Problem Solving and Advanced Kinematics/Dynamics

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Problem Solving and Advanced Kinematics/Dynamics

TL;DR

You'll learn how to pull numbers off distance-time and velocity-time graphs, turn them into speed, acceleration and force using the right equations, and chain these together in multi-step problems. You'll also see how Newton's second law explains stopping distances, speeding cars, and why a plane's wings keep it in the air. By the end, you can plan a solution path before you even touch a calculator.

1. The Mental Model

Every motion problem is really the same three questions asked in different orders: how fast, how is that speed changing, and what force is causing the change. A graph shows you how fast and how fast that's changing (its gradient); Newton's second law connects the changing bit to force. Once you see motion problems as a chain — graph → acceleration → force, or force → acceleration → distance — the "advanced" problems are just longer chains of the same links.

2. The Core Material

2.1 Reading Motion Graphs Properly

A distance-time graph tells you position over time. The gradient (slope) of that graph is speed:

$$v = \frac{\Delta d}{\Delta t}$$

A straight diagonal line means constant speed. A curved line means the speed is changing — the object is accelerating. You can't read acceleration directly off a distance-time graph; you have to look at how the gradient itself is changing, which is awkward. That's why a velocity-time graph is more useful for acceleration problems.

On a velocity-time graph:
- The gradient gives acceleration: $a = \dfrac{\Delta v}{\Delta t}$
- The area under the line gives distance travelled.

Example: a cyclist's velocity rises from 0 to 8 m/s in 4 seconds, then stays constant at 8 m/s for 6 seconds.

Acceleration during the first phase:
$$a = \frac{8 - 0}{4} = 2 \text{ m/s}^2$$

Distance in that phase (area of a triangle):
$$d = \frac{1}{2} \times 4 \times 8 = 16 \text{ m}$$

Distance in the constant-speed phase (area of a rectangle):
$$d = 8 \times 6 = 48 \text{ m}$$

Total distance = 64 m. This "split the shape into triangles and rectangles" trick works for almost any velocity-time graph you'll meet.

2.2 Newton's Second Law and Combining Forces

Once you know acceleration, Newton's second law connects it to force:

$$F = ma$$

where $F$ is the resultant (net) force in newtons, $m$ is mass in kg, and $a$ is acceleration in m/s².

The key idea examiners love to test: $F$ must be the resultant of all forces acting, not just one force you're told about. If forces are balanced (resultant = 0), acceleration is zero — the object moves at constant velocity or stays still. If forces are unbalanced, there's a resultant force and the object accelerates in the direction of that resultant.

This is exactly how a plane flies. Four forces act on it: lift (up, from air moving faster over the curved top of the wing, lowering pressure there versus underneath — a pressure difference that pushes the wing upward), weight (down, from gravity), thrust (forward, from engines), and drag (backward, from air resistance). For level, constant-speed flight, lift = weight and thrust = drag — forces are balanced on both axes, so there's no acceleration. To climb, the pilot increases thrust or wing angle so lift exceeds weight, creating an unbalanced upward force.

Worked force example: a car of mass 1200 kg has an engine driving force of 3000 N and a resistive (drag + friction) force of 600 N.

Resultant force:
$$F_{net} = 3000 - 600 = 2400 \text{ N}$$

Acceleration:
$$a = \frac{F}{m} = \frac{2400}{1200} = 2 \text{ m/s}^2$$

flowchart TD
    A["What does the question give you?"] --> B{"Graph given?"}
    B -->|"Distance-time"| C["Find gradient = speed"]
    B -->|"Velocity-time"| D["Find gradient = acceleration<br/>Find area = distance"]
    B -->|"No graph, just numbers"| E{"Looking for acceleration or force?"}
    E -->|"Have v, u, t"| F["a = (v - u) / t"]
    E -->|"Have mass and force"| G["a = F / m"]
    F --> H["Use a to find F = m x a"]
    G --> I["Use a with SUVAT to find distance"]
    C --> J["Answer"]
    D --> J
    H --> J
    I --> J

2.3 Stopping Distances — the Classic Multi-Step Problem

This is where graphs, acceleration, and force all meet in one real-world problem. Stopping distance has two separate parts:

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

Thinking distance is how far the car travels during the driver's reaction time, before the brakes are even applied. Since the car moves at constant speed during this time:

$$d_{think} = v \times t_{reaction}$$

Braking distance is how far the car travels once the brakes are on, decelerating to a stop. This depends on speed, friction between tyres and road, and the car's mass — and it does not scale linearly with speed, because kinetic energy (and therefore the energy the brakes must remove) scales with $v^2$.

Worked numbers: a car travels at 20 m/s. The driver's reaction time is 0.7 s. Braking deceleration is 5 m/s².

Thinking distance:
$$d_{think} = 20 \times 0.7 = 14 \text{ m}$$

Braking distance, using $v^2 = u^2 + 2as$ rearranged (final velocity $v=0$):
$$0 = 20^2 + 2(-5)s \implies s = \frac{400}{10} = 40 \text{ m}$$

Total stopping distance = 14 + 40 = 54 m.

Now double the speed to 40 m/s with the same reaction time and deceleration:

Thinking distance: $40 \times 0.7 = 28$ m (doubled, as expected — it's linear in $v$).

Braking distance: $s = \dfrac{40^2}{2 \times 5} = \dfrac{1600}{10} = 160$ m (quadrupled, not doubled, because of the $v^2$ term).

Total stopping distance = 188 m — more than three times further, for only double the speed. This is the reason speeding is so dangerous: small increases in speed cause disproportionately large increases in the distance needed to stop, and in the energy that must be absorbed in a crash. Wet roads, worn tyres, or poor reaction time (tiredness, phone use) all make the braking or thinking part worse individually, but it's the speed-squared relationship that does the real damage.

3. Worked Example

Question: A delivery van of mass 1500 kg is travelling at 15 m/s along a flat road when the driver sees a hazard. Reaction time is 0.6 s. Once the brakes are applied, a constant braking force of 4500 N acts on the van (ignore air resistance during braking). Find: (a) the thinking distance, (b) the deceleration during braking, (c) the braking distance, (d) the total stopping distance.

Step 1 — Thinking distance. The van moves at constant 15 m/s during the reaction time:
$$d_{think} = v \times t = 15 \times 0.6 = 9 \text{ m}$$

Step 2 — Deceleration from the braking force. Use Newton's second law, rearranged:
$$a = \frac{F}{m} = \frac{4500}{1500} = 3 \text{ m/s}^2$$
This is a deceleration, so in SUVAT work we treat it as $a = -3$ m/s².

Step 3 — Braking distance. The van decelerates from 15 m/s to 0 m/s. Use $v^2 = u^2 + 2as$:
$$0 = 15^2 + 2(-3)s$$
$$0 = 225 - 6s$$
$$s = \frac{225}{6} = 37.5 \text{ m}$$

Step 4 — Total stopping distance.
$$d_{total} = d_{think} + d_{brake} = 9 + 37.5 = 46.5 \text{ m}$$

Notice the chain of reasoning: speed → thinking distance (direct multiplication), then force and mass → deceleration (Newton's second law), then deceleration and speed → braking distance (SUVAT). Three different tools, one connected problem. If the question had instead given you a velocity-time graph of the van's braking phase, the braking distance would just be the area of the triangle under the line — same answer, different route.

4. Key Takeaways

4.1 Most Important Concepts

• Gradient of a distance-time graph = speed; gradient of a velocity-time graph = acceleration.
• Area under a velocity-time graph = distance travelled — split it into triangles/rectangles.
• $F = ma$ uses the resultant force, not a single individual force acting on the object.
• Balanced forces mean zero acceleration (constant velocity or rest); unbalanced forces always cause acceleration in the direction of the resultant.
• Stopping distance = thinking distance + braking distance, and they depend on different things (reaction time vs. braking force/friction).
• Braking distance grows with $v^2$, not $v$ — doubling speed roughly quadruples braking distance.
• Flight is a direct application of balanced/unbalanced forces: steady flight needs lift = weight and thrust = drag.

4.2 Common Misconceptions

• "A steeper distance-time graph line always means acceleration." No — a straight, steep line still means constant (high) speed. Only a curving line shows acceleration.
• "F = ma means I just plug in any force given in the question." You must find the resultant of all forces first; using one force alone gives the wrong acceleration.
• "Doubling your speed doubles your stopping distance." False — thinking distance doubles, but braking distance roughly quadruples, so total stopping distance more than doubles.
• "Lift exists only because a wing pushes air downward." Downward air deflection does contribute, but the main GCSE-level explanation is the pressure difference from air moving at different speeds over the curved top versus the flatter bottom of the wing.

4.3 Compare & Contrast

5. Now Try It

Exam-style question: A car of mass 900 kg is travelling at 18 m/s. The driver's reaction time is 0.5 s. When the brakes are applied, a constant resultant braking force of 3600 N acts on the car. Calculate (a) the thinking distance, (b) the braking deceleration, (c) the braking distance, (d) the total stopping distance, and (e) explain, using the idea of kinetic energy or the $v^2$ relationship, why this total would more than double if the car's speed were 36 m/s instead.

A top-band answer shows: correct working for thinking distance (9 m), correct use of $a = F/m$ (4 m/s²), correct SUVAT rearrangement for braking distance (40.5 m), a correctly summed total (49.5 m), and a clear written explanation linking the $v^2$ term in $v^2 = u^2+2as$ (or kinetic energy $\tfrac{1}{2}mv^2$) to the disproportionate increase in braking distance at higher speed.

Success looks like: all four numerical answers correct with units, plus a one-to-two sentence explanation correctly citing the squared relationship between speed and braking distance.