English:Forces and motion calculations - Science Year 10
aiMOOCForces and motion calculations - Science Year 10
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Introduction
Interpret force diagrams and signed motion, calculate speed, acceleration and resultant force, and audit piecewise velocity records. Year 10 placement is editorial within the GCSE pathway.
Distinguish route distance, displacement, speed and velocity
Choose a reference direction before describing motion. In the original straight-line examples here, right is positive, left is negative, position is measured from a stated origin and elapsed time is measured in seconds. Distance travelled is the accumulated length of the route and cannot decrease as the journey continues. Displacement is final position minus initial position; it can be positive, negative or zero. Speed is a scalar; velocity includes direction and, in one dimension, can be represented with a sign.
An object goes six metres right and then two metres left. Its route length is eight metres but its displacement is four metres right. Over a four-second interval, average speed is two metres per second while average velocity is one metre per second right. These are different calculations with the same duration; returning to the origin could make average velocity zero even when substantial distance has been travelled.
Here the first ratio means average signed velocity; average speed uses total distance travelled instead of displacement. Acceleration is rate of change of velocity, not simply a synonym for going fast. Changing direction can mean acceleration even at constant speed. On a straight line, a negative acceleration can slow a right-moving object or speed up a left-moving object: compare the signs of velocity and acceleration before using the everyday word slowing.
In that particular picture the object is about ten metres from the start at six seconds, stays there until about eleven seconds and returns to the origin by about eighteen seconds. The last curved section means the rate changes; do not read one fixed return speed from its whole curved shape. The magnitude of the position-time gradient gives speed in this straight-line interpretation, while its sign indicates direction. The flat section indicates no change of position, rather than motion with the vertical-axis value as speed.
Before calculating, read what both axes represent and what units they use. A gradient on a position-time graph represents velocity; a gradient on a velocity-time graph represents acceleration. The area under a velocity-time graph represents signed displacement. Area under a position-time graph is not route distance. Axis shape alone does not tell you which of these quantities is being plotted.
Convert units before using a formula. For the supplied value 36.0 kilometres per hour, multiply by one thousand metres per kilometre and divide by 3600 seconds per hour to obtain 10.0 metres per second. A change from 2.00 to 8.00 metres per second over 3.00 seconds gives acceleration 2.00 metres per second squared; the squared second means change in velocity per second, not a distance measured in square seconds.
Pause at each calculation and identify distance, time and speed with their units before substituting. Explain why a route-distance triangle does not replace the need to distinguish signed displacement and velocity. If unavailable, use the original four-second out-and-return example and the explicit equations in this chapter. Original: FuseSchool - Global Education, Speed Distance Time | Forces & Motion | Physics | FuseSchool. The written explanations and native tasks work without sound or the external player.
Combine forces and explain acceleration on one body
Use the whole force diagram for one chosen object. A force is an interaction and has direction. A free-body diagram shows forces acting on the selected body, with an arrow for each relevant interaction. The resultant is their vector sum. In a horizontal one-dimensional case, subtract opposed forces after choosing a positive direction; do not add their magnitudes simply because both arrows appear on the diagram.
Use this original cart case on a level track. A 4.00 kg cart has a 12.0 N horizontal force to the right and a 4.00 N opposing force to the left. Take right as positive and include no other horizontal force in this model. The resultant is positive 8.00 N. For constant mass in the ordinary non-relativistic model, resultant force equals mass times acceleration.
The twelve-newton applied force is not the resultant. The eight-newton net force and two-metres-per-second-squared acceleration point right. If the cart were initially moving left, the same rightward acceleration would first reduce its leftward speed; the force does not have to point in the direction of the existing velocity.
For the vertical account, stipulate no vertical acceleration and no other vertical interaction beyond weight and the track’s normal contact force. With supplied g equal to 9.80 newtons per kilogram, weight is 39.2 N down and the normal contact force is 39.2 N up. Their vector sum is zero. This equality follows from the stated level-track conditions; normal force is not universally equal to weight in lifts, on slopes or with additional vertical forces.
The friction arrow in this chosen drawing is upslope, consistent with opposing a tendency to slide down the slope. Friction opposes relative sliding or its tendency at the contact; it is not always opposite the object’s overall velocity in every situation. The diagram does not by itself give a measured coefficient, acceleration or numerical angle, so none is inferred for our questions. Trigonometric resolution is an optional interpretation of the labels, not a prerequisite calculation here.
A zero resultant gives zero acceleration in this model: velocity stays constant. That includes rest and straight-line motion at constant non-zero velocity. It does not mean the forces have vanished. If the horizontal opposing force on the cart becomes 12.0 N while the applied force stays 12.0 N, a cart already moving right can continue at its current velocity. To reduce that rightward velocity requires a leftward resultant during the reduction.
Newton’s third-law pair acts on different bodies. Earth pulls the cart down and the cart pulls Earth up through the same gravitational interaction. The track pushes the cart up and the cart pushes the track down through contact. Weight and normal contact both act on the cart, so they are not each other’s third-law pair, even when their magnitudes happen to match.
Mass measures inertia in this account and is measured in kilograms; weight is a gravitational force in newtons and depends on field strength. A 4.00 kg cart does not become a 39.2 kg cart when its weight is calculated. A spring balance reading and an acceleration record also represent different measurements. State assumptions and uncertainty rather than treating an ideal diagram as a full equipment test.
Calculate from piecewise motion and stopping models
Build a motion account from a velocity-time record. The following original data are invented for a cart; they are not measurements from either image or the video. Assume velocity varies linearly between adjacent rows and remains along the chosen straight line. The two-second sampling interval and this interpolation rule are essential: a table alone does not prove what happened between measurements.
| Elapsed time in s | Signed velocity in metres per second |
|---|---|
| 0 | 0 |
| 2 | 4 |
| 4 | 8 |
| 6 | 8 |
| 8 | 4 |
| 10 | 0 |
From zero to four seconds, velocity increases uniformly from zero to eight metres per second, so acceleration is positive two metres per second squared. From four to six seconds it stays eight and acceleration is zero. From six to ten seconds it falls uniformly to zero, giving acceleration negative two. At ten seconds the cart has stopped momentarily; the table does not tell us whether it remains stopped afterwards.
Displacement equals the signed area under the velocity-time graph. The first four seconds form a triangle with area sixteen metres; the two-second constant section is a rectangle of sixteen metres; the final four seconds form another triangle of sixteen metres. Total displacement is forty-eight metres right. Velocity is never negative, so total distance also equals forty-eight metres. That equality depends on the absence of a reversal, not on distance and displacement always being the same.
This expression applies to each interval of constant acceleration, where u and v are its initial and final signed velocities. It does not justify replacing the entire record by a single ten-second constant-acceleration interval. Using just the first and last velocities, both zero, would wrongly predict zero displacement because acceleration did not stay constant throughout. Divide a piecewise record at the changes in its model.
Now analyse a separate original reversal record with the same assumption of straight lines between rows. This time the velocity becomes negative after four seconds; a negative area is leftward displacement, rather than negative distance travelled.
| Elapsed time in s | Signed velocity in metres per second |
|---|---|
| 0 | 0 |
| 2 | 4 |
| 4 | 0 |
| 6 | -4 |
| 8 | 0 |
The first four seconds give positive eight metres of area; the last four give negative eight. Total displacement and average velocity over the eight seconds are zero. Total distance is sixteen metres and average speed two metres per second. Between two and six seconds acceleration is negative two throughout: the object first slows while moving right, reaches zero velocity at four seconds, then speeds up left. Calling all negative acceleration braking would miss this change.
For a final original model, suppose a vehicle travels at 12.0 metres per second during a supplied 0.750 s thinking interval, then brakes uniformly at 3.00 metres per second squared to rest. Thinking distance is nine metres. Braking time is four seconds; with average braking velocity six metres per second, braking distance is twenty-four metres. Total is thirty-three metres. These supplied numbers are a calculation exercise, not safe stopping guidance or an official road-test value.
At a doubled initial speed with the same assumed reaction time and deceleration magnitude, thinking distance doubles while braking distance quadruples. The resulting total does not simply double. Real grip, slope, condition, attention and braking force can change the assumptions, so the fictional prediction cannot certify a safe following distance. A sound answer separates given values, physical assumptions, equations, units, signs and the scope of its conclusion.
Interactive Tasks
Quiz: Test Your Knowledge
A route goes six metres right and two left over four seconds. What is its average velocity? (One metre per second right from four metres displacement) (!Two metres per second right from all eight route metres) (!Zero because the object changed direction) (!Eight metres per second because distance and velocity are identical)
What can the falling part of the retained distance-time picture represent? (Decreasing separation from the start in a straight-line return) (!A decrease in total route length already travelled) (!A negative accumulated distance travelled) (!A velocity-time area without reading the axes)
For the supplied cart, what horizontal force enters the acceleration calculation? (Eight newtons right, the resultant of twelve right and four left) (!Sixteen newtons right by adding both arrow magnitudes) (!Twelve newtons right while ignoring the opposing force) (!Thirty-nine point two kilograms down)
What acceleration follows for the supplied four-kilogram cart? (Two metres per second squared right) (!Three metres per second squared because applied force alone is used) (!Eight metres per second right because force equals velocity) (!Zero whenever an object was previously at rest)
Why do the normal force and weight match in the stated level-track model? (There is no vertical acceleration and no other vertical force in that model) (!Normal contact is always weight on every surface) (!They are necessarily a third-law pair on the same body) (!Mass has been converted into a different unit of mass)
What do the dashed sine and cosine arrows in the slope picture mean? (Components of the same weight, not additional forces to count twice) (!Two extra gravitational forces to add to the full weight) (!A measured friction coefficient and numerical acceleration) (!A guarantee that all friction always acts against overall velocity)
What does zero resultant mean for a cart already moving straight? (Its velocity stays constant under the stated model) (!Its velocity must instantly become zero) (!No interactions can be acting at all) (!Its mass must have vanished)
Which description correctly identifies a third-law pair? (The track pushes the cart and the cart pushes the track) (!Weight and normal force both acting on the cart) (!Applied force and friction simply because they have opposite arrows) (!Acceleration and velocity of the same object)
What are the distance and displacement in the original ten-second non-reversing record? (Both forty-eight metres, with displacement to the right) (!Both zero because the initial and final velocities are zero) (!Distance negative forty-eight metres and displacement positive) (!Eighty metres because the peak velocity lasts for all ten seconds)
What does the separate reversal record give over eight seconds? (Sixteen metres distance and zero displacement) (!Zero distance because the final position equals the start) (!Negative sixteen metres of distance travelled) (!Eight metres displacement because only the first half counts)
Memory Game
| Distance | Accumulated route length, a scalar that does not decrease |
| Displacement | Final position minus initial position, with direction |
| Velocity | Rate of change of position with direction |
| Acceleration | Rate of change of velocity |
| Resultant force | Vector sum of the forces on one chosen body |
| Inertia | Resistance to a change in velocity associated with mass |
Drag and Drop
| A position-time trace slopes downward on a straight-line return | Interpret a negative velocity, not a loss of route distance |
| A velocity-time interval is horizontal above zero | Use constant positive velocity and zero acceleration |
| Equal opposed forces act on a moving cart | Keep its velocity constant rather than force it to stop |
| A diagram resolves mg into two dashed components | Use the components instead of counting them again with full mg |
| A velocity-time trace crosses through zero into negative values | Separate signed displacement from total distance travelled |
Crossword Puzzle
| SCALAR | What kind of quantity has magnitude without a direction? |
| VECTOR | What kind of quantity includes direction? |
| INERTIA | What property is associated with resistance to velocity change? |
| RESULTANT | What single force represents the vector sum on a body? |
| VELOCITY | What motion quantity includes direction and rate of position change? |
| GRADIENT | What feature of a velocity-time graph gives acceleration? |
LearningApps
Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.
Cloze Text
Open Tasks
Easy
- Read the reference direction: Explain the original right-left route, its distance and its displacement.
- Audit the pictured graph: Explain why its falling values cannot be cumulative route distance and state the out-and-return interpretation.
- Read a free-body diagram: Identify weight, normal contact, friction and the alternative weight components without double counting.
- Distinguish mass and weight: Use the supplied cart to explain kilograms, newtons and the stated gravitational field strength.
Standard
- Calculate the horizontal resultant: Find the cart acceleration, then explain what happens if the opposing force rises to twelve newtons.
- Interpret the first motion record: Find all acceleration phases and calculate displacement by separate triangle and rectangle areas.
- Calculate a reversal: Find signed areas, total distance, average speed and average velocity for the eight-second record.
- Calculate the stopping model: Find thinking distance, braking time, braking distance and total from the supplied assumptions.
Advanced
- Test a misleading whole-record formula: Explain why averaging only the first and last velocities fails for the complete ten-second record.
- Analyse negative acceleration: Describe the two-to-six-second interval of the reversal case before and after zero velocity.
- Distinguish force balance and third law: Give separate body-labelled gravitational and contact pairs and explain why weight and normal are not one pair.
- Evaluate a doubled-speed prediction: Derive the different thinking and braking scaling, calculate the new total and state why it is not certified driving guidance.
Learning Check
- Distance, displacement and averages: Use the stated origin, right-positive convention and common duration.
- Force and acceleration: Use the vector sum on the cart and compatible SI quantities.
- Balance, friction and paired interactions: State which body each force acts on.
- Piecewise velocity record: Use the stated linear interpolation and split at phase changes.
- Reversal record: Keep signed area and route distance separate.
- Stopping and doubled speed: Calculate under the supplied reaction time and uniform deceleration assumptions.
Worked checks and assessment guidance
- Six metres right then two left gives eight metres distance and four metres right displacement. In four seconds average speed is two metres per second; average velocity is one metre per second right. The graph’s falling vertical values must be interpreted as separation from the start, not decreasing accumulated route length. A position-time gradient gives signed velocity; read axes before assigning a physical meaning.
- Twelve newtons right minus four left gives eight right. Divide by the four-kilogram mass for acceleration two metres per second squared right. Vertical weight is 39.2 newtons down and normal force 39.2 up only under the stipulated no-vertical-acceleration and no-other-vertical-force assumptions. Mass remains four kilograms. Components in the inclined-plane picture replace the weight vector for resolution rather than add extra weight.
- A zero resultant permits rest or constant non-zero straight-line velocity, not mandatory stopping. In the shown slope case friction is upslope to oppose a down-slope sliding tendency; this is not a rule that friction always opposes overall velocity. Track-on-cart and cart-on-track are a third-law pair on different bodies, as are Earth-on-cart and cart-on-Earth. Normal force and weight both on the cart are not one such pair.
- Acceleration is positive two from zero to four seconds, zero from four to six, and negative two from six to ten. Areas are sixteen, sixteen and sixteen metres, total forty-eight right. Velocity never reverses so distance is also forty-eight. Treating the whole ten seconds as constant acceleration and averaging its zero endpoint velocities would wrongly give zero; the assumption is false for that whole interval.
- The positive first-half area is eight metres and the negative second-half area is negative eight. Displacement and average velocity are zero; route distance is sixteen metres and average speed two metres per second. Between two and six seconds acceleration is negative two: positive velocity decreases to zero at four, then negative velocity grows in magnitude. Negative acceleration is therefore not always a decrease in speed.
- At twelve metres per second with reaction time 0.750 seconds, thinking distance is nine metres. Deceleration magnitude three gives four seconds braking, average braking velocity six and braking distance twenty-four, total thirty-three metres. At twenty-four metres per second, thinking distance is eighteen, braking time eight and braking distance ninety-six, total 114 metres. Doubling speed doubles the thinking part but quadruples the braking part in this model. Real conditions can change the assumptions; this is not safety-certified road guidance.
Learning evidence
Retain signed journey calculations, body-labelled force and interaction diagrams, independently added motion-table areas and a bounded stopping-model comparison. State units, interpolation and constant-acceleration assumptions, distinguish cumulative route distance from position, and explain what the unchanged illustrations do not measure.
OER and Wikipedia
Original cart, right-left journey, two velocity-time tables, stopping model, questions and worked feedback. All three exact Commons originals and the original FuseSchool preview were visually inspected; selected files also resolved in the native English renderer before authoring. The diagram’s weight components and the pictured graph’s decreasing distance label are explicitly qualified.
DfE GCSE combined science, forces and motion scope. OpenStax Physics relative motion, distance and displacement, acceleration and Newton’s second law, primary concept and equation context. Numerical cases, tasks and worked explanations here are original, not copied source exercises. Sources checked 13 November 2026.
Free body1.3.svg, Jer S, CC BY-SA 4.0. Unchanged original; caption and learning activities are original.
Distance-time graph example.svg, Sjlegg at English Wikibooks, Public domain. Unchanged original; caption and learning activities are original.
Stoppuhr.jpg, Ulfbastel, Public domain. Unchanged original; caption and learning activities are original.
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