01 · Falling segment
Find velocity from t=5 to t=9 s in the model.
Hint
Use (−1−11)/(9−5).
Worked solution
Velocity=−12/4=−3 m/s: leftward at speed 3 m/s.
Understand · explore · practise
Reading displacement–time graphs
Read signed positions, calculate velocities from straight-line slopes and interpret turning points and tangents on displacement–time graphs.
Before you startSigned coordinates, gradients and velocity versus speed.
01 / Read the axes before describing motion
The graph is not a picture of the road.
Gradient on a displacement–time graph gives signed velocity.
A rising graph means increasing position along the chosen axis. A falling graph means negative velocity. The vertical coordinate tells you where the particle is relative to the origin, not how fast it moves.
Read a segment, then its slopeExplore
Position x is measured in metres to the right of a fixed origin; t is elapsed seconds. The line joins (0,2), (3,11), (5,11), (9,−1) and (10,2). Choose a straight segment.
02 / Calculate change in position over change in time
Use differences of coordinates, not a single coordinate ratio.
On the first segment, position changes from 2 m at t=0 to 11 m at t=3 s.Worked example
Change in position = 11−2 = 9 m
The particle did not start at the origin.
Elapsed time = 3−0 = 3 s
Use the segment duration.
Velocity = 9/3 = +3 m/s
The positive sign means rightward motion on this axis.
02 · Last segment
Find velocity from t=9 to t=10 s.
Hint
The position rises from −1 to2.
Worked solution
Velocity=(2−(−1))/(10−9)=+3 m/s.
03 · Wrong ratio
Why is 11/3 not the first segment velocity?
Hint
Account for the initial position2.
Worked solution
It divides final position by time and ignores the initial position. The displacement during this interval is9m, so velocity is3m/s.
03 / A horizontal graph represents rest
A particle can be stationary away from the origin.
04 · Waiting
Describe the model from t=3 to t=5 s.
Hint
Compare the two positions.
Worked solution
Position remains11m, so velocity is0 throughout this interval. The particle is stationary11m right of the origin.
05 · Height versus speed
Does the graph’s highest straight horizontal section represent its greatest speed?
Hint
Slope, not vertical height, gives velocity.
Worked solution
No. That section has zero slope and zero speed. Its vertical coordinate represents position.
06 · Equal steepness
Compare speeds in the first and third segments.
Hint
Take the magnitudes of their slopes.
Worked solution
Both speeds are3m/s. Their velocities are+3 and−3m/s, so their directions differ.
04 / Crossing the origin does not mean stopping
Zero position and zero velocity are different statements.
07 · Origin on return
Find when the particle crosses x=0 on the segment beginning at t=5 s.
Hint
Write x=11−3(t−5).
Worked solution
Set11−3(t−5)=0, giving t=5+11/3=26/3s, about8.67s. Velocity is−3m/s there; it is not at rest.
08 · Negative position
At t=9.2 s, is the particle moving left because x is negative?
Hint
Use the final segment slope.
Worked solution
No. x=−1+3(0.2)=−0.4m, but v=+3m/s. It is left of the origin and moving right.
05 / Use a tangent for instantaneous velocity
A curved graph has a changing slope.
Watch: the tangent turns horizontal, then negative
Pause, replay or seek freely. The notes explain the same idea and stay in view.
For x=10−(t−2)² m on 0≤t≤4 s:Worked example
At t=1, the tangent slope is +2 m/s
Position is increasing.
At t=2, the tangent is horizontal
Velocity is zero at this turning point.
At t=3, the tangent slope is −2 m/s
Position is decreasing. The equal speeds do not mean equal velocities.
09 · Peak
Find the position at t=2 and interpret the horizontal tangent.
Hint
Substitute into x; distinguish position from velocity.
Worked solution
x=10m. The particle is instantaneously at rest and changes from rightward to leftward motion in this model.
10 · Secant
Calculate average velocity from t=0 to t=4 in the curved model. Does it imply rest throughout?
Hint
The endpoint positions are both6m.
Worked solution
Average velocity=(6−6)/4=0. It moves out and back, so it was not stationary throughout. A secant over an interval is not every instantaneous tangent slope.
06 / A sharp corner is an idealisation
Instantaneous changes in velocity need care.
11 · Corner at9s
State the velocities just before and just after t=9 in the piecewise model.
Hint
Use the neighbouring straight segments.
Worked solution
Just before:−3m/s. Just after:+3m/s. There is no single ordinary derivative at the sharp corner; it idealises a very rapid reversal.
12 · Smooth versus sharp
Would a smooth physical reversal usually have a horizontal tangent on its position-time graph?
Hint
Continuous velocity must pass through zero.
Worked solution
Yes, for a smooth reversal with continuous velocity. Do not claim a sharp-corner model itself has a unique zero derivative there.
07 / Check signs and units in every reading
A graph of position against time is not a distance accumulator.
13 · Negative gradient
A graph falls by6km in15minutes. Find velocity in km/h.
Hint
Fifteen minutes is0.25h.
Worked solution
Velocity=−6/0.25=−24km/h; speed24km/h.
14 · Impossible distance graph
Why could a falling line be valid for displacement but not for cumulative distance travelled?
Hint
Think about what can decrease.
Worked solution
Signed displacement can decrease as an object moves in the negative direction. Total distance already travelled cannot decrease.
Use a stated origin and positive direction. A graph’s slope can change even while its vertical coordinate stays positive. Curvature describes changing velocity; a negative slope alone does not say whether speed is increasing.
08 / Position is height; velocity is slope
Use a tangent at an instant and a secant over an interval.
For a straight segment, divide the signed position change by elapsed time. A horizontal segment is rest; a downward segment is negative velocity. Crossing the origin is not stopping. Curved graphs require tangent slopes for instantaneous velocity, and sharp corners require an idealisation caveat.
Section 1 of 8 · Read the axes before describing motion