B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Motion Graphs: From Position to Velocity and Acceleration

B2.1About 65 minutes

Learning goals

I can sketch velocity-time and acceleration-time graphs from a position-time graph.
I can describe a motion in words from its velocity-time graph.
I can connect concavity and inflection points of a position graph to acceleration.
I can plan a walk that produces a given motion-sensor graph.

Words to know

motion sensor
A device that measures the distance to an object many times per second (usually with ultrasound echoes) and graphs position against time.

concave up
Bending upward like a cup; on an s-t graph it means a > 0.

concave down
Bending downward like a cap; on an s-t graph it means a < 0.

point of inflection
A point where a graph changes concavity; on an s-t graph, a = 0 and v has a local max or min.

corner
A sharp point on a graph where the slopes from the left and right disagree, so the derivative does not exist.

Walk the graph

Many Ontario classrooms use a motion sensor: a small device that sends out ultrasound pulses and times their echoes to measure how far away you are, many times per second. Software then draws your position-time graph as you walk. A classic challenge is "match the graph": the screen shows a graph, and you must walk so that your graph matches it.

To win, you need to read a graph's slopes. In this lesson you sketch velocity and acceleration graphs from a position graph, and you go the other way, from a velocity graph to a story about the motion.

Reading one graph from another

Since v = s′ and a = v′:

The height of the v-t graph at any time is the slope of the s-t graph at that time.
The height of the a-t graph is the slope of the v-t graph.

So:

Where the s-t graph rises, v > 0. Where it falls, v < 0. At a peak or valley (horizontal tangent), v = 0.
A straight s-t graph means constant velocity, so a = 0.
Where the s-t graph is concave up (bending up like a cup), v is increasing, so a > 0. Where it is concave down, a < 0.
At a point of inflection of the s-t graph, the slope is steepest (or least steep), so v has a local maximum or minimum and a = 0.

These are the same connections between f, f′ and f″ that you used in Curve Sketching: Connecting f, f′ and f″, now with a physical meaning.

Position-time graph	Velocity	Acceleration
rising	positive	—
falling	negative	—
horizontal tangent (peak or valley)	zero	—
straight line	constant	zero
concave up	increasing	positive
concave down	decreasing	negative
point of inflection	local max or min	zero (changes sign)

How features of the s-t graph show up in v and a.

Worked example

Example 1: A motion-sensor walk

A student starts 1 m from a motion sensor. She walks steadily away for 3 s until she is 4 m away, stands still for 2 s, then walks steadily back to 1 m from the sensor in 1.5 s. Sketch the position-time and velocity-time graphs and describe the acceleration.

From t = 0 to 3: s goes from 1 m to 4 m, so the slope is (4 − 1) ÷ 3 = 1 m/s.
From t = 3 to 5: s stays at 4 m, so v = 0.
From t = 5 to 6.5: s goes from 4 m to 1 m, so v = (1 − 4) ÷ 1.5 = −2 m/s (moving toward the sensor).
Each piece of the v-t graph is horizontal, so a = 0 within each piece. At t = 3 and t = 5 the s-t graph has a corner: the slope from the left and from the right do not agree, so v is not defined there. (In real life she changes speed quickly but not instantly, so the corners are slightly rounded and a is briefly large.)

v = 1 m/s, then 0, then −2 m/s. a = 0 on each piece, with sudden changes in v at the corners.

Left: the position-time graph from the motion sensor. Right: its velocity-time graph. Each straight piece of s gives a constant v; the steeper the piece, the farther v is from 0.

Worked example

Example 2: A smooth position graph

An electric scooter's position is s(t) = −t³ + 6t² metres for 0 ≤ t ≤ 4 s. Describe its s-t graph, then sketch v and a.

v(t) = −3t² + 12t = −3t(t − 4). v = 0 at t = 0 and t = 4, and v > 0 in between, so the scooter always moves forward during the trip. s rises from s(0) = 0 to s(4) = −64 + 96 = 32 m.
a(t) = −6t + 12, zero at t = 2. For t < 2, a > 0 (s is concave up, the scooter speeds up). For t > 2, a < 0 (concave down, it slows down).
At t = 2 the s-t graph has a point of inflection: it is steepest there. v(2) = −12 + 24 = 12 m/s is the maximum velocity.
The v-t graph is a downward parabola with vertex (2, 12). The a-t graph is a falling line through (2, 0).

The scooter speeds up to a top velocity of 12 m/s at t = 2 s, then slows to a stop at t = 4 s, 32 m from the start.

For Example 2, the three graphs line up at t = 2. Where s is steepest, v is highest and a crosses zero.

Worked example

Example 3: From a velocity graph to a story

The graph below shows the velocity of a robot vacuum on a straight hallway. Describe its motion and give its acceleration on each piece.

From 0 to 2 s, v rises steadily from 0 to 4 m/s: it speeds up with a = (4 − 0) ÷ 2 = 2 m/s².
From 2 to 6 s, v = 4 m/s: it cruises at constant velocity, a = 0.
From 6 to 9 s, v falls from 4 to −2 m/s: a = (−2 − 4) ÷ 3 = −2 m/s². It slows down until v = 0. Solve 4 − 2(t − 6) = 0: t = 8 s. That is where it turns around.
From 8 to 9 s, v < 0 and a < 0: it is moving backward and speeding up.

Speeds up, cruises, slows, stops at t = 8 s, then backs up faster and faster until t = 9 s.

The velocity-time graph for Example 3. The robot turns around where the graph crosses the t-axis (t = 8 s), not where it reaches its lowest point.

Reading a velocity graph

Ask three questions. Above or below the axis? That gives the direction. Rising or falling? That gives the sign of a. Moving toward or away from the axis? That tells you whether the object is slowing down or speeding up.

Just for fun

When you try a motion-sensor "match the graph" challenge, the hardest graph to match is a perfectly vertical line in s. You would have to be in two places at once. (Physics says no; so does your teacher.)

Hands-on activity

Match the graph at home

You need: a hallway at least 4 m long, masking tape, a phone stopwatch and a helper. (If you have a free sensor app such as phyphox, its acoustic or acceleration tools can stand in for a motion sensor, but you do not need one.)

Put tape marks at 0 m, 1 m, 2 m, 3 m and 4 m from a wall. The wall plays the motion sensor.
Draw these three target position-time graphs on paper, each over 6 s: (a) a straight line from (0, 1) to (6, 4); (b) flat at 3 m for 2 s, then a straight line down to (6, 1); (c) a curve that starts flat at 1 m and gets steeper and steeper, reaching 4 m at t = 6.
For each target, plan your walk in words (direction, speed, speeding up or slowing down). Then walk it while your helper calls out each second and checks your position against the graph.
Sketch the velocity-time graph for each target.

What to look for: (a) walk away at a steady 0.5 m/s, so v is a horizontal line at 0.5. (b) stand still for 2 s (v = 0), then walk toward the wall at 0.5 m/s (v = −0.5). (c) start from rest and speed up steadily as you walk away: v is a line rising from 0, and a is positive.

Your turn

An s-t graph is a straight line from (0, 2) to (4, 10). Describe the v-t and a-t graphs.

An s-t graph has a peak at t = 3. What is v(3)? What is the sign of a near t = 3?

An s-t graph is concave up and falling. What are the signs of v and a? Speeding up or slowing down?

A v-t graph is a horizontal line at −3 m/s. Describe the motion.

A v-t graph is a straight line from (0, 6) to (3, 0). Find a, and describe the motion.

s(t) = t³ − 3t² + 5. At what time is the s-t graph steepest going downward, and what is v then?

A v-t graph crosses the t-axis from above to below at t = 5. What happens to the object at t = 5, and what does the s-t graph look like there?

More practice: warm-up

The height of a v-t graph equals the ___ of the s-t graph.

The slope of a v-t graph equals the ___.

A straight s-t graph means the acceleration is…

A concave-down s-t graph means a is…

Where the s-t graph has a valley, v equals…

More practice: on your own

s(t) = 2t³ − 9t² + 12t. Find when the s-t graph has horizontal tangents and when it has its inflection point.

A runner's v-t graph rises from 0 to 8 m/s in 4 s along a curve that is concave down. Is her acceleration increasing or decreasing? Explain.

Sketch (describe) the a-t graph for s(t) = 4t − t² on 0 ≤ t ≤ 4.

A walker's s-t graph from a motion sensor: s = 3 m from t = 0 to 2, then a straight line to s = 0.5 m at t = 7. Find v on each piece.

v(t) = t² − 4t + 3 (m/s). Describe the motion on 0 ≤ t ≤ 4, including turning times.

Challenge

A position graph s(t) = t³ − 6t² + 12t has one point where the tangent is horizontal. Show that the object does not turn around there.

A cyclist's velocity is v(t) = 10 − 10e^(−0.5t) m/s. Find a(t), explain why the cyclist always speeds up, and describe what happens to the velocity for large t.

On a motion sensor, a student's graph is s(t) = 2 + 1.5 sin(πt/2) metres. What is the farthest she gets from the sensor, and what is her greatest speed?

Check your understanding

Quiz

On a position-time graph, the object is moving fastest where…

A v-t graph is below the t-axis and rising toward it. What is happening?

An s-t graph has a point of inflection at t = 4. What is true at t = 4?

A motion-sensor graph is s = 2 m for 0 ≤ t ≤ 3. What is the velocity?

Where does an object turn around, if you are given only its v-t graph?

For s(t) = −t³ + 6t², at what time is the velocity greatest?

For parents and teachers

Big idea (B2.1): velocity is the slope of the position-time graph, and acceleration is the slope of the velocity-time graph. Your teen should be able to move between the three graphs and tell a story about the motion.

How to help: draw a simple position-time graph and ask your teen to walk it across the kitchen while you watch. Then swap: you walk, and they sketch your graph.

Common difficulties:

Reading the height of an s-t graph as speed (height is position; slope is velocity).
Saying an object turns around at the lowest point of a v-t graph (it turns where v = 0).
Forgetting that a concave-up s-t graph means positive acceleration even when the graph is falling.