimensions

Position-Time Graphs: Slope, Uniform Motion and Instantaneous Velocity

B2.2 B3.1 B2.7 A1.6About 70 minutes

Learning goals

I can describe an object's motion from a position-time graph.
I can calculate velocity as the slope of a position-time graph.
I can find average velocity from a secant and instantaneous velocity from a tangent on a curved graph.
I can use graphs or equations to find where and when two moving objects meet.

Words to know

position-time graph (d–t graph)
A graph with time on the horizontal axis and position on the vertical axis.

slope
Rise ÷ run. On a d–t graph, slope = Δd ÷ Δt = velocity.

secant
A straight line joining two points on a curve. Its slope gives the average velocity between those times.

tangent
A straight line that touches a curve at one point and has the same steepness as the curve there. Its slope gives the instantaneous velocity.

uniform motion
Motion at constant velocity. It gives a straight-line d–t graph.

A picture of a whole trip

A sentence like "the bus went 3 m/s for 4 s, stopped, then reversed" takes effort to imagine. A position-time graph shows the whole trip at once: when the object moved, how fast, which way, and when it stopped. Physicists, engineers and coaches read these graphs the way you read a map.

Slope is velocity

On a position-time (d–t) graph, time goes on the horizontal axis and position on the vertical axis. Between any two points,

slope = rise ÷ run = Δd ÷ Δt = vₐᵥ

So the slope of a d–t graph is the velocity. Here is how to read the shape:

Straight, sloping line: constant velocity (uniform motion). The steeper the line, the greater the speed.
Positive slope (up to the right): moving in the positive direction.
Negative slope (down to the right): moving in the negative direction.
Horizontal line (slope 0): at rest; the position is not changing.
Where the line crosses the time axis: the object passes the reference point (d = 0). This is not where it stops.

A toy car's trip in three segments (east positive). A: moves east at 3 m/s. B: at rest at 12 m [E]. C: moves west at 4 m/s, passes the reference point at t = 9 s and ends at 4 m [W].

Worked example

Example 1: Reading the toy-car graph

Use the graph above. Find (a) the velocity in each segment, (b) the displacement for the whole 10 s, and (c) the average speed and average velocity for the whole 10 s.

(a) A: slope = (12 − 0) ÷ (4 − 0) = +3.0 m/s, so 3.0 m/s [E]. B: slope = (12 − 12) ÷ (6 − 4) = 0 m/s (at rest). C: slope = (−4 − 12) ÷ (10 − 6) = −16 ÷ 4 = −4.0 m/s, so 4.0 m/s [W].
(b) Δd = d₂ − d₁ = (−4 m) − 0 = −4 m = 4 m [W].
(c) Distance = 12 m (going east) + 16 m (coming back from 12 to −4) = 28 m. Average speed = 28 m ÷ 10 s = 2.8 m/s.
vₐᵥ = −4 m ÷ 10 s = −0.4 m/s = 0.4 m/s [W].

A: 3.0 m/s [E]; B: 0; C: 4.0 m/s [W]. Δd = 4 m [W]; average speed 2.8 m/s; average velocity 0.4 m/s [W].

Curved graphs: non-uniform motion

If the d–t graph curves, the slope is changing, so the velocity is changing. A curve that gets steeper means the object is speeding up; one that flattens out means it is slowing down.

Two straight lines help you read a curve:

A secant joins two points on the curve. Its slope is the average velocity between those two times.
A tangent touches the curve at a single point and matches its steepness there. Its slope is the instantaneous velocity at that time.

To draw a tangent by hand, place a ruler so it touches the curve at the point without cutting through it, and so the curve bends away from the ruler equally on both sides. Then pick two points far apart on the ruler line (not on the curve) and calculate rise ÷ run.

A ball rolling down a ramp, speeding up. The secant from t = 1.0 s to 4.0 s gives the average velocity over that interval. The tangent at P gives the instantaneous velocity at t = 3.0 s.

Worked example

Example 2: Tangent and secant

Use the curved graph above. Find (a) the average velocity from t = 1.0 s to t = 4.0 s and (b) the instantaneous velocity at t = 3.0 s.

(a) Secant points: (1.0 s, 1.0 m) and (4.0 s, 16 m). vₐᵥ = (16 − 1.0) ÷ (4.0 − 1.0) = 15 ÷ 3.0 = 5.0 m/s [E].
(b) Use two points on the tangent line, far apart: (1.5 s, 0 m) and (5.0 s, 21 m).
v = (21 − 0) ÷ (5.0 − 1.5) = 21 ÷ 3.5 = 6.0 m/s [E].
The instantaneous velocity at 3.0 s (6.0 m/s) is larger than the average from 1.0 s to 4.0 s (5.0 m/s). That makes sense: the ball is speeding up the whole time, so it moves slower than the average early in the interval and faster than the average later, and 3.0 s is in the later part.

(a) 5.0 m/s [E]; (b) 6.0 m/s [E].

Worked example

Example 3: When and where do they meet?

At t = 0, a cyclist is at the reference point and rides at a constant 2.0 m/s [E]. At the same time a jogger is 30 m [E] of the reference point and runs at a constant 3.0 m/s [W]. When and where do they meet?

East positive. Constant velocity means d = d₁ + vt for each person (a straight-line d–t graph with slope v and vertical intercept d₁).
Cyclist: d = 0 + 2.0t. Jogger: d = 30 − 3.0t.
They meet when their positions are equal (where the two lines cross): 2.0t = 30 − 3.0t, so 5.0t = 30 and t = 6.0 s.
Position: d = 2.0 × 6.0 = 12 m [E]. Check with the jogger: 30 − 3.0 × 6.0 = 12 m ✓.

They meet at t = 6.0 s, 12 m [E] of the reference point.

Graphing data from an experiment (A1.6)

When you graph your own data: put the independent variable (usually time) on the horizontal axis; give each axis a title with units; choose a scale that uses most of the grid; plot points carefully; and draw a line or smooth curve of best fit rather than joining the dots. Real data scatter a little, so the best-fit line will not pass through every point. Calculate slopes from points on the best-fit line, not from data points.

Just for fun

A position-time graph of my cat on a Saturday: one long horizontal line... and then a single, extremely steep section at the exact moment the treat bag rustles.

Hands-on activity

Make your own position-time graph

You need: a long hallway or driveway, a measuring tape, 8 small markers (coins, socks or beanbags), a stopwatch, a partner and graph paper.

Your partner starts the stopwatch and says "drop" every 2.0 s. Walk in a straight line from the start. Each time you hear "drop", drop a marker beside your foot.
Trial 1: walk at a steady pace. Trial 2: start slowly and speed up.
Measure each marker's position from the start. Make a table of t and d for each trial.
Graph both trials on the same d–t axes. Draw a best-fit line for Trial 1 and a smooth curve for Trial 2.
Find your velocity in Trial 1 from the slope. For Trial 2, draw a tangent near the end and find your instantaneous velocity there.

What to look for: Trial 1 gives nearly equally spaced markers and a nearly straight graph (slope around 1 to 1.5 m/s for a normal walk). Trial 2 gives growing gaps and a curve that gets steeper. A tangent near the end should be steeper than the Trial 2 secant over the whole walk.

Your turn

On a d–t graph, what does a horizontal line mean?

A d–t graph is a straight line through (0 s, 5.0 m) and (4.0 s, 25.0 m). Find the velocity (north positive).

A straight d–t graph goes from (2.0 s, 18 m) to (8.0 s, 0 m). Find the velocity (east positive) and describe the motion.

Two lines on the same d–t graph have slopes of +2 m/s and −5 m/s. Which object is faster?

On a curved d–t graph, how do you find the instantaneous velocity at t = 4 s?

A d–t graph curves and gets flatter as time goes on. Describe the motion.

More practice: warm-up

What does the slope of a d–t graph represent?

What does a straight, sloping d–t line tell you about the velocity?

Slope from (0 s, 0 m) to (5 s, 20 m)?

What does a negative slope mean (east positive)?

Which line gives average velocity on a curve: a secant or a tangent?

More practice: on your own

A d–t graph passes through (0 s, −10 m), (5 s, −10 m) and (9 s, 14 m) with straight segments between. Describe each segment and give the velocities (east positive).

For the same graph, find the average velocity from 0 to 9 s.

A truck at 40 m [E] drives at 4.0 m/s [W]. A car at 0 m drives at 6.0 m/s [E]. Both start at t = 0. When and where do they meet?

On a curved d–t graph, the tangent at t = 2.0 s passes through (0 s, 3.0 m) and (4.0 s, 11.0 m). Find the instantaneous velocity at 2.0 s.

Sketch (in words) the d–t graph of a student who walks to the corner at a steady pace, waits for the light, then jogs back home.

Challenge

A ball's position is d = 2.0t² (d in m, t in s). Find the average velocity from t = 1.0 s to 3.0 s. Then estimate the instantaneous velocity at t = 2.0 s using a very short secant from 1.99 s to 2.01 s.

Can a d–t graph ever be a vertical line? Explain.

Car A leaves a town at 9:00 driving 80 km/h [E]. Car B leaves the same town at 9:30 driving 100 km/h [E]. When does B catch A, and how far from town?

Check your understanding

Quiz

On a position-time graph, the slope gives:

A straight d–t line through (1.0 s, 4.0 m) and (3.0 s, 12.0 m) shows a velocity of:

A d–t graph that curves upward and gets steeper shows an object that is:

Where a d–t line crosses the time axis, the object:

The instantaneous velocity at a point on a curved d–t graph is the slope of the:

Two objects' d–t lines cross at (6 s, 15 m). This means:

For parents and teachers

Big idea: The slope of a position-time graph is velocity (B2.2). A secant gives average velocity and a tangent gives instantaneous velocity (B3.1). Solving "when do they meet?" connects to Grade 9 work on points of intersection (B2.7).

How to help: Ask your teen to tell the story of a d–t graph out loud, segment by segment: "Here I walk away slowly, here I stop, here I run back." Telling the story catches most misreadings.

Common difficulties:

Reading a d–t graph as a picture of the path (a hill on the graph is not a hill on the road).
Thinking the object stops where the line crosses the time axis.
Using data points instead of points on the tangent line to find a tangent's slope.
Forgetting units on slopes (m/s).