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B1.1 · Analyse a technology that applies kinematics

Learn to analyse a technology that applies kinematics through clear examples and targeted practice.

Ontario Grade 11 Physics

Kinematics

Analysing a technology through motion, evidence, and design limits

Kinematics describes motion using quantities such as position, displacement, time, velocity, and acceleration. It does not explain what causes the motion. Many technologies use kinematics to measure, predict, or respond to motion. This lesson analyses a speed camera that estimates a vehicle’s average speed from images taken at known locations and times. The calculations use simplified, stated information. They are examples, not measurements from a real investigation.

What you will learn

1. Prerequisite bridge: describing motion

A scalar has magnitude only. Time and distance are scalars. A vector has both magnitude and direction. Displacement and velocity are vectors. For example, a displacement of 20 m east includes a size and a direction.
Distance is the total path length travelled. Displacement is the change in position from the starting point to the ending point. They are not always equal: a vehicle that drives around a block and returns to its start has travelled a distance, but its displacement is zero.
Choose a system before describing motion. Here, the system is one vehicle. Choose the direction along the road as positive. Motion in the opposite direction is negative. The position coordinate, xx, is measured in metres; time, tt, in seconds. A speed is a scalar, in metres per second. Velocity is a vector, also in metres per second, and includes direction.
Average velocity is displacement divided by the time interval. Average speed is distance divided by the time interval. Acceleration describes how velocity changes over a time interval. Its SI unit is metres per second squared. These average quantities can describe a whole interval without telling us every detail of the motion within it.
Δx=xf−xi,vavg=ΔxΔt,aavg=ΔvΔt\Delta x=x_f-x_i,\quad v_{\mathrm{avg}}=\frac{\Delta x}{\Delta t},\quad a_{\mathrm{avg}}=\frac{\Delta v}{\Delta t}

2. Physical situation and camera model

Imagine two cameras or image points along a straight road. The distance between them is known. A vehicle passes the first point, then the second. The system is the vehicle; the positive direction is the vehicle’s direction of travel. The technology records the times when the vehicle reaches each point.
A simple position–time diagram shows the measurement locations. The vertical axis represents position along the road, and the horizontal axis represents time. The line between the two recorded positions has a slope: position change divided by time change. That slope gives average velocity for the interval.
Position–time sketch: xx (m) increases upward. At t1t_1, the vehicle is at x1x_1; at the later time t2t_2, it is at x2>x1x_2>x_1. The straight segment joining (t1,x1)(t_1,x_1) and (t2,x2)(t_2,x_2) represents the average motion over that interval.
For travel in the positive direction, the camera’s average speed is the known separation divided by the elapsed time. If the vehicle travels in the opposite direction, the displacement and average velocity are negative under this convention, but speed remains positive. A camera system can use the direction of travel when it interprets the result.
The result is an average over the measured section. If the vehicle speeds up or slows down between the points, this value does not show its speed at every instant. The system also depends on accurate distance information, reliable timing, and correctly matching the two images to the same vehicle.
vavg=ΔxΔtv_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}

3. Analysing the technology

To analyse a technology, connect its physical principle to its purpose. The camera applies kinematics because it uses positions and times to calculate motion. It converts image records into an estimate of average speed. Its useful output depends on the quality of the inputs.
The known distance must represent the separation along the vehicle’s path. The timing must correspond to the vehicle crossing each measurement point. If either value is inaccurate, the calculated speed may be inaccurate. If the images are matched to different vehicles, the result is not a valid motion calculation for one vehicle.
A speed camera can assess motion across a road section rather than relying on a single position reading. That is useful when the purpose is to estimate average speed over that section. However, the value cannot by itself reveal whether the vehicle maintained a constant speed, accelerated, or slowed down during the interval. More detailed motion information would require additional position-and-time records.
Keep measured evidence separate from a proposed procedure. In this lesson, the numbers in the examples are given for practice. They are not claimed as camera measurements. A real evaluation would need records from the specific system, such as its stated distance and timing method, and evidence about its accuracy. A proposed test is not itself evidence that the technology works.
average speed=distance travelledelapsed time\text{average speed}=\frac{\text{distance travelled}}{\text{elapsed time}}

4. Reading results and checking reasonableness

A motion result should be reported with a value, SI unit, and direction when it is a vector. Average speed has no direction. Average velocity does. The sign of a velocity depends on the direction chosen as positive; it is not an extra physical speed.
Before accepting a calculated result, check the units. Dividing metres by seconds gives metres per second. Check the direction against the chosen coordinate system. Finally, ask whether the value makes sense for the distance and time: a longer distance in the same time means a greater speed, while a longer time for the same distance means a lower speed.
For an analysis of the camera, also check what the result can and cannot support. A correct average-speed calculation supports a statement about the measured interval. It does not, without additional records, establish the vehicle’s speed at every point along the road.
1 m/s=3.6 km/h1\ \mathrm{m/s}=3.6\ \mathrm{km/h}

Worked example

Finding average speed from timed camera points

A camera system records a car at two points 84 m apart. The car takes 6.0 s to travel between them. Find its average speed and average velocity. The car travels in the positive direction.
  1. Define the system and direction
    The system is the car. The road direction of travel is positive. The known values are a displacement of 84 m and an elapsed time of 6.0 s. The unknowns are average speed and average velocity.
  2. Choose the relationship
    The car moves between two points along a straight section without reversing, so its distance and the magnitude of its displacement are both 84 m. Divide displacement by elapsed time for average velocity; the speed is the magnitude of that velocity.
    vavg=ΔxΔtv_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}
  3. Substitute and report
    The quotient is positive because the car moves in the chosen positive direction. Report two significant figures, consistent with the given values.
    vavg=84 m6.0 s=14 m/sv_{\mathrm{avg}}=\frac{84\ \mathrm{m}}{6.0\ \mathrm{s}}=14\ \mathrm{m/s}
Answer: The car’s average speed is 14 m/s. Its average velocity is 14 m/s in the positive road direction.
Check: The units reduce to m/s. A distance of 84 m in 6.0 s gives a plausible average speed, and the positive sign agrees with the chosen direction.

Worked example

Interpreting travel in the opposite direction

A car travels from position 310 m to position 190 m in 8.0 s. The positive direction is east. Determine its average velocity and average speed.
  1. Define the system and direction
    The system is the car. East is positive, so westward displacement will be negative. The initial and final positions and the elapsed time are known.
  2. Find displacement
    Displacement is final position minus initial position. The negative result means the car moved west, opposite to the positive east direction.
    Δx=xf−xi=190 m−310 m=−120 m\Delta x=x_f-x_i=190\ \mathrm{m}-310\ \mathrm{m}=-120\ \mathrm{m}
  3. Calculate velocity and speed
    Average velocity keeps the negative sign to show direction. The car does not reverse, so its distance is 120 m; average speed is the positive distance divided by time.
    vavg=−120 m8.0 s=−15 m/s,average speed=120 m8.0 s=15 m/sv_{\mathrm{avg}}=\frac{-120\ \mathrm{m}}{8.0\ \mathrm{s}}=-15\ \mathrm{m/s},\quad \text{average speed}=\frac{120\ \mathrm{m}}{8.0\ \mathrm{s}}=15\ \mathrm{m/s}
Answer: Average velocity is 15 m/s west, or −15 m/s using east as positive. Average speed is 15 m/s.
Check: Both results have units of m/s. The negative velocity agrees with westward motion. The speed is positive because speed has no direction.

Worked example

Using acceleration to interpret a changing speed

A test vehicle moves east at 12 m/s and, 5.0 s later, moves east at 20 m/s. East is positive. Find its average acceleration.
  1. Define the system and known values
    The system is the test vehicle. East is positive. Initial velocity is 12 m/s, final velocity is 20 m/s, and the time interval is 5.0 s. The unknown is average acceleration.
  2. Apply the average-acceleration relationship
    Average acceleration is the change in velocity divided by elapsed time. Both velocities are positive because both point east.
    aavg=vf−viΔta_{\mathrm{avg}}=\frac{v_f-v_i}{\Delta t}
  3. Substitute and interpret
    The positive result means the velocity change is in the positive east direction. Report two significant figures.
    aavg=20 m/s−12 m/s5.0 s=1.6 m/s2a_{\mathrm{avg}}=\frac{20\ \mathrm{m/s}-12\ \mathrm{m/s}}{5.0\ \mathrm{s}}=1.6\ \mathrm{m/s^2}
Answer: The vehicle’s average acceleration is 1.6 m/s² east.
Check: The units are (m/s)/s, or m/s². The velocity increases eastward, so positive acceleration is reasonable. This average does not show exactly how the velocity changed at every moment.

Common mistakes and how to avoid them

Treating speed and velocity as interchangeable.
Correction: Speed is a scalar and has no direction. Velocity is a vector and includes direction; its sign depends on the chosen positive direction.
Using distance when calculating displacement.
Correction: Displacement is final position minus initial position. Distance is the total path length and cannot be negative.
Claiming an average-speed camera gives the vehicle’s speed at every instant.
Correction: It estimates average speed over the measured section. The vehicle could have sped up or slowed down within that interval.
Reporting a numerical answer without checking units or direction.
Correction: Carry SI units through the calculation and state the direction for vector quantities. Check that the sign matches the coordinate choice.

Lesson summary

Check your understanding

Question 1

A vehicle travels 150 m in the positive direction in 10 s. What are its average speed and average velocity?
  1. 15 m/s speed; 15 m/s in the positive direction
  2. 15 m/s speed; 15 m/s in the negative direction
  3. 1.5 m/s speed; 1.5 m/s in the positive direction
  4. 1500 m/s speed; 1500 m/s in the positive direction
Show answer and explanation
15 m/s speed; 15 m/s in the positive direction
The distance and positive displacement are both 150 m. Dividing by 10 s gives 15 m/s; the velocity is positive because the motion is in the positive direction.

Question 2

A camera reports an average speed for a road section. Which conclusion is supported by that value alone?
  1. The vehicle kept exactly the same speed throughout the section.
  2. The vehicle’s average speed over the measured section has been estimated.
  3. The vehicle’s speed at every position on the road is known.
  4. The vehicle accelerated at a constant rate.
Show answer and explanation
The vehicle’s average speed over the measured section has been estimated.
The calculation uses the section’s distance and elapsed time. It gives an average for that interval, not a complete record of how speed changed.

Question 3

East is positive. A vehicle’s average velocity is −8.0 m/s. Which statement is correct?
  1. Its average speed is −8.0 m/s and it moved east.
  2. Its average speed is 8.0 m/s and it moved west.
  3. Its average speed is 8.0 m/s and it moved east.
  4. Its average speed is zero and it moved west.
Show answer and explanation
Its average speed is 8.0 m/s and it moved west.
The negative velocity indicates motion opposite to east, so the vehicle moved west. Speed is the magnitude of velocity and is positive: 8.0 m/s.

Key terms

Kinematics
The description of motion using quantities such as position, time, velocity, and acceleration.
Scalar
A quantity with magnitude only, such as distance, time, or speed.
Vector
A quantity with both magnitude and direction, such as displacement, velocity, or acceleration.
Displacement
The change in position from the starting point to the ending point, including direction.
Average velocity
Displacement divided by the elapsed time.
Average speed
Distance travelled divided by the elapsed time.
Acceleration
The change in velocity divided by the elapsed time.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B1.1. 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.

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