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Learnics

The night before Minor 2

Motion graphs — cheat sheet

Shape describes the motion, slope gives the rate, area gives the displacement.

① Read the y-axis first.

Every rule depends on which graph it is. Most lost marks start with treating a V–T graph like a D–T graph.

Displacement–timeVelocity–time
y-axisDisplacement (m)Velocity (m/s)
Horizontal lineAt rest — v = 0Constant velocity — a = 0 (moving!)
Straight line sloping upConstant velocity, positive directionVelocity increasing — positive acceleration
Straight line sloping downConstant velocity, negative directionNegative acceleration (not negative velocity!)
CurveVelocity changing (accelerating)Acceleration changing (not on Minor 2)
Gradient =velocity v=Δd/Δtv = \Delta d/\Delta tacceleration a=Δv/Δta = \Delta v/\Delta t
Area under =nothing usefuldisplacement (below the axis = negative)
At rest shows asa horizontal sectionthe line on v = 0
Changes direction whenthe gradient changes sign (peak / trough)the line crosses the time axis

Equations & units

  • v=ΔdΔt=d2−d1t2−t1v = \dfrac{\Delta d}{\Delta t} = \dfrac{d_2 - d_1}{t_2 - t_1}m/s
    gradient of D–T
  • a=ΔvΔt=v2−v1t2−t1a = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1}m/s²
    gradient of V–T
  • Δd=signed area under V–T\Delta d = \text{signed area under V–T}m
    above +, below −
  • distance=∑∣areas∣\text{distance} = \sum |\text{areas}|m
    all areas positive
  • vavg=total displacementtotal timev_{avg} = \dfrac{\text{total displacement}}{\text{total time}}m/s
    include time at rest

Area shapes

  • RectangleA=b×hA = b \times h
  • TriangleA=12×b×hA = \tfrac{1}{2} \times b \times h
  • TrapeziumA=12(h1+h2)×bA = \tfrac{1}{2}(h_1 + h_2) \times b

Speeding up or slowing down?

Compare the signs of v and a.

+v, +a
speeding up
+v, −a
slowing down
−v, −a
speeding up
−v, +a
slowing down

Same signs → speeding up. Opposite signs → slowing down.

D–TGraph 3 — Autonomous Vehicle Test
0246810120246810121416Time (s)Displacement (m)Δt = 2 sΔd = −9 mgradient = −4.5 m/s
Steeper means faster: −4.5 m/s is faster than +3 m/s. The sign is only the direction.
V–TGraph 2 — Train Motion
18 m6 m−4 m−8 m−4 m0246810−4−20246Time (s)Velocity (m/s)
Displacement = 18 + 6 − 4 − 8 − 4 = +8 m. Distance = 18 + 6 + 4 + 8 + 4 = 40 m.

Full marks pattern — don't lose the halves

v = Δd / Δt = (d₂ − d₁) / (t₂ − t₁) = (8 − 0) / (4 − 0) = 8 / 4 → v = +2 m/s
  • −½ missing/wrong equation
  • −½ missing/wrong unit
  • −½ missing sign / direction

Top traps

  • Horizontal V–T line = at rest. On a velocity–time graph a horizontal line means constant velocity (a = 0). It is only at rest if the line is on v = 0.
  • Negative slope = negative velocity. On a V–T graph a negative slope means negative acceleration. The sign of the velocity comes from being above (+) or below (−) the time axis.
  • Negative velocity is 'smaller'. The sign only gives the direction. Compare the size (magnitude): −4.5 m/s is faster than +3 m/s.
  • Area below the axis counted as positive. For displacement, area below the time axis is negative. Only use magnitudes when finding total distance.
  • Distance = displacement. Distance adds up every part of the path. Displacement is just final − initial position, with a direction.
  • Instant of rest read as an interval. If a V–T line only touches or crosses v = 0, the object is at rest for an instant, not for a period of time.
  • Direction change at a V–T peak. On a V–T graph the direction changes where the line crosses the time axis (v changes sign), not at a peak.
  • Average velocity = average of velocities. Average velocity = total displacement ÷ total time (include the time spent at rest).
  • 'Stopped' means 'changed direction'. Stopping is not reversing. The direction changes only when the gradient (D–T) or velocity (V–T) changes sign.

“Right is Right” sentences

  • From __ s to __ s the object moves in the positive / negative direction at a constant velocity of __ m/s.
  • From __ s to __ s the object is stationary at a displacement of __ m (v = 0).
  • From __ s to __ s the object speeds up / slows down from __ m/s to __ m/s with an acceleration of __ m/s².
  • The acceleration is __ m/s² because acceleration is the gradient of a velocity–time graph.