HomeDraw Velocity-Time Graphs from Motion Descriptions: Slope as Acceleration Free original lessons
Draw Velocity-Time Graphs from Motion Descriptions: Slope as Acceleration
Practice plotting segmented velocity-time graphs from written motion scenarios and reading slope as acceleration, with worked example, rubric, and four self-contained exercises.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
A velocity-time graph shows how fast an object moves and how that speed changes. The vertical axis is velocity in metres per second (m/s); the horizontal axis is time in seconds (s). A horizontal segment means constant velocity, so acceleration is zero. A straight sloped segment means constant acceleration; its slope equals acceleration. In this lesson you will translate written motion descriptions into segmented velocity-time graphs, label each segment, and compute acceleration from slope. Use a consistent convention: plot velocity on the vertical axis and time on the horizontal axis, mark each segment endpoint clearly, and connect segments with straight lines unless the description says otherwise. All numerical values here are hypothetical teaching examples, not equipment specifications.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Axes, scale, and segment conventions
Before plotting, choose a scale that fits the largest time and velocity values in the scenario. Label both axes with quantity and unit, for example velocity (m/s) and time (s). Mark equal intervals on each axis and state the value of one grid square. A segment is a straight line between two time endpoints; at each endpoint write the time and velocity so the graph can be checked without guessing.
Use a simple convention for direction: positive velocity means motion in the chosen forward direction; negative velocity means motion opposite that direction. Acceleration is the slope of the velocity-time line: slope = change in velocity divided by change in time. If velocity increases, slope is positive; if velocity decreases, slope is negative; if velocity is constant, slope is zero.
- Read the scenario and list each time interval with its starting and ending velocity.
- Choose axis scales that include all listed times and velocities, including negative values if present.
- Plot each endpoint as a point, then connect consecutive points with straight segments.
- Label each segment with its time interval and compute its slope as acceleration.
- Check that the graph returns to the stated velocity at each boundary time.
Computing acceleration from slope
For a straight segment from time t1 to t2 with velocities v1 and v2, acceleration a = (v2 - v1) / (t2 - t1). Keep units consistent: velocity in m/s and time in s gives acceleration in m/s². A segment from 0 s to 4 s with velocity rising from 0 m/s to 8 m/s has slope (8 - 0) / (4 - 0) = 2 m/s². A horizontal segment from 4 s to 7 s at 8 m/s has slope 0 m/s².
When a scenario says an object slows to rest, the final velocity is 0 m/s. When it reverses direction, velocity passes through zero and becomes negative; the slope over that segment still gives acceleration. Do not confuse the height of the graph (velocity) with its steepness (acceleration).
- Identify the two endpoint coordinates (t1, v1) and (t2, v2) for the segment.
- Subtract velocities to find change in velocity, and subtract times to find change in time.
- Divide change in velocity by change in time and include units m/s².
- State whether the slope is positive, negative, or zero and what that means for the motion.
Checking a completed graph
A correct graph matches every stated time and velocity pair, uses straight segments where acceleration is constant, and has slopes that agree with the computed accelerations. Check each boundary: at a shared time, the end of one segment and the start of the next must have the same velocity unless the description states an instantaneous change. Also check that horizontal segments have zero slope and that steeper segments correspond to larger magnitude of acceleration.
Because letter styles and handwriting vary, accept any clear labelling that identifies axes, units, scale, endpoints, and segment slopes. A graph drawn with a different but stated scale is valid if all plotted points and slopes are correct.
- Verify axis labels and units are present and the scale is stated.
- Check each endpoint against the scenario list.
- Recompute each slope from the plotted endpoints and compare with the stated acceleration.
- Confirm horizontal segments are level and sloped segments are straight.
Worked example
A cart starts at rest. From 0 s to 3 s it accelerates uniformly to 6 m/s. From 3 s to 8 s it travels at constant 6 m/s. From 8 s to 10 s it slows uniformly to 2 m/s. Draw the velocity-time graph and state the acceleration for each segment.
Worked answer
Plot points (0,0), (3,6), (8,6), (10,2). Connect with straight segments. Segment 1 slope = (6-0)/(3-0)=2 m/s². Segment 2 slope = (6-6)/(8-3)=0 m/s². Segment 3 slope = (2-6)/(10-8)=-2 m/s². The graph rises from 0 to 6 m/s over 3 s, stays level at 6 m/s for 5 s, then falls to 2 m/s over 2 s.
Four practice exercises
Exercise 1
A cyclist moves at 4 m/s from 0 s to 5 s, then accelerates uniformly to 10 m/s from 5 s to 9 s, then stays at 10 m/s from 9 s to 12 s. Draw the velocity-time graph and give the acceleration for each segment.
Show answer and check
Points: (0,4), (5,4), (9,10), (12,10). Segment 1 slope 0 m/s²; segment 2 slope (10-4)/(9-5)=1.5 m/s²; segment 3 slope 0 m/s².
Why: List endpoints from the description, plot them, connect with straight lines, and compute slope as change in velocity over change in time for each segment.
Check: Recompute each slope from the plotted endpoints; confirm the graph is level from 0 to 5 s and from 9 to 12 s, and rises steadily from 4 to 10 m/s between 5 s and 9 s.
Exercise 2
A lift starts at 0 m/s at 0 s, speeds up uniformly to 3 m/s at 2 s, holds 3 m/s until 6 s, then slows uniformly to 0 m/s at 8 s. Draw the velocity-time graph and state each acceleration.
Show answer and check
Points: (0,0), (2,3), (6,3), (8,0). Accelerations: (3-0)/(2-0)=1.5 m/s²; 0 m/s²; (0-3)/(8-6)=-1.5 m/s².
Why: The lift has three constant-acceleration phases, so three straight segments. Slope of the first is positive, second zero, third negative.
Check: Check that the graph starts and ends at 0 m/s, is level between 2 s and 6 s, and that the two sloped segments have equal magnitude but opposite sign.
Exercise 3
A toy car moves forward at 2 m/s from 0 s to 3 s, then accelerates uniformly to 8 m/s from 3 s to 7 s, then brakes uniformly to rest at 9 s. Draw the graph and compute all accelerations.
Show answer and check
Points: (0,2), (3,2), (7,8), (9,0). Accelerations: 0 m/s²; (8-2)/(7-3)=1.5 m/s²; (0-8)/(9-7)=-4 m/s².
Why: Three segments: constant velocity, positive acceleration, negative acceleration. Plot endpoints and connect straight.
Check: Verify the final velocity is 0 m/s at 9 s and that the last slope is steeper than the middle slope in magnitude, matching -4 versus 1.5 m/s².
Exercise 4
A ball rolls forward at 5 m/s from 0 s to 2 s, slows uniformly to 0 m/s at 4 s, then continues to -3 m/s at 6 s. Draw the velocity-time graph and give the acceleration for each segment.
Show answer and check
Points: (0,5), (2,5), (4,0), (6,-3). Accelerations: 0 m/s²; (0-5)/(4-2)=-2.5 m/s²; (-3-0)/(6-4)=-1.5 m/s².
Why: The ball reverses direction, so velocity passes through zero at 4 s and becomes negative. Each segment is straight because acceleration is constant.
Check: Confirm the graph crosses the time axis at 4 s, the vertical axis includes negative values, and both sloped segments have negative slopes with the computed magnitudes.
Common mistakes
- Plotting time on the vertical axis and velocity on the horizontal axis. Use velocity on the vertical axis and time on the horizontal axis so slope equals acceleration.
- Treating the height of the graph as acceleration. Acceleration is the slope, not the velocity value; compute change in velocity divided by change in time.
- Drawing curved segments when the description gives constant acceleration. Constant acceleration produces a straight velocity-time segment; connect endpoints with a straight line.
- Forgetting negative velocity when the object reverses direction. Extend the vertical axis below zero and plot negative velocities; slope still equals acceleration.
Self-check rubric
- Axes are labelled velocity (m/s) vertical and time (s) horizontal, with a stated scale.
- Every stated time-velocity endpoint is plotted correctly, including zero and negative values if present.
- Segments are straight where acceleration is constant and horizontal where velocity is constant.
- Each segment's slope is computed and labelled with correct units m/s².
- Graph is clear enough to check without guessing; alternative scales and handwriting styles are acceptable if labels are unambiguous.
Questions and answers
What does a horizontal segment on a velocity-time graph mean?
Velocity is constant, so acceleration is zero for that time interval.
How do I find acceleration from the graph?
Pick two points on a straight segment and divide the change in velocity by the change in time; the result is the slope in m/s².
Can velocity be negative on the graph?
Yes. Negative velocity means motion opposite the chosen positive direction; the slope over that segment still gives acceleration.
Do I need to draw curved lines for changing acceleration?
Only if the description states that acceleration itself changes. For constant acceleration, use straight segments.