Forces and motion

This first part of the sub-topic is about describing motion — how far, how fast, and how quickly the speed changes — before any forces are brought in.

Four quantities do the work, and they come in two pairs. Distance and displacement both say how far; speed and velocity both say how fast. In each pair the second one includes a direction and the first does not, and that single difference is what the first three sections below are about.

Throughout this section on forces and motion you should be able to use ratios and proportional reasoning to convert units and to compute rates.

Distance is how far an object moves. Distance does not involve direction. Distance is a scalar quantity.

Displacement includes both the distance an object moves, measured in a straight line from the start point to the finish point, and the direction of that straight line. Displacement is a vector quantity.

You should be able to express a displacement in terms of both the magnitude and direction.

The difference shows up whenever a path is not straight. Walk 3 km north then 4 km east and the distance travelled is 7 km, but the displacement is 5 km in a north-easterly direction — the straight line from start to finish. Walk a complete lap of a track and the distance is a lap, but the displacement is zero.

Speed does not involve direction. Speed is a scalar quantity.

The speed of a moving object is rarely constant. When people walk, run or travel in a car their speed is constantly changing. The speed at which a person can walk, run or cycle depends on many factors, including age, terrain, fitness and distance travelled.

You should be able to recall typical values of speed, so these are worth learning as numbers.

You should also be able to recall typical values of speed for different types of transportation systems — a car on a motorway is roughly 30 m/s, a train perhaps 55 m/s, and a passenger aeroplane around 250 m/s.

For an object moving at constant speed, the distance travelled in a specific time can be calculated using the equation:

distance travelled = speed × time s = v t

distance, s, in metres, m; speed, v, in metres per second, m/s; time, t, in seconds, s.

You should be able to recall and apply this equation, to make measurements of distance and time and then calculate speeds of objects, and to calculate average speed for non-uniform motion.

The velocity of an object is its speed in a given direction. Velocity is a vector quantity.

You should be able to explain the vector–scalar distinction as it applies to displacement, distance, velocity and speed. Distance and speed are scalars, with magnitude only; displacement and velocity are vectors, with a direction as well.

Higher tier students should be able to explain qualitatively, with examples, that motion in a circle involves constant speed but changing velocity.

An object going round a circular track at a steady 10 m/s has a constant speed — the number never changes. But its direction is changing continuously, and velocity includes direction, so the velocity is changing all the time. A changing velocity means the object is accelerating, which means there must be a resultant force acting on it, directed towards the centre of the circle. A satellite in orbit and a car on a roundabout are both examples.

If an object moves along a straight line, the distance travelled can be represented by a distance–time graph.

The speed of an object can be calculated from the gradient of its distance–time graph.

You should be able to draw distance–time graphs from measurements, extract and interpret lines and slopes of distance–time graphs, translating information between graphical and numerical form, and determine speed from a distance–time graph.

Higher tier: if an object is accelerating, its speed at any particular time can be determined by drawing a tangent and measuring the gradient of the distance–time graph at that time.

Draw the tangent so that it just touches the curve at the time asked for, extend it well across the graph, and take a large triangle from the tangent itself.

The average acceleration of an object can be calculated using the equation:

acceleration = change in velocity / time taken a = ∆v / t

acceleration, a, in metres per second squared, m/s²; change in velocity, ∆v, in metres per second, m/s; time, t, in seconds, s.

You should be able to recall and apply this equation. An object that slows down is decelerating, which gives a negative acceleration.

You should be able to estimate the magnitude of everyday accelerations — a car pulling away is a few m/s², and a passenger lift perhaps 1 m/s².

The acceleration of an object can be calculated from the gradient of a velocity–time graph.

You should be able to draw velocity–time graphs from measurements and interpret lines and slopes to determine acceleration.

The following equation applies to uniform acceleration:

final velocity² − initial velocity² = 2 × acceleration × distance v² − u² = 2 a s

final velocity, v, in m/s; initial velocity, u, in m/s; acceleration, a, in m/s²; distance, s, in metres, m.

This equation is given on the Physics equation sheet. Use it when a question gives you two velocities and a distance but no time.

Near the Earth's surface any object falling freely under gravity has an acceleration of about 9.8 m/s².

Higher tier: the distance travelled by an object, or its displacement, can be calculated from the area under a velocity–time graph.

You should be able to interpret enclosed areas in velocity–time graphs to determine distance travelled or displacement, and to measure the area under a velocity–time graph by counting squares where appropriate.

Counting squares is the method to use when the line is curved and the area cannot be split into neat triangles and rectangles. Work out what one square represents from the axes, count the whole squares, then count part-squares as a half each.

An object falling through a fluid initially accelerates due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity.

Physics only students should be able to draw and interpret velocity–time graphs for objects that reach terminal velocity, and interpret the changing motion in terms of the forces acting.

The sequence of forces, which is what the marks are for:

1. At the moment of release the only significant force is weight, so the resultant force is large and the acceleration is greatest. 2. As the object speeds up, air resistance increases, because it depends on speed. 3. The resultant force is therefore smaller, so the acceleration decreases — the object is still speeding up, but less quickly. 4. Eventually air resistance equals weight. The resultant force is zero, so by Newton's First Law the object continues at a constant velocity — its terminal velocity.

On a velocity–time graph this appears as a line that starts steep, gradually becomes shallower, and finally levels off horizontally.

The three laws that follow answer three different questions, and it helps to know which is which before reading them.

• The First Law asks what happens when the resultant force is zero — the answer is that the velocity does not change. • The Second Law asks what happens when it is not zero — the answer is that the object accelerates, and by how much. • The Third Law is not about one object at all. It is about what two interacting objects do to each other.

A question that mentions balanced forces or a steady speed is a First Law question; one that gives you a force and a mass is a Second Law question; one that asks about a pair of objects pushing on each other is a Third Law question.

Newton's First Law: if the resultant force acting on an object is zero and:

• the object is stationary, the object remains stationary • the object is moving, the object continues to move at the same speed and in the same direction — so the object continues to move at the same velocity.

So when a vehicle travels at a steady speed the resistive forces balance the driving force. And the velocity — speed and/or direction — of an object will only change if a resultant force is acting on the object.

You should be able to apply Newton's First Law to explain the motion of objects moving with a uniform velocity, and objects where the speed and/or direction changes.

Higher tier: the tendency of objects to continue in their state of rest or of uniform motion is called inertia.

Inertia is not a force. It is the name for the behaviour the First Law describes — the fact that an object will not change its velocity by itself. It is why a passenger continues forwards when a car brakes sharply, and why a heavy trolley is hard to get moving and, once moving, hard to stop.

Newton's Second Law: the acceleration of an object is proportional to the resultant force acting on the object, and inversely proportional to the mass of the object.

You should recognise and be able to use the symbol for proportionality, ∝, so this is a ∝ F and a ∝ 1/m.

As an equation:

resultant force = mass × acceleration F = m a

force, F, in newtons, N; mass, m, in kilograms, kg; acceleration, a, in metres per second squared, m/s².

You should be able to recall and apply this equation.

You should also be able to estimate the speed, accelerations and forces involved in large accelerations for everyday road transport, and recognise and use the symbol that indicates an approximate value or approximate answer, ~.

Higher tier students should be able to explain that:

• inertial mass is a measure of how difficult it is to change the velocity of an object • inertial mass is defined as the ratio of force over acceleration.

That second point is just the Second Law rearranged: m = F ÷ a. Reading it that way makes the first point obvious — if a large force produces only a small acceleration, the ratio is large, and the object is hard to get moving or to stop.

Newton's Third Law: whenever two objects interact, the forces they exert on each other are equal and opposite.

You should be able to apply Newton's Third Law to examples of equilibrium situations.

The crucial detail is that the two forces act on different objects. A book on a table: the book pushes down on the table, and the table pushes up on the book. Those are a Third Law pair. The weight of the book and the table's upward push are not a Third Law pair, even though they are equal and opposite — because both act on the same object, the book.

Bringing a vehicle to a stop is where the rest of this sub-topic is put to work: reaction times decide how far it travels before the brakes are touched, and forces and energy decide how far it travels afterwards.

The four sections that follow split that into the total stopping distance, the driver's reaction time, the conditions that affect braking, and what braking does to the vehicle.

The stopping distance of a vehicle is the sum of the distance the vehicle travels during the driver's reaction time — the thinking distance — and the distance it travels under the braking force, the braking distance. For a given braking force, the greater the speed of the vehicle, the greater the stopping distance.

stopping distance = thinking distance + braking distance

Physics only students should be able to estimate how the distance for a vehicle to make an emergency stop varies over a range of speeds typical for that vehicle, and will be required to interpret graphs relating speed to stopping distance for a range of vehicles.

Reaction times vary from person to person. Typical values range from 0.2 s to 0.9 s.

A driver's reaction time can be affected by tiredness, drugs and alcohol. Distractions may also affect a driver's ability to react.

You should be able to explain methods used to measure human reaction times and recall typical results, interpret and evaluate measurements from simple methods to measure the different reaction times of students, and evaluate the effect of various factors on thinking distance based on given data.

The usual method is the ruler drop test. One person holds a ruler vertically between another's open fingers and releases it without warning; the distance the ruler falls before it is caught is converted to a reaction time using a conversion table or the equations of motion.

Repeating it and taking a mean reduces the effect of random variation, and it can be repeated with a distraction — music or conversation — to measure the effect. Points worth making when evaluating the method: the person catching may anticipate the drop if there is any pattern to the timing, and reading the ruler at the top of the fingers rather than the bottom introduces a systematic error.

The braking distance of a vehicle can be affected by adverse road and weather conditions and poor condition of the vehicle.

Adverse road conditions include wet or icy conditions. Poor condition of the vehicle is limited to the vehicle's brakes or tyres — so worn brake pads or bald tyres, and nothing else.

You should be able to explain the factors which affect the distance required for road transport vehicles to come to rest in emergencies, and the implications for safety, and estimate how the distance required for road vehicles to stop in an emergency varies over a range of typical speeds.

The implication for safety is the reason speed limits are lower in towns and why stopping distances are published: a driver cannot shorten their reaction time much, and cannot change the weather, so the one factor under their control is speed — and it is the factor that affects both parts at once.

When a force is applied to the brakes of a vehicle, work done by the friction force between the brakes and the wheel reduces the kinetic energy of the vehicle, and the temperature of the brakes increases.

That is work done applied to a car: work done against friction raises the temperature, and the energy comes from the kinetic store of the vehicle.

The greater the speed of a vehicle, the greater the braking force needed to stop the vehicle in a certain distance. The greater the braking force, the greater the deceleration of the vehicle. Large decelerations may lead to brakes overheating and/or loss of control.

You should be able to explain the dangers caused by large decelerations. Overheated brakes work less well, so the car may not stop as quickly as the driver expects; and a large braking force can lock the wheels or break the grip between tyres and road, so the driver loses steering control and the vehicle skids.

Higher tier students should be able to estimate the forces involved in the deceleration of road vehicles in typical situations on a public road. The route is F = m a, using an estimated mass for the vehicle and an acceleration worked out from a typical speed and stopping time.