Problem Solving and Advanced Kinematics/Dynamics

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Problem Solving and Advanced Kinematics/Dynamics

TL;DR

You'll learn to chain together motion graphs, F = ma, and stopping-distance equations to solve multi-step physics problems. You'll see why braking distance and lift both grow with speed squared, not speed itself. By the end, you can work through a full "speeding car" or "aircraft in flight" question without getting stuck on which formula to use first.

1. The Mental Model

Every problem here is really the same three-step chain: find the acceleration, find the force behind it, then find the distance or time it produces. Motion graphs tell you what happened; forces tell you why. The skill isn't memorising more equations — it's spotting which two things you know so you can pick the one equation that connects them. If you know any three of u, v, a, s, t, you can always find the other two.

2. The Core Material

2.1 Reading motion graphs and finding acceleration

On a distance-time graph, the gradient is speed. A curved line means changing speed (acceleration); a straight diagonal line means constant speed; a flat horizontal line means stationary.

On a velocity-time graph, the gradient is acceleration, and the area under the line is distance travelled. This is the graph you want for acceleration questions.

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

where $u$ = initial velocity, $v$ = final velocity, $t$ = time taken. Example: a car speeds up from 10 m/s to 26 m/s in 8 seconds:

$$a = \frac{26 - 10}{8} = 2.0 \text{ m/s}^2$$

When you don't know the time but you know the distance, switch to:

$$v^2 = u^2 + 2as$$

This equation is the quiet hero of this whole topic — it's what links speed directly to distance without needing time, and it's the one that explains why stopping distances explode at high speed (more below).

2.2 Forces, balanced forces, and Newton's second law

Newton's second law ties force, mass, and acceleration together:

$$F = ma$$

If a 1200 kg car decelerates at 5 m/s², the braking force needed is:

$$F = 1200 \times 5 = 6000 \text{ N}$$

Balanced forces mean the resultant (net) force is zero. This does not mean nothing is moving — it means velocity isn't changing. A plane cruising at a constant 250 m/s in a straight line has balanced forces acting on it, exactly like a parked car. Unbalanced forces are what cause acceleration (speeding up, slowing down, or changing direction).

Flight is the clearest example of this. Four forces act on an aircraft: lift (up), weight (down), thrust (forward), drag (backward). Lift happens because the wing's shape forces air to move faster over the top than underneath, lowering pressure above the wing (and by Newton's third law, the wing also pushes air downward, so the air pushes the wing up). For level, constant-speed flight:

$$\text{Lift} = \text{Weight} \quad \text{and} \quad \text{Thrust} = \text{Drag}$$

Both force pairs are balanced — that's why the plane doesn't climb, dive, speed up, or slow down. To climb, the pilot makes lift exceed weight; that's an unbalanced force causing upward acceleration.

Here's the elegant link to the rest of this topic: lift increases with the square of airspeed ($L \propto v^2$), for the same reason braking distance increases with the square of speed. Doubling your speed doesn't double the effect — it quadruples it.

Simple machines (levers, pulleys, gears) don't create force from nothing — they trade force for distance. For a lever:

$$\text{effort} \times \text{effort distance} = \text{load} \times \text{load distance}$$

A gearbox in a car works the same way: low gear gives you more turning force (useful for acceleration) at the cost of speed; high gear gives more speed at the cost of force. Mechanical advantage = load force ÷ effort force — a bigger number means less effort needed, but you push through a longer distance to get it.

2.3 Stopping distances and speeding cars

Stopping distance has two separate parts, and mixing them up is the single most common exam error:

$$\text{stopping distance} = \text{thinking distance} + \text{braking distance}$$

Thinking distance is how far the car travels during your reaction time, before you even touch the brake:

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

This is proportional to speed — double the speed, double the thinking distance.

Braking distance comes from $v^2 = u^2 + 2as$, setting final velocity to zero (the car stops):

$$d_{brake} = \frac{u^2}{2a}$$

This is proportional to speed squared — double the speed, and braking distance quadruples. This is why speed limits matter so much more than intuition suggests: a car going twice as fast doesn't need twice the stopping distance, it needs roughly four times as much.

graph TD
A["List the known quantities"] --> B{"Have u, v, t?"}
B -- Yes --> C["a = (v - u) / t"]
B -- No --> D{"Have u, v, s?"}
D -- Yes --> E["v^2 = u^2 + 2as"]
D -- No --> F{"Have mass and acceleration?"}
F -- Yes --> G["F = ma"]
F -- No --> H{"Asked for stopping distance?"}
H -- Yes --> I["thinking distance + braking distance"]

3. Worked Example

A car travels at 28 m/s (about 100 km/h). The driver's reaction time is 0.7 s, and once braking, the car decelerates at 6 m/s². Car mass is 1400 kg. Find: (a) thinking distance, (b) braking distance, (c) total stopping distance, (d) the braking force.

(a) Thinking distance:
$$d_{think} = v \times t = 28 \times 0.7 = 19.6 \text{ m}$$

(b) Braking distance — use $v^2 = u^2 + 2as$, with final velocity $v = 0$:

$$0 = 28^2 - 2(6)s \implies s = \frac{784}{12} = 65.3 \text{ m}$$

(c) Total stopping distance:
$$d_{total} = 19.6 + 65.3 = 84.9 \text{ m}$$

(d) Braking force:
$$F = ma = 1400 \times 6 = 8400 \text{ N}$$

Notice the shape of this answer: thinking distance (19.6 m) is small compared to braking distance (65.3 m) at this speed, because braking distance scales with $v^2$. If this driver had been going at 14 m/s instead (half the speed), thinking distance would halve to 9.8 m, but braking distance would drop to a quarter: $14^2 / 12 = 16.3$ m. That's the real-world reason speed cameras exist on fast roads — the braking distance penalty for speeding is brutal, not gentle.

4. Key Takeaways

4.1 Most Important Concepts

• Pick the equation from what you know, not what you remember first — if you have $u$, $v$, $t$ use $a = (v-u)/t$; if you have $u$, $v$, $s$ use $v^2 = u^2 + 2as$.
• Balanced forces mean constant velocity, not zero velocity — a cruising plane and a parked car both have zero resultant force.
• Braking distance scales with speed squared, thinking distance scales with speed — they are not the same relationship.
• Lift grows with $v^2$ too — wings and brakes share the same square-law physics, just in opposite directions (one force you want, one you want to remove).
• $F = ma$ connects force and motion directly — no acceleration means no resultant force, whatever the speed.
• Simple machines trade force for distance — mechanical advantage is never "free"; you pay with a longer effort distance.
• Gradient and area have different meanings on different graphs — gradient on a v-t graph is acceleration, but on a d-t graph it's velocity.

4.2 Common Misconceptions

• "No motion means no force." Wrong — balanced forces can hold a stationary object OR keep a moving object at constant velocity.
• "Doubling speed doubles stopping distance." Wrong — only thinking distance doubles; braking distance quadruples, so total stopping distance roughly triples or more.
• "A steeper distance-time graph means acceleration." Wrong — a straight line, however steep, means constant speed. Acceleration needs a curve.
• "Lift is caused only by air hitting the underside of the wing." Incomplete — the main contribution is the pressure difference from airflow shape over the top and bottom surfaces, plus the downward deflection of air (Newton's third law).

4.3 Compare & Contrast

5. Now Try It

A cyclist travels at 12 m/s. Their reaction time is 0.5 s, and once braking they decelerate at 3 m/s². Combined mass of cyclist and bike is 80 kg.

Calculate: (a) the thinking distance, (b) the braking distance, (c) the total stopping distance, (d) the braking force needed. Then answer in one sentence: if the cyclist had been going twice as fast, would the stopping distance also just double? Explain why or why not using the equations above.

Success looks like: you get $d_{think} = 6$ m, $d_{brake} = 24$ m, $d_{total} = 30$ m, $F = 240$ N, and you correctly explain that doubling speed would more than double stopping distance because braking distance depends on $v^2$.

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