Motion & Forces: kinematics and Newtonian dynamics
Key idea: Connect motion representations to the force and momentum models that explain changes in motion.
Before you start: Kinematics GraphsNewton’s Laws, Vectors and Problem Solving
By the end, you can
- Define and distinguish position, distance, displacement, speed, velocity and acceleration.
- Interpret position–time and velocity–time graphs using gradients and areas, including non-uniform acceleration.
- Derive and apply the equations of uniformly accelerated motion, including vertical motion without air resistance.
- Use inertia, momentum and Newton’s laws to connect resultant force to change in momentum.
Starting-point self-check
1. Check your starting point
Attempt both alignment groups without notes and mark the first representation or law you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Kinematics outcomes 3(a)–(e)
Question 1
A runner completes one 400 m lap in 50 s and finishes at the starting point. State the distance, displacement, average speed and average velocity.
Check the model response
Distance = 400 m; displacement = 0; average speed = 400/50 = 8.0 m s⁻¹; average velocity = displacement/time = 0.
Question 2
On a position–time graph, what does the gradient represent? On a velocity–time graph, what do its gradient and signed area represent?
Check the model response
Position–time gradient is velocity. Velocity–time gradient is acceleration, and signed area between the graph and time axis is displacement.
Question 3
A car starts from rest and accelerates uniformly at 3.0 m s⁻² for 4.0 s. Find its final velocity and displacement.
Check the model response
v = u + at = 0 + 3.0(4.0) = 12 m s⁻¹. s = ut + ½at² = 0 + 0.5(3.0)(4.0²) = 24 m.
repair
2. Repair the five common breaks
Use only the repair matching an error, then redraw the graph or force model.
Kinematics outcomes 3(a)–(e)
Check this idea
Misconception: Distance and displacement, or speed and velocity, are interchangeable.
Repair: Distance and speed are scalar path measures. Displacement and velocity include a declared direction and may be zero even after motion.
Check this idea
Misconception: The height of any motion graph gives the object’s acceleration.
Repair: Read the axis labels first. Acceleration is the velocity–time gradient; velocity is the position–time gradient. Velocity–time area gives displacement.
Check this idea
Misconception: The uniformly accelerated equations apply whenever an object is moving.
Repair: They require constant acceleration over the chosen interval. For vertical motion without air resistance, choose a positive direction and use a constant acceleration of magnitude g with the correct sign.
worked example
3. Follow four worked models
Follow how each solution selects a representation, sign convention and physical law.
Kinematics outcomes 3(a)–(e)
Model 1
A velocity rises linearly from 4.0 to 16 m s⁻¹ in 6.0 s, then falls linearly to 10 m s⁻¹ over 3.0 s. Find the displacement and acceleration in each interval.
Check the model response
Displacement is velocity–time area: first interval = ½(4.0 + 16)(6.0) = 60 m; second = ½(16 + 10)(3.0) = 39 m; total = 99 m. Accelerations are (16 − 4)/6 = 2.0 m s⁻² and (10 − 16)/3 = −2.0 m s⁻².
Model 2
Derive v² = u² + 2as from constant-acceleration definitions.
Check the model response
Use a = (v − u)/t, so t = (v − u)/a. Constant acceleration gives average velocity (u + v)/2, hence s = (u + v)t/2. Substitution gives s = (u + v)(v − u)/(2a) = (v² − u²)/(2a), so v² = u² + 2as.
guided practice
4. Guided practice
Use each hint only to set up axes, graph quantities or the chosen system.
Kinematics outcomes 3(a)–(e)
Question 1
A cyclist’s position changes from −5 m to +19 m in 8.0 s along a straight axis. Find displacement and average velocity. Can distance be found from this information alone?
Hint: Use final position minus initial position.
Check the model response
Displacement = 19 − (−5) = +24 m; average velocity = +24/8.0 = +3.0 m s⁻¹. Distance cannot be found without knowing the path or whether direction changed.
Question 2
A stone is thrown vertically upward at 14 m s⁻¹. Neglect air resistance and use g = 9.8 m s⁻². Find the time to maximum height and that height above release.
Hint: Take upward positive, so a = −9.8 m s⁻² and v = 0 at the top.
Check the model response
0 = 14 − 9.8t gives t = 1.43 s. Using v² = u² + 2as: 0 = 14² − 2(9.8)s, so s = 10.0 m.
independent practice
5. Independent practice
Solve without repair notes and show axes, graph interpretation and force-system boundaries.
Kinematics outcomes 3(a)–(e)
Question 1
Sketch in words a velocity–time graph for motion that starts at 2 m s⁻¹, accelerates non-uniformly to 8 m s⁻¹, then continues at 8 m s⁻¹. Explain how acceleration and displacement are obtained.
Check the model response
The first section is a curve rising from 2 to 8 m s⁻¹, with changing gradient; the second is horizontal at 8 m s⁻¹. Instantaneous acceleration is the tangent gradient and total displacement is the signed area under both sections.
Question 2
A train moving at 25 m s⁻¹ brakes uniformly to rest over 200 m. Find its acceleration and stopping time.
Check the model response
From v² = u² + 2as: 0 = 25² + 2a(200), so a = −1.5625 m s⁻². From v = u + at: t = (0 − 25)/(−1.5625) = 16 s.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Kinematics outcomes 3(a)–(e)
Question 1
A vehicle’s velocity changes uniformly from −4.0 to +8.0 m s⁻¹ in 6.0 s. Find its acceleration and signed displacement, and explain what the sign change means.
Check the model response
Acceleration = [8 − (−4)]/6 = 2.0 m s⁻². Signed displacement = average velocity × time = [(-4 + 8)/2]6 = 12 m. The velocity crosses zero, so the vehicle reverses direction.
Question 2
Starting from rest, an object falls 45 m with negligible air resistance. Use g = 9.8 m s⁻² to find its speed and fall time.
Check the model response
Taking downward positive, v² = 0 + 2(9.8)(45), so v = 29.7 m s⁻¹. Then v = gt gives t = 29.7/9.8 = 3.03 s.
Re-test practice
7. Delayed re-test practice
Return after three days and solve these fresh prompts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Kinematics outcomes 3(a)–(e)
Question 1
A particle travels 30 m east then 18 m west in 12 s. Find distance, displacement, average speed and average velocity.
Check the model response
Distance = 48 m; displacement = 12 m east; average speed = 48/12 = 4.0 m s⁻¹; average velocity = 12/12 = 1.0 m s⁻¹ east.
Question 2
A car accelerates uniformly from 6.0 to 18 m s⁻¹ over 72 m. Find its acceleration and elapsed time.
Check the model response
From v² = u² + 2as: 18² − 6² = 2a(72), so a = 2.0 m s⁻². From v = u + at: t = (18 − 6)/2.0 = 6.0 s.