Overview
In this topic you must be able to:
- Use and read distance–time and velocity–time graphs.
- Calculate average speed and acceleration.
- Describe and calculate the effect of forces, including friction and weight.
- Use the formulas F = m a, W = m g, and v² = u² + 2as.
- Understand momentum, Newton’s third law and moments.
Exam idea: markers love questions where you combine more than one of these ideas in the same problem (for example graphs + F = m a + stopping distance).
1. Motion Graphs & Basic Formulas
Distance–Time Graphs
- Plot and explain distance–time graphs.
- Describe what a flat, rising or steeper line means.
On a distance–time graph:
- Gradient (slope) = speed.
- Flat line = object is stationary (distance not changing).
- Steeper line = higher speed.
Average Speed
vavg = d / t
Always convert minutes to seconds (and hours to seconds) before using the formula.
Velocity–Time Graphs
- Plot and explain velocity–time graphs.
- Determine acceleration from the gradient.
- Determine distance travelled from the area under the graph.
Acceleration
a = (v − u) / t
u = initial velocity, v = final velocity.
Linking Speed, Acceleration and Distance
v² = u² + 2 a s
Quick Check
Q1. A car travels 600 m in 30 s. What is its average speed?
Show answer
v = d / t = 600 / 30 = 20 m/s
Q2. A cyclist speeds up from 4 m/s to 10 m/s in 3 s. Calculate the acceleration.
Show answer
a = (v − u)/t = (10 − 4) / 3 = 6 / 3 = 2 m/s²
2. Forces, Friction & Elastic Behaviour
Types of Force
- Describe the effects of forces between bodies such as changes in speed, shape or direction.
- Identify different types of force such as gravitational or electrostatic.
- Understand how vector quantities differ from scalar quantities.
- Understand that force is a vector quantity.
- Calculate the resultant force of forces that act along a line.
Friction & Stretching
- Know that friction is a force that opposes motion.
- Describe practicals: investigate how extension varies with applied force for helical springs, metal wires and rubber bands.
- Know that the initial linear region of a force–extension graph is associated with Hooke’s law.
- Describe elastic behaviour as the ability of a material to recover its original shape after the forces causing deformation have been removed.
Quick Check
Q3. A 6 N force to the right and a 4 N force to the right act on a trolley. What is the resultant force?
Show answer
Both forces are in the same direction, so resultant = 6 + 4 = 10 N to the right.
3. F = m a, Weight & Stopping Distance
Unbalanced Force & Acceleration
F = m a
Weight
W = m g
Stopping Distance & Falling Objects
- Know that the stopping distance of a vehicle is the sum of the thinking distance and the braking distance.
- Describe the factors affecting vehicle stopping distance, including speed, mass, road condition and reaction time.
- Describe the forces acting on falling objects and explain why falling objects reach a terminal velocity.
Quick Check
Q4. A 1 200 kg car has an unbalanced braking force of 4 800 N. What is its deceleration?
Show answer
a = F / m = 4 800 / 1 200 = 4 m/s² (deceleration).
4. Momentum & Newton’s Third Law
Momentum
p = m v
m1u1 + m2u2 = m1v1 + m2v2
- Use the idea of momentum to explain safety features (air bags, crumple zones, seat belts).
- Apply initial momentum = final momentum to calculate the mass, velocity or momentum of objects in collisions and explosions.
Force & Change in Momentum
F = (m v − m u) / t
- Demonstrate an understanding of Newton’s third law: for every action, there is an equal and opposite reaction.
Quick Check
Q5. A 1 000 kg car moving at 12 m/s crashes to a stop in 0.4 s. What average force acts on it?
Show answer
Change in momentum Δp = m v − m u = 1 000 × 0 − 1 000 × 12 = −12 000 kg m/s.
F = Δp / t = −12 000 / 0.4 = −30 000 N (opposite to the motion).
5. Moments & Turning Effects
Moment of a Force
- Know that the weight of a body acts through its centre of gravity.
Principle of Moments
- Use the principle of moments for a simple system of parallel forces acting in one plane (clockwise moments = anticlockwise moments).
- Understand how the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam.
Quick Check
Q6. A 300 N child sits 2.0 m from the pivot of a seesaw. What moment does the child produce?
Show answer
Moment = F × d = 300 × 2.0 = 600 N m.
What Next?
When you are comfortable with these sections, try:
- Past-paper questions on: motion graphs, F = m a, momentum and moments.
- Explaining each formula in your own words, including units and when to use it.
- Teaching a friend one sub-topic (for example, stopping distance or Hooke’s law).