8867 / 2027
Motion and Forces overview

Lesson 2 of 6

Read slopes, areas and changing acceleration

  • I can represent distance, displacement, speed, velocity and acceleration graphically.Syllabus

    Syllabus 8867, 3(b). Represent distance, displacement, speed, velocity and acceleration graphically.

  • I can use position or displacement gradients, and velocity gradients and areas, including non-uniform acceleration.Syllabus

    Syllabus 8867, 3(c). Use position or displacement gradients, and velocity gradients and areas, including non-uniform acceleration.

Make a guess. You are not marked.

On a velocity-time graph, the line rises but its slope gets steadily smaller. What is happening to the object?

  1. Its acceleration is increasing, because the velocity keeps rising.
  2. It is slowing down, because the line is flattening.
  3. It speeds up, but its acceleration is decreasing.
Show the answer

It speeds up, but its acceleration is decreasing.

The velocity is still rising, so the object speeds up. The gradient is the acceleration, and a flattening line means the acceleration is getting smaller.

Read the quantity and unit on each axis before interpreting a graph. A height, a gradient and an area describe different things.

For the reversal model, choose right as positive, with starting position 2.0 m, initial velocity +6.0 m/s and constant acceleration -2.0 m/s2 over 0 to 5 s. With t in seconds, the numerical position in metres is x = 2 + 6t - t2; displacement from the start is s = 6t - t2.

Position and displacement have the same gradient

The constant 2 m changes the vertical origin, not the gradient. Position-time and displacement-time graphs therefore give the same velocity at a given time. Cumulative distance is different: it continues increasing after a reversal.

Three different vertical quantities for the same journey

Each graph covers the same 0 to 5 s interval. Read its vertical quantity and units before using a gradient or area.

Position: measured from the fixed origin

Position: measured from the fixed originFor zero to five seconds, position is two plus six t minus t squared metres. The six positions are 2, 7, 10, 11, 10 and 7 metres. A dashed chord joins the first and last positions and has gradient positive one metre per second. The purple tangent at four seconds, position ten metres, passes through tangent points three seconds and twelve metres, and five seconds and eight metres. Its gradient is negative two metres per second. The tangent touches the curve at four seconds; its two chosen gradient points need not lie on the curve. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.01234502468101214Position x / mTime t / sTC

The solid blue curve is position. Chord C connects (0 s, 2 m) and (5 s, 7 m), so its gradient is the whole-interval average velocity, +1.0 m/s.

Tangent T touches at 4 s. Its gradient from (3 s, 12 m) to (5 s, 8 m) is (8 - 12)/(5 - 3) = -2.0 m/s. These two gradient points lie on the tangent.

Displacement: measured from the start

Displacement: measured from the startDisplacement from the start is six t minus t squared metres. The six values at whole seconds are 0, 5, 8, 9, 8 and 5 metres. It is two metres below the position graph at every time and has the same gradients. The purple tangent touches at one second and five metres. It passes through tangent points zero seconds and one metre, and two seconds and nine metres, giving positive four metres per second. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.01234502468101214Displacement s / mTime t / sT

The purple tangent T touches at 1 s. Using its points (0 s, 1 m) and (2 s, 9 m) gives (9 - 1)/(2 - 0) = +4.0 m/s. Position and displacement have the same gradients despite their different starting values.

Cumulative distance: add both parts of the journey

Cumulative distance: add both parts of the journeyCumulative distance is six t minus t squared until three seconds, then nine plus t minus three squared. At whole seconds the distances are 0, 5, 8, 9, 10 and 13 metres. It becomes flat momentarily at three seconds but does not decrease when the body reverses. Its gradient is speed, never negative in this model. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.01234502468101214Distance d / mTime t / s

Distance reaches 9 m at the turn, then increases to 13 m. Its gradient is speed. The graph becomes momentarily flat at 3 s and continues increasing afterwards.

The three vertical axes represent different quantities. The position graph starts at 2 m; displacement and cumulative distance start at zero. Tangents give local velocity, while the whole-interval chord gives average velocity.
Displacement and distance from t = 0
Time / sDisplacement / mDistance / m
000
155
288
399
4810
5513

A chord joins two points on a curve. Its gradient Δx/Δt gives average velocity over that interval. Here the chord from t = 0 to 5 s has gradient (7 - 2)/5 = +1.0 m/s.

A tangent follows the curve's direction at one contact point. Its gradient gives instantaneous velocity there. The model's tangent at t = 1 s has gradient +4 m/s; at t = 4 s it has gradient -2 m/s.

v = dx/dt

a = dv/dt

These are local rate notations: dx/dt means the rate of change of position with time, and dv/dt the rate of change of velocity. You can obtain them from graph tangents. A finite Δx/Δt or Δv/Δt instead describes an average over the stated interval.

Velocity, speed and acceleration need separate axes

Read velocity, speed and acceleration separately

Each graph covers the same 0 to 5 s interval. Read its vertical quantity and units before using a gradient or area.

Velocity: the sign gives direction

Velocity: the sign gives directionVelocity is six minus two t metres per second, a straight line from positive six at zero seconds to negative four at five seconds. It crosses zero at three seconds. The shaded positive triangle has signed area positive nine metres. The shaded triangle below zero has signed area negative four metres. Signed areas give displacement positive five metres; adding their magnitudes gives distance thirteen metres. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.012345-4-20246Velocity v / m s-1Time t / s+9 m-4 m

Signed area gives displacement: +9 + (-4) = +5 m. The line's gradient is -2.0 m/s2 throughout; being below zero describes velocity's direction.

Speed: the magnitude of velocity

Speed: the magnitude of velocitySpeed is the magnitude of six minus two t: six, four, two, zero, two and four metres per second at whole seconds. Its V-shaped graph never goes below zero. The two nonnegative area triangles represent nine and four metres, giving total distance thirteen metres. Its corner at three seconds does not mean the acceleration is undefined; acceleration is obtained from the velocity graph. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.0123450246Speed / m s-1Time t / s9 m4 m

Both areas are nonnegative: 9 + 4 = 13 m travelled. The slope of this speed graph is not the signed acceleration throughout the event; use the velocity graph for that.

Acceleration: unchanged through the reversal

Acceleration: unchanged through the reversalAcceleration is a horizontal line at negative two metres per second squared for the whole interval zero to five seconds. It remains negative two when velocity is zero at three seconds. Negative acceleration first reduces positive velocity, then makes the negative velocity increasingly negative. Every graph uses fifty drawing units per second horizontally. The plotted functions and tangents are exact within the supplied mathematical model.012345-3-2-101Acceleration a / m s-2Time t / s

Acceleration remains -2.0 m/s2 even at the turn. It first reduces rightward speed and then increases leftward speed.

Velocity crosses zero at 3 s, while speed stays nonnegative. Acceleration remains -2 m/s squared, including at the turning instant. Signed velocity-time areas distinguish displacement from total distance.
Gradient of position-time or displacement-time
Velocity, with units m/s. A falling graph means negative velocity in the chosen direction.
Gradient of cumulative distance-time
Speed, with units m/s. The graph does not decrease while distance is being accumulated.
Gradient of velocity-time
Acceleration, with units (m/s)/s = m/s2. Being below the time axis means negative velocity, not necessarily negative acceleration.
Signed area under velocity-time
Displacement, with units (m/s) × s = m. Areas below the time axis count negatively.
Area under speed-time
Distance, since speed is nonnegative. Equivalently, add the magnitudes of the positive and negative velocity-time areas.

Signed area through a turn

Keep the below-axis area negative

From 0 to 3 s, the positive triangle has area ½(3)(6) = +9 m. From 3 to 5 s, the signed area is -½(2)(4) = -4 m.

Displacement = 9 - 4 = +5 m.
Distance = 9 + 4 = 13 m.

Check your understanding

A velocity-time graph has area +9 m above the time axis and signed area -4 m below it. What are the displacement and total distance?

  1. Displacement +5 m; distance 5 m.
  2. Displacement 13 m; distance 13 m.
  3. Displacement +5 m; distance 13 m.
Show the answer

Displacement +5 m; distance 13 m.

Signed area gives displacement: +9 - 4 = +5 m. Adding the magnitudes gives total distance: 9 + 4 = 13 m.

A curved velocity graph has changing acceleration

Consider a separate smooth model for 0 to 4 s. With t in seconds, its numerical velocity in m/s is v = 1 + 0.5t2. The table gives values of this supplied model.

Changing velocity in the supplied smooth model
Time / sVelocity / m s-1
01.0
11.5
23.0
35.5
49.0

A curved velocity graph needs a local gradient and an area estimate

The supplied model is v = 1 + 0.5t2, with t in seconds and v in m/s. These are calculated model values.

A true tangent at 3 s

Instantaneous acceleration is the tangent gradientThe velocity curve passes through zero seconds and one metre per second, one and 1.5, two and three, three and 5.5, and four and nine. The purple straight tangent is v equals three t minus 3.5. It touches the curve at three seconds and 5.5 metres per second. Its labelled points A and B are two seconds and 2.5 metres per second, and four seconds and 8.5 metres per second. These are points on the tangent used to measure its gradient, not two observations on the curve. The six-metres-per-second rise divided by two seconds gives instantaneous acceleration three metres per second squared.012340246810Velocity v / m s-1Time t / st = 3 sAB

A = (2 s, 2.5 m/s) and B = (4 s, 8.5 m/s) are on the tangent. Its gradient is (8.5 - 2.5)/(4 - 2) = 3.0 m/s2. The actual curve readings at 2 s and 4 s are 3.0 and 9.0 m/s.

The whole-interval average acceleration is (9.0 - 1.0)/4.0 = 2.0 m/s2. It answers a different question from the local tangent.

Four 1 s trapezia approximate the displacement

Trapezium tops lie above the convex velocity curveThe velocity curve passes through zero seconds and one metre per second, one and 1.5, two and three, three and 5.5, and four and nine. The shaded estimate consists of four one-second trapezia with areas 1.25, 2.25, 4.25 and 7.25 metres, totalling fifteen metres. The straight brown tops join the supplied curve samples and lie above the curve between them. Fifteen metres is an overestimate of the curved area, not an exact displacement.012340246810Velocity v / m s-1Time t / s

Estimated area = 1.25 + 2.25 + 4.25 + 7.25 = 15.0 m. The trapezium tops sit above the curve. Using 0.5 s strips gives a closer estimate of 14.75 m.

The endpoint shortcut (u + v)t/2 would give 20 m. It assumes a straight velocity graph and does not apply to this curve.

The tangent at 3 s describes a local gradient. The area panel uses straight trapezium tops to approximate the curved area; the tangent's marked points are not extra observations on the curve.

Instantaneous and average acceleration

Choose points on the tangent

The tangent at t = 3 s passes through (2 s, 2.5 m/s) and (4 s, 8.5 m/s). Its gradient is (8.5 - 2.5)/(4 - 2) = 3.0 m/s2.

The whole-interval average acceleration is instead (9.0 - 1.0)/4.0 = 2.0 m/s2. It does not describe every instant of this motion.

Check your understanding

A tangent to a curved velocity-time graph at t = 3 s passes through (2 s, 2.5 m/s) and (4 s, 8.5 m/s). What does its gradient give?

  1. Instantaneous acceleration at 3 s: 3.0 m/s2.
  2. Instantaneous velocity at 3 s: 3.0 m/s.
  3. Average acceleration over the whole motion: 3.0 m/s2.
Show the answer

Instantaneous acceleration at 3 s: 3.0 m/s2.

The tangent gradient is (8.5 - 2.5)/(4 - 2) = 3.0 m/s2. The two supplied points are on the tangent; they need not be observations on the curve.

Bound the area, then improve the estimate

Velocity increases throughout the interval. Four 1 s rectangles using the left endpoint heights lie below the curve; using right endpoint heights puts them above it:

Lower sum = (1.0 + 1.5 + 3.0 + 5.5)(1) = 11 m

Upper sum = (1.5 + 3.0 + 5.5 + 9.0)(1) = 19 m

11 m < displacement < 19 m

This is a useful plausibility bound before detailed calculation. Since velocity is positive throughout, displacement and distance have the same numerical value here.

For a trapezium, area = mean of its two endpoint velocities × the time interval. Four 1 s trapezia give 1.25 + 2.25 + 4.25 + 7.25 = 15.0 m. Repeating with 0.5 s intervals gives 14.75 m.

Both values are numerical estimates, not exact areas. Here the straight trapezium tops lie above the upward-curving graph, so they overestimate the area. Smaller intervals follow the curve more closely.

Using (initial velocity + final velocity) × time / 2 would give (1 + 9)(4)/2 = 20 m. That assumes a straight velocity-time line and is invalid here; it even exceeds the 19 m upper bound. An average acceleration does not make an event a constant-acceleration event.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A ball is thrown vertically upwards at 15 m s-1 and caught at the same height. Air resistance is negligible. Taking upwards as positive, sketch the velocity-time graph from release to catch, with values on both axes. [3]

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