Question pattern: Position–time graph analysis
Original question 1 of 2 in this pattern
Piecewise position–time journey
A position–time graph is formed by straight-line segments joining the points in the table. Positive position is east.
a) Determine the velocity during each of the three time intervals.
b) State when the object is stationary.
c) Calculate the displacement and the total distance travelled from t = 0 s to t = 11 s.
| Time / s | Position / m |
|---|---|
| 0 | 0 |
| 4 | +12 |
| 7 | +12 |
| 11 | −4 |
Worked answer
Solution
- Step 1
From 0 to 4 s:
v = (12 - 0) / (4 - 0) = +3.0 m s⁻¹. - Step 2
From 4 to 7 s:
v = (12 - 12) / (7 - 4) = 0 m s⁻¹. - Step 3
From 7 to 11 s:
v = (-4 - 12) / (11 - 7) = -4.0 m s⁻¹. - Step 4
The object is stationary from 4 s to 7 s because the position–time graph is horizontal.
- Step 5
Displacement:
-4 - 0 = -4 m, or 4 m west. - Step 6
Distance:
|12 - 0| + |12 - 12| + |-4 - 12| = 12 + 0 + 16 = 28 m.
Final answer
Velocities: +3.0 m s⁻¹, 0 m s⁻¹ and −4.0 m s⁻¹; stationary from 4–7 s; displacement = −4 m; distance = 28 m.
Common mistake
Using the final position as the distance travelled.
Concepts tested
- position–time graph
- gradient
- velocity
- distance
- displacement
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