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A1.12 · Use numeric, symbolic, graphical, and diagram representations

Learn to use numeric, symbolic, graphical, and diagram representations through clear examples and targeted practice.

Ontario Grade 11 Physics

Scientific Investigation Skills and Career Exploration

Using numeric, symbolic, graphical, and diagram representations

One physical situation can be shown in several ways. Numbers give specific values. Symbols show relationships. Graphs make changes visible. Diagrams can show positions and directions. These are different representations of information about the same situation. This lesson uses a cart as its system. A system is the object being studied. We choose the track's origin as position zero and define right as positive. Left is negative. Keep these choices consistent in every representation.

What you will learn

1. Bridge: quantities, units, and direction

A scalar has size but no direction. Time is a scalar. A vector has both size and direction. Position and velocity are vectors. For example, a velocity of 2.0 m/s2.0\ \mathrm{m/s} to the right differs from a velocity of the same size to the left.
Position, written as xx, tells where an object is relative to an origin. Its SI unit, the standard science unit, is the metre (m\mathrm{m}). Time, written as tt, is measured in seconds (s\mathrm{s}). Velocity, written as vv, describes change in position over time and includes direction. Its unit is metres per second (m/s\mathrm{m/s}).
A sign convention is a stated choice of which direction is positive. Here, right is positive and left is negative. So v=−1.2 m/sv=-1.2\ \mathrm{m/s} means the cart moves left at 1.2 m/s1.2\ \mathrm{m/s}. The minus sign gives direction; it does not mean the cart has a negative speed.
Before representing a situation, identify the system, origin, positive direction, known quantities, and unknown quantity. Label quantities with units. Then make sure that signs in equations and arrows in diagrams follow the same direction choice.
v=ΔxΔtv=\frac{\Delta x}{\Delta t}

2. Four ways to represent one motion

A numeric representation gives values with units. For example, a cart can start at 1.0 m1.0\ \mathrm{m} and move at 2.0 m/s2.0\ \mathrm{m/s} to the right. A symbolic representation uses letters to show a relationship. For motion at constant velocity, the final position equals the starting position plus velocity multiplied by elapsed time. Here, x0x_0 is starting position, and tt is elapsed time.
A graphical representation plots one quantity against another. A position–time graph has time on the horizontal axis and position on the vertical axis. The plotted points below show the cart starting at 1.0 m1.0\ \mathrm{m} and moving right at constant velocity. Each star represents one time and its matching position. The graph is labelled with quantities and units.
Position–time graph: xx in metres; tt in seconds

x (m)x\ (\mathrm{m})

7⋆7\quad\quad\quad\quad\quad\quad\quad\quad\star

5⋆5\quad\quad\quad\quad\quad\quad\star

3⋆3\quad\quad\quad\quad\star

1⋆1\quad\star

0123t (s)0\quad\quad 1\quad 2\quad 3\quad t\ (\mathrm{s})

The rising line means position increases as time passes. With right defined as positive, that shows motion to the right. A falling position–time graph shows motion in the negative direction. A horizontal graph means position is not changing.
A diagram representation can show direction directly. Draw the cart as a dot and an arrow for its velocity. The arrow points in the direction of motion. If arrows are drawn to scale, a longer arrow represents a greater velocity magnitude. Label the arrow and state the positive direction.
Representations should agree. A negative velocity should match a graph that falls as time passes and an arrow pointing left when right is positive. If they do not agree, check the sign convention, labels, and calculations.
x=x0+vtx=x_0+vt

3. Translate between representations

To move from a description to an equation, identify each quantity and substitute the values with their units. To make a graph, find the position at selected times and plot each matching pair (t,x)(t,x). Connect the points with a straight line when the stated motion has constant velocity.
To read a graph, first check the axis labels and units. Then identify points and compare the change in position with the change in time. The sign of the position change indicates direction under the chosen sign convention. For constant velocity, the ratio of position change to time change gives velocity.
A vector diagram adds a visual direction check. If the velocity is positive under a right-positive convention, draw the arrow to the right. If it is negative, draw it to the left. Include the magnitude and unit in the label.
Finish by checking that the units fit the quantity, the sign matches the chosen direction, and all representations tell the same story. Use a sensible number of significant figures, which are the digits that reflect the precision of the given values. Also ask whether the result is reasonable for the stated time and motion.
Δx=xf−xi\Delta x=x_f-x_i

Worked example

1. From a description to an equation and value

A cart starts at x0=1.0 mx_0=1.0\ \mathrm{m} and moves right at constant velocity 2.0 m/s2.0\ \mathrm{m/s} for 3.0 s3.0\ \mathrm{s}. Find its final position.
  1. Set the system and direction
    The system is the cart. The origin is the zero point on the track, and right is positive. The known values are the starting position, velocity, and elapsed time. The unknown is final position.
  2. Choose the relationship
    For motion at constant velocity, final position equals starting position plus velocity multiplied by elapsed time. The positive velocity agrees with motion to the right.
    x=x0+vtx=x_0+vt
  3. Substitute and calculate
    Keep units with the values. Metres per second multiplied by seconds gives metres, which can be added to the starting position.
    x=1.0 m+(+2.0 m/s)(3.0 s)=7.0 mx=1.0\ \mathrm{m}+(+2.0\ \mathrm{m/s})(3.0\ \mathrm{s})=7.0\ \mathrm{m}
Answer: The cart's final position is 7.0 m7.0\ \mathrm{m}, to the right of the origin.
Check: The units reduce to metres. The positive result fits the rightward motion from a positive starting position. The cart travels 6.0 m6.0\ \mathrm{m} in 3.0 s3.0\ \mathrm{s}, so the result is reasonable.

Worked example

2. From graph points to velocity

A position–time graph for a cart passes through (0 s,5.0 m)(0\ \mathrm{s},5.0\ \mathrm{m}) and (4.0 s,−3.0 m)(4.0\ \mathrm{s},-3.0\ \mathrm{m}). Find its constant velocity. Right is positive.
  1. Identify the system and information
    The system is the cart. The graph shows an initial position of 5.0 m5.0\ \mathrm{m} and a later position of −3.0 m-3.0\ \mathrm{m}. The unknown is velocity, including direction.
  2. Find position and time changes
    Subtract each initial value from its final value. The negative position change indicates motion to the left with this sign convention.
    Δx=−3.0 m−5.0 m=−8.0 m,Δt=4.0 s−0 s=4.0 s\Delta x=-3.0\ \mathrm{m}-5.0\ \mathrm{m}=-8.0\ \mathrm{m},\quad \Delta t=4.0\ \mathrm{s}-0\ \mathrm{s}=4.0\ \mathrm{s}
  3. Calculate velocity
    For constant velocity, divide position change by elapsed time. The negative sign indicates leftward motion.
    v=ΔxΔt=−8.0 m4.0 s=−2.0 m/sv=\frac{\Delta x}{\Delta t}=\frac{-8.0\ \mathrm{m}}{4.0\ \mathrm{s}}=-2.0\ \mathrm{m/s}
Answer: The cart's velocity is −2.0 m/s-2.0\ \mathrm{m/s}, or 2.0 m/s2.0\ \mathrm{m/s} to the left.
Check: Metres divided by seconds gives m/s\mathrm{m/s}. The graph falls as time increases, matching the negative sign. A change of 8.0 m8.0\ \mathrm{m} over 4.0 s4.0\ \mathrm{s} gives a reasonable speed of 2.0 m/s2.0\ \mathrm{m/s}.

Worked example

3. From a description to a vector diagram

A cart is at x=−2.0 mx=-2.0\ \mathrm{m} and moves left at 1.5 m/s1.5\ \mathrm{m/s}. Represent its position and velocity with signs and a labelled direction diagram. Right is positive.
  1. Write signed values
    The system is the cart. Since right is positive, a leftward velocity is negative. The stated position is also negative because it is left of the origin.
    x=−2.0 m,v=−1.5 m/sx=-2.0\ \mathrm{m},\quad v=-1.5\ \mathrm{m/s}
  2. Show the direction in a diagram
    Place the cart on the negative side of the origin. Draw its velocity arrow leftward and label the arrow with the signed velocity. This makes the diagram agree with the numeric representation.
    left← v=−1.5 m/s∙ (x=−2.0 m)right\text{left}\quad\leftarrow\ v=-1.5\ \mathrm{m/s}\quad\bullet\ (x=-2.0\ \mathrm{m})\quad\text{right}
Answer: The cart is at −2.0 m-2.0\ \mathrm{m} and moves left at 1.5 m/s1.5\ \mathrm{m/s}.
Check: Position is measured in metres and velocity in metres per second. The negative position places the cart left of the origin, and the negative velocity and left-pointing arrow agree. The diagram uses the stated positive direction.

Common mistakes and how to avoid them

Using a negative sign without stating which direction is positive.
Correction: State the positive direction first. With right positive, a negative velocity means motion left.
Leaving units off values or graph axes.
Correction: Label each quantity with its unit, including graph axes and final answers.
Putting time on the vertical axis of a position–time graph.
Correction: Put time on the horizontal axis and position on the vertical axis. Label both axes.
Drawing an arrow opposite to the direction shown by the sign.
Correction: Compare the arrow with the chosen positive direction. When right is positive, a negative velocity arrow points left.
Reporting more digits than the given values support.
Correction: Use a sensible number of significant figures and check that the rounded answer still fits the situation.

Lesson summary

Check your understanding

Question 1

Right is positive. A cart changes position from −1.0 m-1.0\ \mathrm{m} to +3.0 m+3.0\ \mathrm{m} in 2.0 s2.0\ \mathrm{s}. What is its constant velocity?
  1. +2.0 m/s+2.0\ \mathrm{m/s}, to the right
  2. −2.0 m/s-2.0\ \mathrm{m/s}, to the left
  3. +4.0 m/s+4.0\ \mathrm{m/s}, to the right
  4. −4.0 m/s-4.0\ \mathrm{m/s}, to the left
Show answer and explanation
+2.0 m/s+2.0\ \mathrm{m/s}, to the right
The position change is +3.0 m−(−1.0 m)=+4.0 m+3.0\ \mathrm{m}-(-1.0\ \mathrm{m})=+4.0\ \mathrm{m}. Dividing by 2.0 s2.0\ \mathrm{s} gives +2.0 m/s+2.0\ \mathrm{m/s}. The positive direction is right.

Question 2

On a position–time graph, position decreases steadily as time increases. With right defined as positive, what does this show?
  1. The object moves left.
  2. The object moves right.
  3. The object remains at rest.
  4. The graph gives no information about direction.
Show answer and explanation
The object moves left.
Position decreases over time, so the change in position is negative. With right defined as positive, this represents motion to the left.

Key terms

Representation
A way to show information, such as numbers, an equation, a graph, or a diagram.
Scalar
A quantity with size but no direction.
Vector
A quantity with both size and direction.
Sign convention
A stated choice of which direction is positive and which is negative.
Position
An object's location relative to a chosen origin.
Velocity
Change in position per unit time, including direction.
Significant figures
Digits that show the precision of a measured or stated value.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation A1.12. 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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