Problem Solving and Advanced Kinematics/Dynamics

<<>>

Problem Solving and Advanced Kinematics/Dynamics

TL;DR

You'll learn to read motion from distance-time graphs, calculate acceleration and resultant force using F = ma, and work out stopping distances from thinking and braking components. You'll also see how balanced and unbalanced forces explain flight and steady speed. By the end you can solve multi-step force and motion problems with correct units.

1. The Mental Model

Every motion problem is really just three questions: how fast, how much change, and what's pushing or pulling. Speed tells you distance over time; acceleration tells you how speed changes over time; force tells you why acceleration happens at all. If you can identify what's changing and what's causing the change, you can solve almost any motion problem.

2. The Core Material

Distance-Time Graphs and Acceleration

A distance-time graph plots distance (y-axis) against time (x-axis). The gradient of the line equals speed:

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{\Delta d}{\Delta t}$$

A straight line means constant speed. A curved line means the speed is changing — the object is accelerating. A steeper curve means faster speed at that moment; a flat horizontal section means the object has stopped.

Acceleration itself is calculated differently — from a speed-time graph, or directly using the equation:

$$a = \frac{v - u}{t}$$

where $v$ is final velocity, $u$ is initial velocity, $t$ is time, and $a$ is acceleration in m/s². On a speed-time graph, the gradient gives acceleration, and the area under the line gives distance travelled. This is a trap students fall into constantly: distance-time graphs give you speed from gradient; speed-time graphs give you acceleration from gradient AND distance from area. Mixing these up is the single most common exam error in this topic.

Forces and Newton's Second Law

Force causes acceleration. The relationship is:

$$F = ma$$

where $F$ is resultant force in newtons (N), $m$ is mass in kilograms (kg), and $a$ is acceleration in m/s². This only works with the resultant (net) force — if multiple forces act on an object, you must add or subtract them first depending on direction.

Balanced forces mean the resultant force is zero, so acceleration is zero — the object either stays still or moves at constant velocity. Unbalanced forces mean there's a resultant force, so the object accelerates (speeds up, slows down, or changes direction).

This is exactly how flight works. A plane flying level at constant speed has four forces balanced in pairs: lift balances weight (vertical), and thrust balances drag (horizontal). Lift is generated because the wing's curved shape forces air moving over the top to travel faster than air underneath. Faster-moving air has lower pressure (this follows from Bernoulli's principle), so there's higher pressure below the wing than above it — that pressure difference pushes the wing upward. When thrust exceeds drag, the plane accelerates forward; when lift exceeds weight, it climbs. Change any one force and the balance breaks, and the plane accelerates in that direction.

flowchart TD
    A["Forces on a level, constant-speed aircraft"] --> B["Lift (up) = Weight (down)"]
    A --> C["Thrust (forward) = Drag (backward)"]
    B --> D["Resultant vertical force = 0"]
    C --> E["Resultant horizontal force = 0"]
    D --> F["Zero acceleration = constant velocity"]
    E --> F

Stopping Distance and Speeding Cars

Stopping distance is the total distance a car travels from the moment a driver spots a hazard to the moment the car fully stops. It has two separate parts, and you must calculate them separately then add them:

$$\text{Stopping distance} = \text{Thinking distance} + \text{Braking distance}$$

Thinking distance is how far the car travels during the driver's reaction time, before the brakes are applied. It depends on speed and reaction time:

$$\text{thinking distance} = \text{speed} \times \text{reaction time}$$

Reaction time is typically around 0.2–0.9 seconds, but gets longer with tiredness, alcohol, or distraction (like phone use) — this increases thinking distance even though speed hasn't changed.

Braking distance is how far the car travels once the brakes are applied until it stops. It depends on speed, road conditions (wet, icy, gravel), tyre condition, and brake condition. Crucially, braking distance is not proportional to speed — it's proportional to speed squared. This comes from the physics of kinetic energy: the brakes must dissipate the car's kinetic energy ($E_k = \frac{1}{2}mv^2$) as heat through friction. Double the speed, and you have four times the kinetic energy to get rid of, so braking distance roughly quadruples, not doubles. This is exactly why speeding is so dangerous — a small increase in speed causes a disproportionately large increase in braking distance.

Simple Machines and Force Calculations

Simple machines (levers, pulleys, inclined planes, gears) let you apply a smaller force over a larger distance to achieve the same amount of work. They don't create energy — they trade force for distance. The key relationship uses mechanical advantage:

$$\text{Mechanical advantage} = \frac{\text{load force}}{\text{effort force}}$$

For a lever, this relates to the distances from the pivot (fulcrum) using the principle of moments:

$$F_{\text{effort}} \times d_{\text{effort}} = F_{\text{load}} \times d_{\text{load}}$$

A longer effort arm means less effort force is needed to lift the same load — this is why a long spanner makes loosening a bolt easier than a short one.

3. Worked Example

Problem: A car is travelling at 20 m/s when the driver spots a hazard. The driver's reaction time is 0.6 s. Once the brakes are applied, the car decelerates uniformly and comes to rest in 4 seconds. The car's mass is 1200 kg.

Part A — Thinking distance:
$$\text{thinking distance} = \text{speed} \times \text{reaction time} = 20 \times 0.6 = 12 \text{ m}$$

Part B — Deceleration during braking:
Using $a = \dfrac{v-u}{t}$, with final velocity $v = 0$, initial velocity $u = 20$ m/s, and $t = 4$ s:

$$a = \frac{0 - 20}{4} = -5 \text{ m/s}^2$$

The negative sign shows deceleration — the car is slowing down.

Part C — Braking distance:
We don't have a direct distance formula here yet, so we use the area under a speed-time graph (a triangle, since velocity goes from 20 m/s to 0 m/s uniformly over 4 s):

$$\text{braking distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 20 = 40 \text{ m}$$

Part D — Total stopping distance:
$$\text{stopping distance} = 12 + 40 = 52 \text{ m}$$

Part E — Braking force:
Using $F = ma$:
$$F = 1200 \times (-5) = -6000 \text{ N}$$

The force is 6000 N acting backward (opposing motion) — this is the resultant braking force the brakes and friction must supply to achieve that deceleration. Notice how every part of this problem chained together: speed and reaction time gave thinking distance; the speed-time relationship gave deceleration; deceleration and time gave braking distance via the graph; and deceleration with mass gave force. This chaining is exactly what multi-step exam questions test.

4. Key Takeaways

4.1 Most Important Concepts

• Gradient of distance-time graph = speed; gradient of speed-time graph = acceleration — these are not interchangeable.
• Area under a speed-time graph = distance travelled — essential when you don't have a clean formula.
• F = ma only works with resultant force — always combine opposing forces first before using the equation.
• Balanced forces = zero acceleration, which explains constant velocity flight and objects at rest.
• Stopping distance = thinking distance + braking distance, and these depend on different variables (reaction time vs. road/speed conditions).
• Braking distance scales with speed squared, not speed, because kinetic energy depends on $v^2$.
• Simple machines trade force for distance, not create energy — mechanical advantage shows that trade-off.

4.2 Common Misconceptions

• "The gradient of any motion graph gives speed." False — it depends on the axes. Distance-time gradient gives speed; speed-time gradient gives acceleration.
• "Doubling speed doubles braking distance." False — braking distance is proportional to speed squared, so doubling speed roughly quadruples it.
• "Balanced forces mean no forces are acting." False — balanced forces means the forces cancel out (sum to zero); they're still present and often large (e.g., lift and weight on a cruising plane).
• "A machine makes a job require less work." False — simple machines reduce the force needed but increase the distance over which it's applied, so the total work (ignoring friction) stays the same.

4.3 Compare & Contrast

5. Now Try It

Exam-style question: A cyclist travelling at 8 m/s brakes and comes to a complete stop in 2 seconds, decelerating uniformly. The cyclist and bike together have a mass of 75 kg.
(a) Calculate the deceleration. (2 marks)
(b) Calculate the braking force. (2 marks)
(c) Using a sketch or calculation of area, find the braking distance. (3 marks)
(d) Explain why this braking distance would more than double if the cyclist had been travelling at 16 m/s instead. (2 marks)

A top-band answer shows full working with units at each step (m/s², N, m), correctly uses $a = \frac{v-u}{t}$ and $F=ma$, calculates area as a triangle, and explains part (d) using kinetic energy's $v^2$ relationship, not just "speed is higher."

Success looks like: you get numerical answers for (a)-(c) with correct units, and a written explanation for (d) that mentions kinetic energy or $v^2$ scaling.