(b) Movement and position
Movement and position links graphs, equations and practical methods so motion can be measured, calculated and explained from evidence reliably.
- Syllabus
- 2024
- Topic
- —
- Level
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Learning objectives
1.3Distance-time graphsPlot and explain distance-time graphs1.4Average speed equationKnow and use average speed = distance moved ÷ time taken.1.5Motion practicalPractical: investigate the motion of everyday objects such as toy cars or tennis balls1.6Acceleration equationKnow and use acceleration = change in velocity ÷ time taken, a = (v − u) ÷ t.1.7Velocity-time graphsPlot and explain velocity-time graphs1.8Acceleration from graph gradientsDetermine acceleration from the gradient of a velocity-time graph1.9Distance from velocity-time graphsDetermine the distance travelled from the area between a velocity-time graph and the time axis1.10Final speed equationUse (final speed)² = (initial speed)² + 2 × acceleration × distance moved, v² = u² + 2as.
Read distance-time graphs
A distance-time graph shows elapsed time on the horizontal axis and distance travelled on the vertical axis. Its gradient is the rate at which distance changes, so it represents speed.
$\text{speed}=\dfrac{\text{change in distance}}{\text{change in time}}$
| Segment feature | Motion it represents |
|---|---|
| straight rising line | constant speed |
| steeper straight line | greater constant speed |
| horizontal line | stationary: distance does not change |
| curve getting steeper | speed is increasing |
| curve becoming less steep | speed is decreasing |
To plot a journey, place each measured pair (t,d) at its coordinate and join points in time order. Explain a segment by comparing its gradient, not simply its vertical position.
A higher point means more distance has been travelled; it does not by itself mean the object is moving faster. A horizontal segment has zero gradient, so the object is stationary, not moving at constant non-zero speed.
Calculate average speed
Average speed describes the whole journey: divide the total distance moved by the total elapsed time. Time spent stationary is part of the elapsed time.
$\text{average speed}=\dfrac{\text{total distance moved}}{\text{total time taken}}$
A toy car travels 18m in 6.0s, pauses for 2.0s, then travels 12m in 4.0s. Total distance =30m and total time =12.0s, so average speed =30/12.0=2.5m/s.
Use compatible units before dividing. For an answer in m/s, convert every distance to metres and every time to seconds; then include m/s in the result.
Do not take the arithmetic mean of two speeds unless the time spent at each speed makes that method valid. The reliable method is always total distance divided by total elapsed time.
Investigate motion reliably
A motion investigation needs a measured distance and time, one deliberately changed independent variable, and other relevant conditions kept constant.
| Role | Example for a toy car on a ramp |
|---|---|
| independent variable | release height or ramp angle |
| dependent variable | average speed over a fixed section |
| controls | same car, surface, start method and measured section |
- Mark a fixed start and finish and measure their separation with a metre rule or tape. 2. Release the same car from rest without pushing. 3. Measure travel time with light gates and a timer/data logger, or use video timestamps. 4. Calculate average speed as distance divided by time. 5. Repeat at least three times at each chosen independent-variable value, identify anomalies and calculate a mean from valid repeats. 6. Test a suitable range of at least five values and record values with units.
Light gates or frame-by-frame video reduce reaction-time uncertainty. A longer measured section increases the travel time and can reduce the percentage timing uncertainty, provided the intended motion is still being measured.
Repeating one uncontrolled run does not make the investigation valid. Reliability comes from consistent repeats; validity also requires changing only the intended independent variable and controlling other factors.
Calculate acceleration from velocity change
Acceleration is the change in velocity per unit time. Let u be initial velocity, v final velocity, t elapsed time and a acceleration.
$a=\dfrac{v-u}{t}$
A car changes velocity from 8.0m/s to 20.0m/s in 4.0s. First find v−u=20.0−8.0=12.0m/s; then a=12.0/4.0=3.0m/s2.
Choose a positive direction and keep velocity signs consistent. If positive velocity falls from 20m/s to 8m/s in 4s, a=(8−20)/4=−3m/s2: the negative sign shows acceleration opposite to the chosen positive direction.
Use the change v−u, not just the final velocity. The unit is (m/s)/s=m/s2, not m/s.
Read velocity-time graphs
A velocity-time graph shows elapsed time on the horizontal axis and velocity on the vertical axis. The vertical coordinate gives velocity at that instant; the gradient gives acceleration.
| Segment feature | Velocity | Acceleration |
|---|---|---|
| horizontal line above the time axis | constant positive velocity | zero |
| straight line rising | velocity increases uniformly | constant positive |
| straight line falling | velocity decreases uniformly | constant negative |
| line on the time axis | zero: stationary | zero if horizontal |
| line below the time axis | motion in the negative direction | set by its gradient |
Plot each measured (t,v) pair in time order. A straight sloping segment represents constant acceleration because equal time intervals produce equal velocity changes; a curve represents changing acceleration.
A horizontal line above the time axis does not mean stationary: its height shows a constant non-zero velocity. Do not confuse graph height (velocity) with graph gradient (acceleration).
Find acceleration from graph gradient
$\text{acceleration}=\text{gradient}=\dfrac{\Delta v}{\Delta t}=\dfrac{v_2-v_1}{t_2-t_1}$
Choose two well-separated points on the same straight segment. If velocity changes from 6.0m/s at 2.0s to 18.0m/s at 8.0s, then a=(18.0−6.0)/(8.0−2.0)=12.0/6.0=2.0m/s2.
For acceleration at one instant on a curved graph, draw a tangent at that point and calculate the tangent's gradient using a large triangle. For mean acceleration over an interval, use the two interval endpoints.
Do not calculate v/t from one arbitrary point unless the straight segment passes through the origin. A falling segment has a negative gradient under the chosen sign convention.
Find distance from area on a velocity-time graph
Distance travelled is found from the area between a velocity-time graph and the time axis. The units confirm the relationship: (m/s)×s=m.
$\text{distance travelled}=\text{area between the velocity-time graph and the time axis}$
A car accelerates uniformly from 4m/s to 10m/s over 3.0s. The area is a rectangle plus a triangle: (4×3.0)+[21×(10−4)×3.0]=12+9=21m. Equivalently, use the trapezium area 21(4+10)×3.0.
Split a piecewise graph at every change of segment, calculate rectangle, triangle or trapezium areas, then add them. Use the axis scale rather than counting visual grid spaces without values.
Gradient gives acceleration, not distance. If velocity is below the time axis, the signed area is negative displacement; distance travelled uses the magnitude of each area between the graph and the axis.
Use the time-free motion equation
For motion with constant acceleration, use the time-free equation when initial speed u, final speed v, acceleration a and distance moved s are involved. Choose one direction as positive before assigning signs.
$v²=u²+2as$
A car moving at 20m/s brakes with constant acceleration −5.0m/s2 over 30m. Taking forward as positive: v2=202+2(−5.0)(30)=100, so v=100=10m/s. Use the positive root because the question asks for speed.
Write the equation, identify the unknown, convert units, substitute signed values, rearrange, and only then take a square root if finding u or v. Check that the result matches the motion: braking should reduce speed.
The equation assumes constant acceleration. During braking, a is negative if forward is positive; using a positive value would incorrectly make v2 increase. Do not report v2 as the final speed.