B2.4 Investigate uniform and non-uniform linear motion

Ontario Grade 11 Physics

Kinematics

Ontario Grade 11 Physics — B2.4

Linear motion is motion along a straight line. In this lesson, the physical system is one moving object, such as a cart. Describe its motion relative to a chosen reference point, such as the starting end of a track. Choose a positive direction along the track before recording positions. Motion in the opposite direction is negative. An investigation can use measurements and graphs to decide whether the object moves uniformly or changes its motion.

What you will learn

Distinguish uniform linear motion from non-uniform linear motion.
Use position, displacement, distance, time, and velocity to describe motion.
Interpret position–time information and use it to investigate motion.
Plan a simple investigation and distinguish proposed procedures from measured evidence.

1. Prerequisite bridge: describing position and motion

A position tells where an object is compared with a reference point. For motion along a track, position can be recorded in metres. The change in position is displacement. Displacement includes direction, so it is a vector. A vector has both magnitude and direction. Distance is the total length of the path travelled. Distance is a scalar: it has magnitude but no direction.

For example, if a cart moves from position

0.0\,\mathrm{m}

to

3.0\,\mathrm{m}

in the positive direction, its displacement is

+3.0\,\mathrm{m}

. If it then returns to

1.0\,\mathrm{m}

, the total distance is

5.0\,\mathrm{m}

, while its displacement from the start is

+1.0\,\mathrm{m}

.

Time is measured in seconds. Average velocity compares displacement with the time taken. Velocity is a vector, so its sign gives direction under the chosen sign convention. Average speed compares total distance with time and is a scalar. For one-way motion in a straight line, speed and the magnitude of velocity are equal. A negative velocity does not mean a negative speed; it means motion in the negative direction.

Uniform linear motion means motion along a straight line with constant velocity. The object covers equal displacements in equal time intervals and does not change direction. Non-uniform linear motion means the velocity changes. The object may change its speed, its direction along the line, or both.

\bar{v}=\frac{\Delta x}{\Delta t}

Choose and state a positive direction before using signed positions or velocities.
Displacement and velocity are vectors; distance and speed are scalars.
Uniform motion has constant velocity. Non-uniform motion does not.

2. Investigating motion with measurements and graphs

A position–time record gives an object's position at selected times. A position–time graph places time on the horizontal axis and position on the vertical axis. A straight, steadily rising or falling line indicates uniform motion: position changes by equal amounts in equal time intervals. A flat line means the object is at rest relative to the reference point. A line that becomes steeper or less steep indicates changing velocity and therefore non-uniform motion.

The slope of a straight section of a position–time graph is the change in position divided by the change in time. Its units are metres per second. A positive slope means motion in the positive direction; a negative slope means motion in the negative direction. A larger slope magnitude means a greater speed. For a curved or changing-slope graph, compare slopes over separate time intervals to see how the motion changes.

A suitable investigation could use a cart on a straight track, a metre scale, and a timer or a motion sensor. First mark a reference point and choose a positive direction. Record the cart's position at known times, or use a sensor to collect position readings. Repeat the run if appropriate, then organize the recorded positions and times in a table and graph them. The procedure is proposed here; no measurements are claimed. Any conclusion must be based on the group's actual recorded evidence.

Keep the track straight and use the same reference point throughout. Record units with every measurement. If timing by hand, reaction time may affect the readings; a motion sensor can reduce this source of timing uncertainty. Do not decide that motion is uniform from one interval alone. Compare several intervals and consider whether differences could be due to measurement uncertainty.

\text{slope}=\frac{\Delta x}{\Delta t}

A position–time graph shows how position changes over time.
The slope of a straight section represents average velocity for that interval.
A proposed procedure is not evidence; conclusions require recorded measurements.

3. Applying the motion model and checking results

For uniform linear motion, the same velocity applies throughout the interval. The displacement can be found by multiplying velocity by elapsed time. If the object starts at position

x_i

, its final position is the initial position plus displacement. Keep the sign of velocity: a negative value gives a displacement in the negative direction.

For non-uniform motion, one average velocity describes the overall displacement over the full time interval, but it does not show every change during the motion. To investigate the changes, compare positions over shorter intervals or examine the changing slope of a position–time graph.

A useful quantity for describing a change in velocity is average acceleration. It compares the change in velocity with the time taken. Acceleration is a vector, measured in metres per second squared. Its sign describes the direction of the velocity change, not automatically whether the object is speeding up or slowing down. For instance, a negative acceleration can mean an object is slowing while moving in the positive direction.

When solving, state the object, reference point, positive direction, known values, and unknown. Write the governing relationship before substituting. Keep units in the calculation, round to a sensible number of significant figures, and state direction. Finally, check that the units fit the quantity and that the result makes sense for the motion described.

\bar{a}=\frac{\Delta v}{\Delta t}

Use signed displacement and velocity for straight-line motion.
Average velocity summarizes an interval; it may hide changes within that interval.
Average acceleration describes how velocity changes over time.

Reading motion from position–time information

Position–time pattern	Motion description
Straight line with constant positive slope	Uniform motion in the positive direction
Straight line with constant negative slope	Uniform motion in the negative direction
Horizontal line	At rest relative to the reference point
Changing slope	Non-uniform motion

Worked example

1. Uniform motion in the positive direction

A cart moves uniformly from position

0.40\,\mathrm{m}

to position

2.20\,\mathrm{m}

in

3.0\,\mathrm{s}

. Find its average velocity. Take the positive direction to be along the track from the start toward the end.

Set the system and direction
The system is the cart, measured relative to the track's starting reference point. The stated direction is positive. The known positions and elapsed time determine the displacement and average velocity.
Find displacement
Subtract initial position from final position. A positive result means the cart moved in the chosen positive direction.
$\Delta x=2.20\,\mathrm{m}-0.40\,\mathrm{m}=+1.80\,\mathrm{m}$
Calculate average velocity
Average velocity is displacement divided by elapsed time. The units reduce to metres per second.
$\bar{v}=\frac{+1.80\,\mathrm{m}}{3.0\,\mathrm{s}}=+0.60\,\mathrm{m/s}$

Answer: The cart's average velocity is

0.60\,\mathrm{m/s}

in the positive direction. Because its motion is stated to be uniform, this is also its velocity during the interval.

Check: The units are metres per second, as required for velocity. The positive sign matches the increasing position. A displacement of about

1.8\,\mathrm{m}

over

3.0\,\mathrm{s}

gives a plausible speed of about

0.60\,\mathrm{m/s}

.

Worked example

2. Detecting non-uniform motion from positions

A cart's recorded positions are

0.0\,\mathrm{m}

at

0.0\,\mathrm{s}

,

1.2\,\mathrm{m}

at

2.0\,\mathrm{s}

, and

3.0\,\mathrm{m}

at

4.0\,\mathrm{s}

. Determine the average velocity in each time interval and decide whether the motion is uniform. Positive is toward increasing position.

Compare the first interval
Use the change in position over the first $2.0\,\mathrm{s}$ . The positive displacement shows motion in the positive direction.
$\bar{v}_1=\frac{1.2\,\mathrm{m}-0.0\,\mathrm{m}}{2.0\,\mathrm{s}-0.0\,\mathrm{s}}=+0.60\,\mathrm{m/s}$
Compare the second interval
Apply the same relationship to the next $2.0\,\mathrm{s}$ . Using equal time intervals makes the change in average velocity easy to compare.
$\bar{v}_2=\frac{3.0\,\mathrm{m}-1.2\,\mathrm{m}}{4.0\,\mathrm{s}-2.0\,\mathrm{s}}=+0.90\,\mathrm{m/s}$
Classify the motion
The average velocities differ, so the position changes by different amounts in equal time intervals. This evidence indicates non-uniform motion over these intervals.
$+0.60\,\mathrm{m/s}\ne+0.90\,\mathrm{m/s}$

Answer: The average velocities are

0.60\,\mathrm{m/s}

and

0.90\,\mathrm{m/s}

, both in the positive direction. The recorded intervals indicate non-uniform motion.

Check: Each velocity has units of metres per second. Both signs agree with increasing position. The conclusion is limited to the recorded data; measurement uncertainty could affect whether the difference is meaningful in a real investigation.

Worked example

3. Describing a change in velocity

A cart's velocity changes from

+0.50\,\mathrm{m/s}

to

+1.70\,\mathrm{m/s}

over

2.0\,\mathrm{s}

. Find its average acceleration. Positive is toward the end of the track.

Identify the velocity change
Both velocities are positive, so the cart moves toward the end of the track at both times. Subtract initial velocity from final velocity to find the signed change.
$\Delta v=+1.70\,\mathrm{m/s}-(+0.50\,\mathrm{m/s})=+1.20\,\mathrm{m/s}$
Calculate average acceleration
Average acceleration is the change in velocity divided by the time interval. The resulting unit is metres per second squared.
$\bar{a}=\frac{+1.20\,\mathrm{m/s}}{2.0\,\mathrm{s}}=+0.60\,\mathrm{m/s^2}$

Answer: The average acceleration is

0.60\,\mathrm{m/s^2}

in the positive direction.

Check: The units are metres per second squared. The positive sign matches the increase in positive velocity. The cart speeds up in the positive direction, consistent with the stated velocities.

Common mistakes and how to avoid them

Treating distance and displacement as interchangeable.

Correction: Distance is the total path length and has no direction. Displacement is the change in position and includes direction.

Calling every negative velocity a negative speed.

Correction: Speed is non-negative. A negative velocity indicates motion in the negative direction under the selected sign convention.

Assuming that a positive acceleration always means an object is speeding up.

Correction: Compare the directions of velocity and acceleration. In the example lesson, a positive velocity and positive acceleration mean the velocity's positive value is increasing.

Concluding that motion is uniform from a single time interval.

Correction: Compare multiple intervals or check whether a position–time graph has a constant slope. Consider measurement uncertainty when interpreting real data.

Lesson summary

Linear motion takes place along a straight line.
Uniform motion has constant velocity; non-uniform motion has changing velocity.
Displacement and velocity include direction. Distance and speed do not.
A position–time graph's slope represents average velocity over an interval.
Investigations require recorded measurements before conclusions can be made.

Check your understanding

Question 1

A position–time graph is a straight line with a constant negative slope. What does this show?

The object is at rest.
The object moves uniformly in the negative direction.
The object moves uniformly in the positive direction.
The object changes direction repeatedly.

Show answer and explanation

The object moves uniformly in the negative direction.

A constant slope indicates constant velocity. A negative slope indicates motion in the negative direction.

Question 2

A student records an object's position every second. The position changes by different amounts in successive one-second intervals. Which conclusion is best supported?

The object's motion is non-uniform over the recorded intervals.
The object must be moving in a circle.
The object is at rest.
The object's speed is constant.

Show answer and explanation

The object's motion is non-uniform over the recorded intervals.

Different position changes in equal time intervals indicate changing average velocity and therefore non-uniform motion in the recorded data.

Key terms

Reference point: A chosen location used to describe an object's position.
Displacement: The change in position, including direction.
Velocity: Displacement divided by elapsed time; a vector with direction.
Uniform linear motion: Motion along a straight line with constant velocity.
Non-uniform linear motion: Motion along a straight line in which velocity changes.
Position–time graph: A graph showing an object's position at different times.
Average acceleration: The change in velocity divided by the time taken for that change.

Continue through SPH3U

View the complete SPH3U Ontario Grade 11 Physics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

B2.4 · Investigate uniform and non-uniform linear motion

Ontario Grade 11 Physics

Ontario Grade 11 Physics — B2.4

What you will learn

1. Prerequisite bridge: describing position and motion

2. Investigating motion with measurements and graphs

3. Applying the motion model and checking results

Reading motion from position–time information

1. Uniform motion in the positive direction

2. Detecting non-uniform motion from positions

3. Describing a change in velocity

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through SPH3U

About this lesson and its review