Grade 10 · Core Mathematics · Lesson 10 of 14

Interpreting Velocity-Time Graphs in Practical Situations

A Grade 10 Core Mathematics CBE lesson from the sub-strand Linear Motion. In it, learners determine velocity at a specified time and find the instant corresponding to a given velocity from a graph.

Strand Measurements and GeometrySub-strand Linear MotionLesson code 10CM-29-L1040 minutes · 4 slides

KICD learning outcome (f)

Interpret velocity-time graph in different situations.

By the end of this lesson, learners should be able to

Determine velocity at a specified time and find the instant corresponding to a given velocity from a graph.
Calculate stage acceleration from the gradient of linear segments on a velocity-time graph.
Calculate total displacement by finding the area under a velocity-time graph.
Construct velocity-time graphs from tabular motion data and interpret motion characteristics.

Key vocabulary

Instant: A precise point in time at which an event or measurement occurs.; The boat reaches a velocity of 25 m/s at the instant t = 3.75 s.
Deceleration: The rate at which an object decreases its velocity over time.; The speed boat experiences a constant deceleration resulting in an acceleration of -4 m/s².
Multi-stage motion: Movement comprising different consecutive phases of changing and uniform velocities.; The bus multi-stage motion includes acceleration, constant speed, and final braking phases.
Displacement: The total change in position measured as the area under a velocity-time graph.; The displacement during the first 25 seconds equals 800 metres.
Data table: A structured arrangement of paired numerical values recorded during an observation.; The recorded data table shows velocity values measured after every 5 seconds.
Linear interpolation: Estimating unknown intermediate values between known data points assuming constant rate of change.; Linear interpolation between data points gives a velocity of 23 m/s at t = 1 s.
Initial velocity: The velocity of an object at the starting instant when measurement begins at time zero.; The rocket has an initial velocity of 100 m/s at time t = 0 s.

Lesson preview

Slide 1 · Interpreting Linear Velocity Reduction in a Speed BoatOpening · 10 min

Interpreting a velocity-time graph involves reading key motion values directly from the grid and determining rates of change. To find the velocity at a given time $t$ , locate $t$ on the horizontal axis and trace vertically to the graph line. To find the instant at which a given velocity is reached, locate that velocity on the vertical axis and trace horizontally. The line slope gives the acceleration.

The horizontal axis represents time in seconds while the vertical axis represents velocity in metres per second.
Reading vertically from a time point gives the exact velocity at that specific instant.
Reading horizontally from a velocity value determines the exact time instant when it occurs.
The gradient of any line segment on a velocity-time graph represents the object's acceleration.

Worked example. Locate

t = 5 s

on the horizontal axis and read the corresponding vertical value

v = 20 m/s

. Locate

v = 25 m/s

on the vertical axis and trace to the line to find time instant

t = 3.75 s

. Calculate acceleration using the gradient formula

a = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{0 - 40}{10 - 0} = - 4 {m/s}^{2}

. Answer: Velocity at $t = 5 s$ is $20 m/s$ ; instant at $25 m/s$ is $t = 3.75 s$ ; acceleration is $- 4 {m/s}^{2}$ ..

Check question: A patrol boat slows down uniformly from $30 m/s$ at $t = 0 s$ to rest at $t = 12 s$ . Find the velocity at $t = 4 s$ and calculate its acceleration.

Velocity is $20 m/s$ ; acceleration is $- 2.5 {m/s}^{2}$
Velocity is $10 m/s$ ; acceleration is $2.5 {m/s}^{2}$
Velocity is $20 m/s$ ; acceleration is $2.5 {m/s}^{2}$
Velocity is $40 m/s$ ; acceleration is $2.5 {m/s}^{2}$

Show answer

Velocity is $20 m/s$ ; acceleration is $- 2.5 {m/s}^{2}$ . Acceleration $a = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{0 - 30}{12 - 0} = - 2.5 {m/s}^{2}$ . The velocity at $t = 4 s$ is found using the equation of the line $v = 30 - 2.5 t$ . At $t = 4 s$ , $v = 30 - 2.5 (4) = 20 m/s$ .

Slide 2 · Analysing Multi-Stage Motion from Velocity-Time GraphsModel · 10 min

Teach the full lesson on the projector

This lesson, and every other Grade 1–10 CBE lesson, comes ready to project with the EAP ClassKit. It runs without electricity or internet.

All 4 slides with diagrams and class activities
Teacher notes, SNE adaptations and assessment checks
Homework and the other 13 lessons in this sub-strand

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Interpreting Velocity-Time Graphs in Practical Situations

By the end of this lesson, learners should be able to

Key vocabulary

Lesson preview

Teach the full lesson on the projector

All 14 lessons in Linear Motion