Define velocity as speed in a given direction

Cambridge IGCSE Physics 0625 – Motion

Objective

Define velocity as “speed in a given direction”, distinguish it from speed, and use the relevant formulas to solve basic quantitative problems (Core 1.2 Motion, Supplement 1.2 Motion).

1. Speed and Velocity – Definitions & Core Formulas

  • Speed (s) – a scalar quantity: magnitude only.


    Average speed (core‑level formula):

    \[

    \bar s = \frac{\text{total distance travelled}}{\Delta t}

    \]

  • Velocity (\(\mathbf{v}\)) – a vector quantity: magnitude + direction.


    Average velocity:

    \[

    \mathbf{\bar v}= \frac{\Delta\mathbf{s}}{\Delta t}

    \qquad\text{where }\Delta\mathbf{s}\text{ is displacement (vector).}

    \]

Both speed and velocity have units of m s⁻¹; for velocity the unit is always accompanied by a direction (e.g. 15 m s⁻¹ north).

Comparison Table

QuantitySymbolNatureFormulaUnits
Speeds or vScalar\(s = \dfrac{\text{distance}}{\Delta t}\)m s⁻¹
Average speed\(\bar s\)Scalar\(\bar s = \dfrac{\text{total distance}}{\Delta t}\)m s⁻¹
Velocity\(\mathbf{v}\)Vector\(\mathbf{v}= \dfrac{\Delta\mathbf{s}}{\Delta t}\)m s⁻¹ (with direction)
Average velocity\(\mathbf{\bar v}\)Vector\(\mathbf{\bar v}= \dfrac{\Delta\mathbf{s}}{\Delta t}\)m s⁻¹ (with direction)

2. Calculating Average Velocity – Step‑by‑Step

  1. Mark the initial and final positions of the object.

  2. Determine the straight‑line displacement \(\Delta\mathbf{s}\) (magnitude + direction).

  3. Measure the elapsed time \(\Delta t\) for that displacement.

  4. Apply \(\mathbf{\bar v}= \dfrac{\Delta\mathbf{s}}{\Delta t}\).

  5. Record the answer with both magnitude and direction (e.g. 5 m s⁻¹ east).

Numerical Illustration

A car travels 150 m north in 10 s.

\[

\mathbf{\bar v}= \frac{150\ \text{m (north)}}{10\ \text{s}} = 15\ \text{m s}^{-1}\ \text{north}

\]

3. Interpreting Motion Graphs

3.1 Distance‑time & Displacement‑time Graphs

Distance‑time graph – straight line from 2 s to 6 s

Distance‑time graph – slope = average speed.

Displacement‑time graph – straight line from 2 s to 6 s, direction shown by arrow

Displacement‑time graph – slope = average velocity (direction indicated).

  • Gradient (slope) = \(\dfrac{\text{change in distance or displacement}}{\Delta t}\).

  • Horizontal line → object at rest (speed = 0, velocity = 0).

3.2 Speed‑time & Velocity‑time Graphs

Speed‑time graph showing a constant speed of 4 m s⁻¹ for 5 s

Speed‑time graph – horizontal line = constant speed; area under the line = distance.

Velocity‑time graph showing a straight line from 0 to 8 m s⁻¹ over 4 s

Velocity‑time graph – gradient = constant acceleration; area = displacement.

  • Horizontal line → constant speed (or constant velocity if the line is on a velocity‑time graph).

  • Straight‑line slope → constant acceleration (or deceleration if the slope is negative).

  • Area under a speed‑time graph = distance travelled (speed × time for each segment).

Worked Example – Reading a Speed‑time Graph

In the speed‑time graph above, the speed is constant at 4 m s⁻¹ for 5 s.

\[

\text{Distance} = \text{area of rectangle} = (4\ \text{m s}^{-1})(5\ \text{s}) = 20\ \text{m}

\]

Worked Example – Acceleration from a Velocity‑time Graph

The velocity‑time graph shows \(\mathbf{v}\) increasing uniformly from 0 to 8 m s⁻¹ in 4 s.

\[

a = \frac{\Delta v}{\Delta t}= \frac{8\ \text{m s}^{-1}}{4\ \text{s}} = 2\ \text{m s}^{-2}

\]

4. Quick‑Check Box (All Three Graph Types)

Identify the type of motion from the graph:

  • Distance‑time – horizontal line = rest; straight line with constant slope = constant speed; curve with increasing slope = accelerating.

  • Speed‑time – horizontal line = constant speed; sloping line = constant acceleration (positive = speeding up, negative = slowing down); area under the line = distance.

  • Velocity‑time – horizontal line = constant velocity; sloping line = constant acceleration; area under the line = displacement (direction matters).

5. Acceleration – Definition, Formula & Free‑Fall

  • Acceleration (a) – a vector quantity describing the rate of change of velocity.

    \[

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

    \]

  • Units: m s⁻².

  • For free fall near the Earth’s surface:

    \[

    g \approx 9.8\ \text{m s}^{-2}\ \text{downwards}

    \]

Example

A car’s speed increases from 5 m s⁻¹ to 15 m s⁻¹ in 2 s.

\[

a = \frac{15-5}{2}=5\ \text{m s}^{-2}

\]

6. Resultant of Perpendicular Vectors

If two velocity vectors are at right angles, the resultant is found using the Pythagorean theorem:

\[

v{\text{result}} = \sqrt{v{x}^{2}+v_{y}^{2}}

\]

Right‑angle vector diagram: 3 m s⁻¹ east and 4 m s⁻¹ north, resultant 5 m s⁻¹ northeast

Graphical construction of the resultant of two perpendicular vectors.

7. Practice Questions

7.1 Algebraic Rearrangement

Given a velocity of 12 m s⁻¹ north for a time interval of 5 s, calculate the displacement \(\Delta\mathbf{s}\).

\[

\Delta\mathbf{s}= \mathbf{v}\,\Delta t = 12\ \text{m s}^{-1}\times 5\ \text{s}=60\ \text{m north}

\]

7.2 Graph Reading – Mixed Table

Read the table below and state the average velocity for each interval.

Time (s)Displacement (m)
0–40 → 16 m east
4–816 m east → 16 m north
8–1216 m north → 0 m (back to start)

  • 0–4 s: \(\mathbf{\bar v}= \dfrac{16\ \text{m east}}{4\ \text{s}} = 4\ \text{m s}^{-1}\) east

  • 4–8 s: \(\mathbf{\bar v}= \dfrac{16\ \text{m north}}{4\ \text{s}} = 4\ \text{m s}^{-1}\) north

  • 8–12 s: \(\mathbf{\bar v}= \dfrac{0\ \text{m}}{4\ \text{s}} = 0\ \text{m s}^{-1}\) (object at rest)

7.3 Area Under a Speed‑time Graph

A speed‑time graph shows a constant speed of 3 m s⁻¹ for 6 s, followed by a constant speed of 5 m s⁻¹ for another 4 s. Find the total distance travelled.

\[

\text{Distance}= (3\ \text{m s}^{-1})(6\ \text{s}) + (5\ \text{m s}^{-1})(4\ \text{s}) = 18\ \text{m} + 20\ \text{m}=38\ \text{m}

\]

8. Classroom Activity (AO3 – Practical Link)

  1. Set up a straight 2 m track and mark a start and finish line.

  2. Using a stopwatch, time a toy car as it moves from start to finish in three separate trials.

  3. Record the times and calculate for each trial:

    • Average speed = distance ÷ time.

    • Average velocity = displacement ÷ time (state “forward” as the direction).

    • Area under a speed‑time graph (draw a simple speed‑time plot for each trial and verify that the area equals the measured distance).

  4. Repeat the experiment with the car returning to the start point after reaching the finish line. Discuss:

    • Why the total displacement is zero even though the distance travelled is 4 m.

    • How the velocity‑time graph now has a positive slope on the outward leg and a negative slope on the return leg, giving a net area (displacement) of zero.

  5. Extension (optional): Use two perpendicular motions (e.g., a toy boat moving east while a wind pushes it north) and construct a vector diagram to find the resultant velocity.

This activity reinforces the definition of velocity, the importance of direction, the distinction between distance and displacement, and the graphical methods required by the syllabus.

9. Key Take‑aways

  • Speed = distance ÷ time (scalar).
    Velocity = displacement ÷ time (vector).

  • Average speed uses total distance; average velocity uses net displacement.

  • Gradient of distance‑time or displacement‑time graphs → speed or velocity.
    Gradient of speed‑time or velocity‑time graphs → acceleration.

  • Area under a speed‑time graph = distance; area under a velocity‑time graph = displacement.

  • Acceleration \(a = \Delta v / \Delta t\); free‑fall acceleration \(g \approx 9.8\ \text{m s}^{-2}\) downwards.

  • Resultants of perpendicular vectors are found using the Pythagorean theorem and can be drawn graphically.