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Draw a Position-Time Graph from a Velocity-Time Graph
Learn to convert a velocity-time graph into a position-time curve by accumulating displacement segment by segment, with clear conventions, worked example, and practice.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
A velocity-time graph tells you how fast and in which direction an object moves; a position-time graph shows where it is. Converting one to the other means accumulating displacement: for each time segment, displacement equals velocity multiplied by elapsed time. This lesson teaches a reliable drawing convention, a segment-by-segment method, and how to check your curve. All numerical values are hypothetical examples chosen for easy arithmetic; measure your own grid if you print it.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Set up a consistent drawing convention
Use the same time axis for both graphs so vertical alignment is easy to check. Label the time axis in seconds and mark equal intervals. On the position axis, choose a scale that fits the total displacement you expect; for example, if velocities are in metres per second and times in seconds, positions are in metres.
Define your convention explicitly before drawing: start position at zero unless told otherwise, positive velocity means increasing position, and a horizontal velocity segment produces a straight position segment whose slope equals that velocity. A change in velocity creates a slope change at the same time value, not a jump in position.
- Draw and label both axes with units and equal time ticks.
- Mark the start position at time zero.
- For each velocity segment, compute displacement as velocity times duration.
- Add the displacement to the previous position to get the next point.
- Connect consecutive points with straight segments; slope equals the segment velocity.
Accumulate displacement segment by segment
Work left to right. For a segment from time t1 to t2 with constant velocity v, displacement is v times (t2 minus t1). Keep a running total of position. If velocity is zero, position stays constant and the position-time segment is horizontal. If velocity is negative, position decreases and the segment slopes downward.
Write each calculation beside the graph so your reasoning is visible. Use consistent units and simple numbers. For instance, a velocity of 2 metres per second for 3 seconds gives 6 metres; starting at 0 metres, the next position is 6 metres.
- List each time segment and its velocity from the velocity-time graph.
- Compute each displacement and record it.
- Add displacements cumulatively to find positions at segment boundaries.
- Plot boundary points and join with straight lines.
- Check that slope signs match velocity signs.
Check slope changes and continuity
The position-time curve must be continuous: position cannot jump instantly. Slope changes occur only at times where velocity changes. Compare each segment's slope with the velocity value; a steeper line means a larger speed. A horizontal line means zero velocity.
A quick check is to compare the sign of each slope with the sign of the corresponding velocity. Also verify that the total change in position equals the sum of all segment displacements. If your curve ends at a different position than your running total, recheck arithmetic or plotting.
- Verify no vertical jumps at segment boundaries.
- Match each slope sign to its velocity sign.
- Confirm final position equals the cumulative displacement.
- Recompute any segment where the slope looks inconsistent.
Worked example
A velocity-time graph shows: from 0 to 2 seconds, velocity 3 metres per second; from 2 to 4 seconds, velocity 0 metres per second; from 4 to 6 seconds, velocity negative 1 metre per second. Start at position 0 metres. Draw the position-time curve.
Worked answer
Segment 1: displacement 3 times 2 equals 6 metres; position goes from 0 to 6 metres. Segment 2: displacement 0 times 2 equals 0 metres; position stays at 6 metres. Segment 3: displacement negative 1 times 2 equals negative 2 metres; position goes from 6 to 4 metres. The position-time curve rises from 0 to 6 metres over 0 to 2 seconds, is horizontal at 6 metres from 2 to 4 seconds, then slopes downward to 4 metres from 4 to 6 seconds.
Four practice exercises
Exercise 1
Velocity-time graph: 0 to 1 second, velocity 4 metres per second; 1 to 3 seconds, velocity 2 metres per second. Start at position 0 metres. Find positions at 1 second and 3 seconds and describe the two position-time segments.
Show answer and check
At 1 second: position 4 metres. From 1 to 3 seconds: displacement 2 times 2 equals 4 metres, so position at 3 seconds is 8 metres. The first segment rises from 0 to 4 metres; the second rises from 4 to 8 metres with a gentler slope.
Why: Displacement equals velocity times duration. First segment: 4 times 1 equals 4 metres. Second segment: 2 times 2 equals 4 metres; cumulative position 4 plus 4 equals 8 metres.
Check: Confirm the slope of the second segment is smaller than the first because 2 is less than 4, and verify final position equals 4 plus 4.
Exercise 2
Velocity-time graph: 0 to 2 seconds, velocity 1 metre per second; 2 to 5 seconds, velocity 0 metres per second. Start at position 2 metres. Find positions at 2 seconds and 5 seconds and describe the curve.
Show answer and check
At 2 seconds: position 2 plus 1 times 2 equals 4 metres. From 2 to 5 seconds: displacement 0 times 3 equals 0 metres, so position stays at 4 metres. The curve rises from 2 to 4 metres, then is horizontal at 4 metres.
Why: Accumulate displacement from the starting position. Zero velocity means no change in position.
Check: Check that the horizontal segment has zero slope and that the final position equals the starting position plus total displacement.
Exercise 3
Velocity-time graph: 0 to 3 seconds, velocity negative 2 metres per second; 3 to 4 seconds, velocity 1 metre per second. Start at position 10 metres. Find positions at 3 seconds and 4 seconds and describe the slope changes.
Show answer and check
At 3 seconds: position 10 plus negative 2 times 3 equals 4 metres. From 3 to 4 seconds: displacement 1 times 1 equals 1 metre, so position at 4 seconds is 5 metres. The curve slopes downward from 10 to 4 metres, then slopes upward from 4 to 5 metres.
Why: Negative velocity decreases position; positive velocity increases it. Compute each displacement and add cumulatively.
Check: Verify the downward segment has negative slope and the upward segment has positive slope, and that the curve is continuous at 3 seconds.
Exercise 4
Velocity-time graph: 0 to 2 seconds, velocity 2 metres per second; 2 to 4 seconds, velocity negative 1 metre per second; 4 to 5 seconds, velocity 0 metres per second. Start at position 0 metres. Find positions at 2, 4, and 5 seconds and describe the full curve.
Show answer and check
At 2 seconds: position 4 metres. From 2 to 4 seconds: displacement negative 1 times 2 equals negative 2 metres, so position at 4 seconds is 2 metres. From 4 to 5 seconds: displacement 0 times 1 equals 0 metres, so position stays at 2 metres. The curve rises to 4 metres, slopes downward to 2 metres, then is horizontal at 2 metres.
Why: Sum displacements segment by segment: 4, then negative 2, then 0. Keep the curve continuous and change slope only at 2 and 4 seconds.
Check: Confirm final position equals 4 minus 2 plus 0 equals 2 metres, and that slope signs match velocity signs for all three segments.
Common mistakes
- Treating the velocity value as a position value and plotting it directly. Accumulate displacement first; plot positions, not velocities.
- Drawing curved segments when velocity is constant. Constant velocity gives a straight position-time segment; reserve curves for changing velocity, which this task does not require.
- Allowing the position curve to jump at a velocity change. Keep the curve continuous; only the slope changes at that time.
- Mixing units or forgetting time durations. Label units and multiply velocity by the segment duration, not by the endpoint time.
Self-check rubric
- Axes are labeled with time and position units and equal time intervals.
- Start position is marked and each segment displacement is computed correctly.
- Position-time curve is continuous with no vertical jumps.
- Each segment is straight and its slope sign matches the velocity sign.
- Final position equals the sum of all segment displacements.
- Alternative valid conventions are accepted if clearly stated and consistently applied.
Questions and answers
Do I need to calculate area under the velocity-time graph?
This task focuses on drawing the position-time curve by accumulating displacement segment by segment. The arithmetic is the same as multiplying velocity by time for each constant-velocity segment.
What if the velocity is negative?
Negative velocity means position decreases, so the position-time segment slopes downward. Add the negative displacement to the running position.
Can the position-time graph have a vertical jump?
No. Position changes continuously over time; only the slope changes when velocity changes.
How do I choose the position axis scale?
Estimate the total displacement by summing segment displacements, then choose a scale that fits that range. Label the scale clearly.