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Motion and Forces overview

Full chapter

Motion and Forces

All 6 topics and the revision summary on one page.

01

Describe motion with a reference

Choose an origin, a positive direction and an interval. Then distinguish where a body is, how far it has travelled and how its velocity is changing.

Position
Location relative to a chosen origin. In one dimension, x = +2 m means 2 m in the positive direction from the origin.
Distance travelled
The accumulated length of the path. It is a scalar, is nonnegative and cannot decrease as the journey continues.
Displacement
The directed change in position. For an interval, s = xfinal - xinitial. It can be positive, negative or zero in a chosen coordinate direction.
Speed and velocity
Average speed is total distance divided by elapsed time. Average velocity is displacement divided by elapsed time. Instantaneous velocity describes the local rate of change of position; instantaneous speed is its magnitude.
Acceleration
The rate of change of velocity. Average acceleration is Δv/Δt; instantaneous acceleration is the local rate at a particular time. A change in direction can produce acceleration even when speed is constant.

Distance and displacement are measured in m; speed and velocity in m/s; acceleration in m/s2. A negative position does not by itself mean leftward motion. Position tells you where the body is; velocity tells you how that position is changing.

A velocity change is a vector difference: final velocity minus initial velocity. A change in speed alone does not include every possible change in velocity.

Follow one motion through a reversal

Choose right as positive. At t = 0, a body is at x = 2.0 m with velocity +6.0 m/s. Its acceleration is constant at -2.0 m/s2 for the next 5.0 s.

Equal time intervals do not mean equal distances

The two rows share the same position scale. Dots are 1 s apart in time; the return row repeats the turning point so you can follow the order.

The body moves right, turns, then moves leftRight is positive and the fixed spatial origin is zero. All positions use twenty-three drawing units per metre. The upper row shows times zero, one, two and three seconds at positions two, seven, ten and eleven metres. The lower row shows three, four and five seconds at eleven, ten and seven metres. Outward gaps are five, three and one metres; return gaps are one and three metres. Motion-path arrowheads show the order, not velocity magnitudes. A separate green leftward arrow labels the constant negative acceleration; acceleration remains negative two metres per second squared even at the turning point.Outward: 0 to 3 s+ right0 s1 s2 s3 sReturn: 3 to 5 s3 s4 s5 s024681012Position x / ma = -2.0 m s-2 throughout

At 3 s, velocity is zero but acceleration is not. At 4 s the position is still positive while the body is moving left. Over the full 5 s, displacement is +5 m and distance is 13 m.

The positions are one second apart in time. Their spacing changes as the body slows, turns and speeds up leftward. Path arrowheads show the order of travel; acceleration remains directed left.
Positions and velocities in the stated model
Time / sPosition / mVelocity / m s-1
02+6
17+4
210+2
3110
410-2
57-4

At 3 s, the body reaches x = 11 m and is instantaneously at rest. Its acceleration is still -2 m/s2, so it reverses rather than staying at rest. At 4 s, it is still to the right of the origin, but its negative velocity means it is moving left.

Worked averages

Distance and displacement answer different questions

The body moves from x = 2 m to 11 m, then back to 7 m.

Distance = (11 - 2) + (11 - 7) = 13 m.
Displacement = 7 - 2 = +5 m.

Over 5.0 s, average speed = 13/5 = 2.6 m/s, while average velocity = 5/5 = +1.0 m/s.

Use both signs to decide whether speed increases

In one dimension, velocity and acceleration of opposite signs mean decreasing speed. Equal signs mean increasing speed. In this model, negative acceleration first slows rightward motion, then speeds up leftward motion after the turn.

Negative acceleration does not always mean slowing down. Acceleration tells you how velocity changes, not the direction in which the body is already moving.

Optional check Taking right as positive, a body has velocity -3.0 m/s and acceleration +2.0 m/s^2 at an instant. How is it moving then?
Taking right as positive, a body has velocity -3.0 m/s and acceleration +2.0 m/s^2 at an instant. How is it moving then?

02

Read slopes, areas and changing acceleration

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.

Optional check 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?
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?

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.

Optional check 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?
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?

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.

03

Derive and use constant-acceleration equations

The familiar motion equations follow from constant acceleration along one straight line. Check that model before substituting numbers.

Start the chosen interval at t = 0. Let u be its initial velocity, v its final velocity, a the constant acceleration, t its duration and s its signed displacement. Use the same positive direction for every signed value. A reversal on the same line is allowed.

Build the equations from a rate and a graph area

  1. Start with acceleration.

    For constant acceleration, a = (v - u)/t. Rearranging gives v = u + at. Velocity therefore changes linearly with time.

  2. Use the straight velocity-time line.

    Its signed area gives displacement: s = (u + v)t/2. The mean of the endpoint velocities is the average velocity because the graph is straight. If it crosses zero, separate positive and negative areas give the same signed result.

  3. Eliminate final velocity.

    Substitute v = u + at into the area equation:
    s = [u + (u + at)]t/2
    = ut + ½at2.

  4. Alternatively, eliminate initial velocity.

    Use u = v - at:
    s = [(v - at) + v]t/2
    = vt - ½at2.

  5. Eliminate time.

    Multiply v - u = at by v + u:
    v2 - u2 = a(v + u)t.
    Since (v + u)t = 2s, this gives v2 = u2 + 2as. This derivation does not require division by a and remains consistent when a = 0.

Choose an equation that connects your known quantities to the required one. For stages with different constant accelerations, start a new interval for each stage and carry its ending position and velocity into the next. Do not use one acceleration across a stage where it changes.

Optional check A body moves along a straight line with a curved velocity-time graph. Why is displacement = (initial velocity + final velocity) x time / 2 generally unsuitable?
A body moves along a straight line with a curved velocity-time graph. Why is displacement = (initial velocity + final velocity) x time / 2 generally unsuitable?

A reversal is still straight-line motion

Displacement need not have the final velocity's sign

Take u = +6.0 m/s, a = -2.0 m/s2 and t = 5.0 s.

v = 6 + (-2)(5) = -4 m/s.
s = 6(5) + ½(-2)(52) = +5 m.

For distance, locate the turn first: 0 = 6 - 2t gives t = 3 s. The body travels 9 m before the turn and 4 m afterwards, for 13 m total. The +5 m result from the equation is displacement.

Keep one sign convention during vertical motion

An object is launched vertically upward at 12.0 m/s from a point 1.50 m above a floor. Take upward as positive and the release point as y = 0. Neglect air resistance and use uniform g = 9.81 m/s2, so a = -9.81 m/s2 throughout the flight.

Keep the release origin and upward-positive convention

Launch upwards from one and a half metres above the floorHeights use forty drawing units per metre, with release at y equals zero and the floor at minus 1.50 metres. The point-like object is launched at positive twelve metres per second. Its highest point is 7.33945 metres above release, at about 1.22 seconds, with instantaneous velocity zero. Acceleration is negative 9.81 metres per second squared throughout. The floor is reached after about 2.57 seconds with downward velocity negative 13.2 metres per second. Velocity and acceleration arrows are drawn beside the one-dimensional path, with schematic lengths; they do not compare unlike quantities. The downward velocity arrow ends beside the floor without depicting any motion after impact.Top: v = 0; t = 1.22 s7.34 mauv1.50 mRelease: y = 0Floor: y = -1.50 mJust before impact: v = -13.2 m/s+ up

The height scale is shared; the coloured arrows show directions schematically. Here u = +12.0 m/s and a = -9.81 m/s2, including at the top. The impact calculation uses displacement -1.50 m, not the total path travelled.

The release point is the position origin, so the floor is at -1.50 m. Velocity is zero at the top, but acceleration remains downward on ascent, at the top and on descent.

Worked vertical motion

Find the height, then choose the impact roots

  1. Time to the top: v = 0, so 0 = 12.0 - 9.81t. Thus t = 1.223... s, or 1.22 s.
  2. Rise above release: 0 = 12.02 + 2(-9.81)s. Thus s = 7.339... m, or 7.34 m. Adding the initial height gives 8.84 m above the floor.
  3. Velocity just before floor impact: s = -1.50 m, so v2 = 12.02 + 2(-9.81)(-1.50) = 173.43 m2/s2. The object is moving downward, so select v = -13.2 m/s.
  4. Time to impact: -1.50 = 12.0t - 4.905t2, or 4.905t2 - 12.0t - 1.50 = 0. The quadratic formula gives t = [12.0 ± √(12.02 + 4(4.905)(1.50))]/9.81. The roots are 2.5657... s and -0.1192... s. The required post-launch event is t = 2.57 s; the negative root is outside the stated t ≥ 0 interval.

Squaring velocity removes its sign; the physical direction selects the root. Zero velocity at the highest point does not mean zero acceleration or zero resultant force.

The model ends just before floor contact. During impact the contact force changes the acceleration. Significant air resistance would also undermine the gravity-only constant-acceleration model.

Optional check An object is launched upward from above a floor. Upward is positive. At its later floor impact, an equation gives v^2 = 173.43 m^2/s^2. Which velocity is appropriate?
An object is launched upward from above a floor. Upward is positive. At its later floor impact, an equation gives v^2 = 173.43 m^2/s^2. Which velocity is appropriate?

04

Inertia and linear momentum

Mass describes a body's resistance to a change in motion. Momentum combines that mass with its velocity, including direction.

Inertia concerns a change in motion

Mass measures inertia. Under the same resultant force, a body of greater constant mass has a smaller acceleration. Inertia is not a resistance to motion itself: a moving body does not need a forward resultant force merely to keep a constant velocity.

Mass is a scalar measured in kg. Weight is a gravitational force measured in N. We treat the laboratory frame as approximately inertial, so the ordinary force laws apply without adding a separate backward "inertia force" to a body's diagram.

Linear momentum is mass times velocity

p = mv

Momentum is a vector in the direction of velocity. In one dimension, retain the velocity's sign. If velocity is given by several components, multiply every component by the scalar mass.

Its unit is kg m/s, also written N s: since N = kg m s-2, multiplying by s gives kg m s-1. Momentum is not a force; N s and N describe different quantities.

Momentum through a reversal

Subtract final and initial vectors in order

A 1.20 kg body changes velocity from +3.0 m/s to -2.0 m/s along the same axis.

Initial momentum = 1.20(+3.0) = +3.6 kg m/s.
Final momentum = 1.20(-2.0) = -2.4 kg m/s.

Change = final - initial:
Δp = -2.4 - (+3.6) = -6.0 kg m/s.

Subtract signed momentum, including the reversal

Positive 3.6 changes to negative 2.4 kilogram metres per secondThree separate arrow rows use forty-five drawing units per kilogram metre per second, with right positive. In the first row initial momentum points right with magnitude 3.6. In the second final momentum points left with magnitude 2.4. The two share a marked zero coordinate. In the third, the change arrow starts at the initial momentum tip coordinate and ends at the final momentum tip coordinate, so it points left with magnitude six. This is final minus initial, not a subtraction of the two magnitudes.Zero coordinateInitial p = +3.6 kg m/sFinal p = -2.4 kg m/sChange = -6.0 kg m/sInitial tipFinal tip

For the 1.20 kg body, Δp = -2.4 - (+3.6) = -6.0 kg m/s. The change is leftwards. If this occurs over 0.50 s, the average resultant force is -12 N; that alone does not establish a constant instantaneous force.

Initial and final momentum arrows refer to different times. Their vector difference points in the negative direction and has magnitude 6.0 kg m/s. It is not the difference of the speed magnitudes.

Momentum can be zero at an instant while changing at that instant. At the top of a vertical flight, velocity and momentum are zero, but gravity still produces downward acceleration and a changing momentum.

Optional check A 1.20 kg body changes velocity from +3.0 m/s to -2.0 m/s. What is its change in momentum?
A 1.20 kg body changes velocity from +3.0 m/s to -2.0 m/s. What is its change in momentum?

05

Apply all three Newton laws

Identify the body before applying a law. The resultant force describes the combined effect of forces on that body; an interaction pair acts on two different bodies.

Use an approximately inertial laboratory frame for these models. A free-body diagram helps keep the recipient of every force explicit.

First law: zero resultant preserves velocity

A body at rest stays at rest, and a moving body continues at constant velocity, unless a resultant external force acts on it.

Zero resultant does not mean that no forces act. For example, the weight and support on a gliding object may balance vertically while horizontal resistance is negligible. It can continue moving horizontally without a forward resultant force.

When a vehicle brakes, a passenger tends to continue with the previous velocity. A restraint or another contact supplies the force needed to change the passenger's motion. The tendency to continue is inertia, not evidence of an extra forward force.

Second law: resultant force changes momentum

The rate of change of momentum is proportional to the resultant force and is in its direction. With consistent SI units, the instantaneous relationship is:

Fresultant = dp/dt

Average resultant force = Δp/Δt

Here dp/dt means the local rate of change of momentum. The interval expression gives the average force over the stated interval; it does not by itself establish the force at every instant.

Worked average force

Use the full momentum change

A 1.20 kg body changes velocity from +3.0 to -2.0 m/s in 0.50 s. Its momentum change is 1.20(-2.0 - 3.0) = -6.0 kg m/s.

Average resultant force = -6.0/0.50 = -12 N.

The negative sign gives the force direction. The body initially moved positively, but its momentum changed negatively. The result does not show that the force was -12 N throughout the interval.

Third law: interaction partners are simultaneous

If body A exerts a force on body B, body B simultaneously exerts a force of equal magnitude and opposite direction on body A. The pair describes one interaction and acts on different bodies.

The partner is not a later response. It can exist whether or not either body is moving. When drawing the forces on B, include A's force on B; B's force on A belongs on A's diagram.

The contact pair acts on two different skaters

These are separate horizontal interaction views. Vertical weight and support forces are omitted. Both contact forces exist at the same instant.

Equal contact forces give different accelerationsThe upper view selects the forty-kilogram skater: the sixty-kilogram skater exerts a 120-newton leftward force on this body. The lower view selects the sixty-kilogram skater: the forty-kilogram skater exerts a 120-newton rightward force on this different body. The two purple force arrows have equal lengths. With other horizontal forces neglected, the forty-kilogram body accelerates left at three metres per second squared and the sixty-kilogram body right at two. The green acceleration arrows use a separate common scale, with lengths in the ratio three to two; they are not extra forces.Force from the 60 kg skater120 N40 kg bodya = 3.0 m/s2 leftForce from the 40 kg skater120 N60 kg bodya = 2.0 m/s2 right

Purple arrows compare forces on one scale; green arrows compare accelerations on another. The pair's forces are equal. The accelerations differ because the masses differ.

The two 120 N contact forces act simultaneously on different skaters. This pair-only view omits their vertical forces. Equal contact forces need not produce equal accelerations when the masses differ.

Equal forces, different accelerations

Two skaters push each other

A 40 kg skater is pushed left and a 60 kg skater right by a pair of 120 N contact forces. Other horizontal forces are negligible.

Their accelerations are 120/40 = 3.0 m/s2 left and 120/60 = 2.0 m/s2 right. The different responses follow from their different masses, not from unequal partner forces.

A stationary book's weight and table support both act on the book. They can balance, but they are not a third-law pair. Earth's pull on the book is paired with the book's pull on Earth. The table's force on the book is paired with the book's force on the table.

Equal and opposite is not enough to identify a third-law pair. Name both source and recipient and check that the forces are the two sides of the same interaction.

Optional check A 40 kg skater pushes a 60 kg skater rightward with 120 N. Other horizontal forces are negligible. Which statement describes the simultaneous interaction?
A 40 kg skater pushes a 60 kg skater rightward with 120 N. Other horizontal forces are negligible. Which statement describes the simultaneous interaction?

06

From resultant force to motion

Build the force model first. For a body of constant mass, the resultant force determines its acceleration.

Momentum is p = mv. When m is constant, its rate of change is dp/dt = m dv/dt = ma. Combining this with the second law gives:

Fresultant = ma

Use the resultant component along the chosen direction, not one selected applied force. The acceleration follows the resultant force; it need not point along the body's current velocity.

Estimate the scale before detailed calculation

Suppose a small cart has mass roughly 1 kg, moves at a speed around 1 m/s, and changes velocity by about 1 m/s in the same direction over about 2 s. These rough assumptions suggest momentum magnitude of order 1 kg m/s = 1 N s and acceleration magnitude about 0.5 m/s2.

If its speed stays around 1 m/s over those 2 s, expect travel of order 2 m. These are scale checks, not precise predictions. They help expose a result such as a kilometre of travel for this short, slow cart run.

A cart pulled along a level track

A 0.80 kg cart is moving right with initial velocity +0.40 m/s. A string pulls it right with 0.60 N and track resistance acts left with 0.20 N. These forces stay constant over the next 2.0 s, during which the cart continues moving right.

Find the resultant from the actual forces on the cart

Four external forces on a cart moving rightThe chosen body is a 0.80-kilogram cart. The string pulls right at 0.60 newtons; track resistance acts left at 0.20 newtons. Their arrow lengths use two hundred drawing units per newton, giving the ratio three to one. The vertical support and weight each have magnitude 7.848 newtons, using a separate shared scale of fifteen drawing units per newton. The horizontal and vertical lengths must not be compared as one force scale. A separate blue rightward motion arrow below the force model indicates that the cart continues moving right. No resultant or inertia force is added as another physical interaction.N = 7.848 NPull: 0.60 N0.20 NResistanceW = 7.848 NCart continues right

Scale note: the horizontal arrows share one scale, and the two vertical arrows share another. Horizontally, 0.60 - 0.20 = +0.40 N. Vertically, support and weight balance. These are the four interactions; the resultant is their sum.

The diagram shows forces on the cart. String pull and track resistance have a rightward resultant; support and weight balance vertically. A resultant is a summary of these forces, not an extra interaction to add.

Worked force-to-motion model

Use the resultant, then justify constant acceleration

  1. Identify the forces: string on cart, track resistance on cart, track support and Earth's gravitational force. With g = 9.81 m/s2, weight and support each have magnitude 0.80(9.81) = 7.848 N, about 7.85 N.
  2. Resolve horizontally: Fx = +0.60 - 0.20 = +0.40 N. Vertical forces balance.
  3. Calculate acceleration: a = 0.40/0.80 = +0.50 m/s2. Constant resultant and constant mass justify constant acceleration.
  4. Predict the ending motion: v = 0.40 + 0.50(2.0) = +1.40 m/s. Displacement s = 0.40(2.0) + ½(0.50)(2.0)2 = +1.80 m.
  5. Check momentum: Δp = 0.80(1.40 - 0.40) = +0.80 kg m/s. Dividing by 2.0 s gives +0.40 N, the same resultant.

Resistance is leftward because the stated cart keeps moving right. If the cart reverses, reconsider the resistance direction. If the pull is removed, calculate a new resultant before using the motion equations.

Optional check A 0.80 kg cart is moving right when its pulling string goes slack. Track resistance remains 0.20 N leftward. What is its acceleration immediately afterwards, taking right as positive?
A 0.80 kg cart is moving right when its pulling string goes slack. Track resistance remains 0.20 N leftward. What is its acceleration immediately afterwards, taking right as positive?

Infer motion from position data

One measurement method is to film a cart beside a length scale in its plane of motion. Fix the camera approximately perpendicular to that plane, establish the time base, calibrate length and identify the same point on the cart in each frame. A suitable position sensor provides another route.

A scale outside the cart's motion plane or a changed camera view can bias the distance conversion. Repeating a run does not repair that calibration. Keep the run within a supported, clear path with enough run-out.

The following are supplied rounded model values for the cart, starting at position +0.200 m. They are used to practise the analysis; they are not laboratory observations.

Cart-position model for the 0 to 2 s interval
Time / sPosition / m
0.0000.200
0.5000.463
1.0000.850
1.5001.363
2.0002.000

For each neighbouring pair, divide displacement by its own time interval. The first gives (0.463 - 0.200)/(0.500 - 0.000) = 0.526 m/s. This is an interval-average velocity.

Under a constant-acceleration model, associate that value with the interval's midpoint time: (0.000 + 0.500)/2 = 0.250 s. This follows because velocity is a straight-line function of time. For more general changing motion, such an assignment is an approximation.

Plotting the interval velocities against their midpoint times allows a straight-line fit. Its gradient estimates acceleration; its intercept estimates the initial velocity for the chosen t = 0. Compare mass × fitted acceleration with the resultant 0.40 N, not the string pull alone.

Choose time intervals short enough to resolve change but not so short that small position differences are dominated by reading noise. Differencing can magnify that noise. A fixed position-zero offset cancels in each difference; a length scale-factor error does not.

Do not plot an interval-average velocity at its later endpoint by default. Also, x against t2 is not automatically a straight line here: the initial velocity is nonzero, so position includes a term proportional to t.

Optional check In a constant-acceleration cart model, x changes from 0.200 m at 0.000 s to 0.463 m at 0.500 s. The interval-average velocity is 0.526 m/s. At what representative time should it be plotted?
In a constant-acceleration cart model, x changes from 0.200 m at 0.000 s to 0.463 m at 0.500 s. The interval-average velocity is 0.526 m/s. At what representative time should it be plotted?
Check the cart velocities and fitted motion

The four midpoint times are 0.250, 0.750, 1.250 and 1.750 s. Their interval velocities are 0.526, 0.774, 1.026 and 1.274 m/s.

A free-intercept straight-line fit gives gradient 0.4992 m/s2 and intercept 0.4008 m/s. These represent approximately a = 0.50 m/s2 and u = 0.40 m/s. The fitted relationship is velocity = initial velocity + acceleration × midpoint time.

Inferred resultant = 0.80 × 0.4992 = 0.39936 N, or about 0.40 N. The small difference from the ideal model follows from the deliberately rounded supplied positions. Extra spreadsheet digits do not establish the precision of a real camera measurement.

Revision summary

State the body, origin, positive direction, interval and model. Keep signed motion quantities distinct from their magnitudes.

Motion quantities

Position and displacement
Position is relative to an origin. Displacement over an interval is final position minus initial position.
Distance and speed
Distance is total path length. Average speed = total distance / elapsed time. Instantaneous speed is the magnitude of velocity.
Velocity and acceleration
Average velocity = displacement / elapsed time. Average acceleration = Δv/Δt. Locally, v = dx/dt and a = dv/dt.

In one dimension, opposite velocity and acceleration signs mean slowing; matching signs mean increasing speed. Zero velocity at a turning instant need not mean zero acceleration.

Graph meanings

  • Position-time or displacement-time gradient: velocity. A chord gives an interval average; a tangent gives a local value.
  • Cumulative distance-time gradient: speed. Cumulative distance does not decrease.
  • Velocity-time gradient: acceleration.
  • Signed velocity-time area: displacement. Add area magnitudes, or use speed-time area, for distance.
  • For a curved graph, estimate local gradients and areas with suitable intervals. Report numerical area approximations as estimates.

Straight-line constant acceleration

v = u + at

s = (u + v)t/2

s = ut + ½at2

s = vt - ½at2

v2 = u2 + 2as

Derive these from constant acceleration and the signed area of a straight velocity-time graph. All variables refer to the same interval. A reversal is allowed, but s is displacement, not automatically total distance.

For vertical motion with upward positive and negligible air resistance, use a = -g throughout flight. Select velocity roots using direction and time roots using the stated time domain. The acceleration remains downward when velocity is zero at the top.

Inertia and momentum

Mass measures resistance to a change in motion. Momentum p = mv is a vector; Δp = pfinal - pinitial. Its units satisfy N s = kg m/s.

All three Newton laws

  1. First: a body stays at rest or at constant velocity unless a resultant external force acts.
  2. Second: momentum-change rate is proportional to the resultant force and in its direction. In SI, Fresultant = dp/dt; interval-average force = Δp/Δt.
  3. Third: interaction partners have equal magnitude and opposite direction, act simultaneously and act on different bodies.

Balanced forces on one body are not automatically a third-law pair. Equal forces on different masses can produce different accelerations.

Use force and data models

For constant mass, Fresultant = ma. Add the actual forces first; a resultant arrow is their summary. A constant resultant and constant mass justify constant acceleration.

Successive position differences divided by their actual time intervals give interval-average velocities. For constant acceleration, plot them at midpoint times; fit velocity against time to estimate acceleration and initial velocity. Preserve raw data and match any claimed precision to the measurement method.

Quantity and unit reference
QuantityUsual symbolsUnits
Distance; displacementd; s, xm
Speed; velocityu, v, w, cm s-1
Acceleration; free-fall accelerationa; gm s-2
Mass; timem; tkg; s
Force; weightF; WN = kg m s-2
Linear momentumpN s = kg m s-1

Symbols depend on context. Here x is position, s is displacement over an interval, and u and v are its initial and final velocities. Other contexts may use w or c for a speed or velocity; read the definition rather than inferring the quantity from the letter alone.

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