Motion graphs

Read the axes before the shape

A motion graph is a description of quantities against time, not a picture of the path. A line going down the page does not necessarily mean downhill motion. First read both axis labels and units, identify the chosen positive direction, and note any scales or offsets. A displacement–time graph and a velocity–time graph can have identical shapes but tell different stories.

Displacement is change in position measured from a reference point, including direction. Distance is total path length and cannot be negative. Velocity is the rate of change of displacement; speed is the magnitude of velocity. In one dimension a negative velocity simply means motion in the negative direction. A negative displacement locates the object relative to the origin; it does not tell you whether it is approaching or leaving that origin.

At any time the graph’s height is the quantity on the vertical axis. A displacement graph crossing zero shows the object passing the reference position. A velocity graph crossing zero shows instantaneous rest and, if the sign changes, reversal of direction. These are different events and may occur at different times.

Worked example

Where is the walker going?

1. Choose east as positive. A walker is at displacement −6 m.
2. Their velocity is +2 m s⁻¹, so they are west of the origin but moving east.
3. They pass the origin after 3 s if that velocity remains constant.

Position and direction of motion must be read separately.

Keep this idea. The axis label gives the meaning of height; the graph’s sign depends on a stated reference direction.

Displacement–time: gradient is velocity

The average velocity over an interval is change in displacement divided by elapsed time. On a displacement–time graph this is the gradient of a chord joining the interval’s endpoints. A horizontal segment has zero gradient, so the object is stationary even if its displacement is far from zero. A straight sloping segment represents constant velocity; a steeper slope means greater speed.

For a curved graph, velocity changes. Find instantaneous velocity by drawing a tangent at the time of interest and calculating that tangent’s gradient using two well-separated points on the tangent. Do not use two arbitrary points on the curve when an instantaneous value is requested. The units of a displacement–time gradient are metres divided by seconds: m s⁻¹.

A curve that becomes less steep while rising describes positive velocity becoming smaller: the object slows down. A curve that slopes down more steeply describes an increasingly negative velocity: the object speeds up in the negative direction. The area below a displacement–time graph has units m s and is not the distance travelled.

Average velocity = Δs / Δt Instantaneous velocity = tangent gradient

Symbols and units

• s is signed displacement in m; t is time in s; Δ means final value minus initial value.
• Velocity v has units m s⁻¹. A secant joins two points; a tangent gives the local slope.

How the equation explains the idea

1. Divide displacement change by elapsed time: vaverage = (s₂ − s₁)/(t₂ − t₁). The graph height s₂ alone cannot supply velocity.
2. For s = t² with s in m and t in s, the chord from 1 s to 3 s gives (9 − 1)/2 = 4 m s⁻¹. The tangent at 1 s instead has slope 2 m s⁻¹.
3. The distinction follows because the gradient changes along a curved graph; only a straight segment has one velocity throughout.

Worked example

A tangent measurement

1. At t = 3.0 s, draw a tangent to the displacement curve.
2. Two points on the tangent are (1.0 s, 2.0 m) and (5.0 s, 14.0 m).
3. Velocity = (14.0 − 2.0)/(5.0 − 1.0) = 3.0 m s⁻¹. The points need not lie on the original curve.

Instantaneous velocity at 3.0 s is approximately +3.0 m s⁻¹.

Keep this idea. A chord gives average velocity; a tangent gives instantaneous velocity.

Velocity–time: gradient is acceleration

The gradient of a velocity–time graph is acceleration: change in velocity divided by elapsed time. A straight line has constant acceleration; a curved line has changing acceleration. A horizontal line at +5 m s⁻¹ means steady forward motion, not rest. Only a horizontal line on v = 0 represents continued rest.

Acceleration has units (m s⁻¹)/s = m s⁻². A negative gradient means acceleration in the negative direction. To decide whether the object is speeding up, compare the signs of velocity and acceleration. A line below the time axis that slopes upward describes negative velocity moving towards zero, so speed is decreasing.

For non-uniform acceleration, use a tangent for the acceleration at one instant, or a chord for average acceleration over an interval. Ideal piecewise straight graphs may have sharp corners where acceleration changes suddenly. At the exact corner there is no single tangent gradient; interpret the intervals on either side. A vertical jump in velocity would imply an unbounded acceleration and is not a realistic finite-force motion.

a = Δv / Δt (use a tangent for an instantaneous value)

Symbols and units

• v is signed velocity in m s⁻¹ and t is time in s.
• a = Δv/Δt is acceleration in m s⁻²; a tangent is required when the slope varies.

How the equation explains the idea

1. Subtract velocities in time order and divide by the positive elapsed time. A fall from +6 to +2 m s⁻¹ over 2 s gives a = (2 − 6)/2 = −2 m s⁻².
2. A negative gradient means negative acceleration. Whether speed increases depends on the velocity sign: negative acceleration speeds up negative motion but slows positive motion.

Worked example

Slowing, stopping, then reversing

1. Velocity decreases linearly from +6 m s⁻¹ at 0 s to −2 m s⁻¹ at 4 s.
2. a = (−2 − 6)/(4 − 0) = −2 m s⁻².
3. The line crosses v = 0 at 3 s. Speed decreases before 3 s and increases afterwards.

Acceleration stays negative throughout, but slowing down does not.

Keep this idea. The sign of acceleration gives its direction; compare it with velocity to decide whether speed rises or falls.

Signed area gives displacement

During a short interval at constant velocity, displacement is velocity × time. This is the area of a rectangle under a velocity–time graph. Adding intervals gives the general rule: the signed area between the graph and the time axis is displacement. Areas above the axis count positive and areas below it negative. Units confirm the result: m s⁻¹ × s = m.

Distance travelled is the total area magnitude: add the positive sizes of all regions, including those below the axis. If the object reverses, distance and the magnitude of displacement usually differ. Split the graph at each crossing of v = 0 before adding areas. For a straight segment that does not cross zero, use a rectangle, triangle or trapezium as appropriate.

For a curved graph, estimate the area by strips or count grid squares. A trapezium estimate uses the mean of the velocities at each strip’s ends multiplied by the strip width. Finer strips usually improve an estimate for a smooth curve. Do not claim exactness from a rough graph; preserve reasonable significant figures and indicate an estimate.

Displacement = signed area under v–t Distance = sum of area magnitudes

Symbols and units

• v is signed velocity in m s⁻¹; t is time in s; s is displacement in m.
• Graph area has units (m s⁻¹) × s = m. Distance is nonnegative path length in m.

How the equation explains the idea

1. For a constant-velocity interval, displacement is vΔt, the signed rectangle area under the graph. Sum such pieces, or integrate, for variable velocity.
2. An interval at +3 m s⁻¹ for 2 s contributes +6 m; an interval at −2 m s⁻¹ for 2 s contributes −4 m. Net displacement is 2 m, while distance is 6 + 4 = 10 m.
3. Split at every zero crossing before adding area magnitudes to find distance.

Worked example

The return journey matters

1. Velocity falls linearly from +4 m s⁻¹ at 0 s to −2 m s⁻¹ at 6 s, crossing zero at 4 s.
2. Positive triangular area = ½ × 4 s × 4 m s⁻¹ = 8 m.
3. Negative triangular area = −½ × 2 s × 2 m s⁻¹ = −2 m.
4. Displacement = 8 − 2 = +6 m. Distance = 8 + 2 = 10 m.

The object travels 10 m but finishes 6 m in the positive direction from its start.

Keep this idea. Split a graph at zero crossings: signed areas give displacement, absolute areas give distance.

Acceleration–time: area is change in velocity

An acceleration–time graph shows acceleration directly as its height. Its signed area gives change in velocity, because m s⁻² × s = m s⁻¹. To find final velocity, add this change to initial velocity. Area alone is not the final velocity unless the object started from rest. A zero-height segment means constant velocity, not necessarily a stationary object.

Connect all three graphs by moving one relationship at a time. Starting from rest with constant positive acceleration gives a horizontal positive acceleration graph, an upward straight velocity graph, and an increasingly steep displacement curve. To determine displacement from an acceleration graph, first establish velocity using the initial condition, then use the velocity–time area. The direct area under acceleration–time is not displacement.

The gradient of an acceleration–time graph has units m s⁻³ and describes how quickly acceleration changes; it is not velocity. The area under displacement–time has no standard motion quantity used here. Units are a useful safeguard against assigning a familiar meaning to the wrong graph. For variable acceleration, tangent gradients and signed-area estimates still apply; constant-acceleration equations apply only within intervals where acceleration is constant.

Δv = signed area under a–t v = u + Δv

Symbols and units

• a is acceleration in m s⁻²; t is time in s; Δv is change of velocity in m s⁻¹.
• u and v are initial and final signed velocities in m s⁻¹.

How the equation explains the idea

1. For constant a over an interval, the signed area aΔt is the velocity change. Adding interval areas gives Δv for a piecewise or curved acceleration graph.
2. Use v = u + Δv, not v = Δv unless the object began at rest. If u = −3 m s⁻¹ and the area is +2 m s⁻¹, the final velocity remains −1 m s⁻¹.

Worked example

Two stages of acceleration

1. Initial velocity is +1 m s⁻¹. Acceleration is +2 m s⁻² for 3 s, then −1 m s⁻² for 2 s.
2. Change in velocity = (2 × 3) + (−1 × 2) = +4 m s⁻¹.
3. Final velocity = 1 + 4 = +5 m s⁻¹. This does not give displacement directly.

The object ends at +5 m s⁻¹.

Keep this idea. Keep the chain straight: displacement gradient → velocity; velocity gradient → acceleration; reverse with areas and initial values.