Why double speed needs more than double space
Which part of stopping distance grows fastest with speed?
Start with a concrete question
Original synthetic motion model: an object moves at constant speed during a response delay, then slows with constant deceleration.
Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.
What do you expect?
Predict what happens to each distance when speed changes from 10 to 20 m/s.
Change one thing. Inspect what happens.
- Set speed to 10, delay to 1, deceleration to 5; record both distance components.
- Double speed without changing anything else. Compare ratios rather than only differences.
- Return speed to 10 and double the delay. Identify which term changes.
- Create a design claim about stopping space and name a condition under which this model would fail.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Compare the current results
Every bar starts at zero and uses the same numeric scale in this chart. The upper bound automatically adjusts to the largest current value (at least 1), so compare the numbers when changing inputs. Units and quantities come from each label; different kinds of quantity are not interchangeable. Displayed values are rounded to at most six significant digits.Follow the calculation
- During the 1 s delay: 15 m/s × 1 s = 15.00 m.
- Braking time: speed / deceleration = 15 / 5 = 3.00 s.
- Average speed while braking is 7.5 m/s; distance = 7.5 × 3.00 = 22.50 m.
- Total = 15.00 + 22.50 = 37.50 m.
Speed affects reaction distance linearly and braking distance quadratically. Doubling speed quadruples only the braking part.
Captures keep your inputs, prediction and observation. Results are recalculated from those inputs. Resetting inputs keeps your captures and challenge history.
Explain the mechanism
What the experiment explains
Original synthetic motion model: an object moves at constant speed during a response delay, then slows with constant deceleration.
During the delay the object has not begun slowing, so distance is speed times time.
With constant deceleration, speed decreases evenly to zero. Average braking speed is half the initial speed. At 15 m/s and 5 m/s², braking takes 3 s and covers 7.5 × 3 = 22.5 m.
The total at the defaults is 15 + 22.5 = 37.5 m. Both reaction and braking contribute, so the total does not generally quadruple when speed doubles.
Read the relationship
d = v × t + v² / (2a). d: total distance in metres; v: initial m/s; t: delay in seconds; a: positive deceleration magnitude in m/s².
Use the trace as evidence
The trace recomputes from your current controls. Change one control at a time, record what changed, and compare the result with your prediction. The bars use the same quantities as the numerical trace; a larger bar is not automatically a better outcome.
Boundaries of the model
- Flat one-dimensional motion, constant deceleration, no traction changes or air resistance.
- A teaching model, not a safe-distance recommendation for driving or equipment operation.
Continue with More current, less useful power? to investigate a different mechanism.
Try a fresh case
An unfamiliar object travels at 12 m/s, responds after 0.5 s, and decelerates at 4 m/s². What total distance does it cover before stopping?
This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.
Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Sketch a speed–time graph and find each distance as an area.
College investigation
Integrate a time-varying deceleration profile and compare it with the constant approximation.
Workplace investigation
Create a requirements brief showing which measurements and uncertainty bounds an engineering review would need.
Keep the evidence, choose your next step
Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.