Question pattern: Position–time graph analysis
Original question 2 of 2 in this pattern
Average and instantaneous velocity
A smooth position–time graph passes through the listed graph points. A tangent drawn to the curve at t = 4.0 s passes through the two tangent-line points given separately.
a) Calculate the average velocity between t = 2.0 s and t = 6.0 s.
b) Calculate the instantaneous velocity at t = 4.0 s.
c) Explain why the two calculations use different pairs of points.
| Time / s | Position / m |
|---|---|
| 2.0 | 5.0 |
| 4.0 | 11.0 |
| 6.0 | 17.0 |
| Time / s | Tangent-line position / m |
|---|---|
| 2.0 | 3.0 |
| 6.0 | 19.0 |
Worked answer
Solution
- Step 1
Average velocity from 2.0 s to 6.0 s:
v_avg = (17 - 5) / (6 - 2) = 3.0 m s⁻¹. - Step 2
Instantaneous velocity is the tangent gradient:
v = (19 - 3) / (6 - 2) = 4.0 m s⁻¹. - Step 3
The interval average uses the two actual endpoints on the curve; the instantaneous velocity uses two convenient points on the tangent at the chosen instant.
Final answer
Average velocity from 2.0–6.0 s = 3.0 m s⁻¹; instantaneous velocity at 4.0 s = 4.0 m s⁻¹.
Common mistake
Using the graph endpoints to calculate the tangent gradient.
Concepts tested
- average velocity
- instantaneous velocity
- secant gradient
- tangent gradient
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