Evaluating a transport safety claim with mechanics

In this lesson you evaluate a transport safety claim not by intuition but by translating it into a mechanical model. First choose the laws: if the impact force matters, use momentum conservation and impulse $\vec{F}\Delta t=\Delta\vec{p}$; if braking geometry matters, use kinematics $v^2=v_0^2+2as$; if a curve is involved, use centripetal acceleration $a_c=v^2/r$. Then connect the words of the claim to quantities: speed, mass, reaction time, coefficient of friction, radius of the curve, banking angle. The calculation yields a number that can be compared with the threshold in the claim, and you conclude whether it is supported, contradicted, or needs refinement. Always state the assumptions — for example, constant braking, no air resistance, an ideal level road — because they set the limits of the conclusion. Finally, check units and limiting cases: increasing speed usually increases stopping distance as $v^2$, not linearly.

Claim: “At $50$ km/h a car comes to a complete stop in $13$ m.” The driver's reaction time is $t_r=0.8$ s, and braking is constant with $a=6.5$ m/s$^2$. Calculate the total stopping distance and check the claim.

A safety article claims: “In a head-on collision where two identical cars each travel at $30$ km/h, each driver experiences the same impact as hitting a rigid wall at $60$ km/h.” Use momentum conservation to evaluate the claim for a perfectly inelastic head-on collision.

A road-safety claim says: “A curve of radius $80$ m banked at $10^\circ$ can be safely taken at $100$ km/h even in rain.” For wet asphalt use $\mu=0.40$; for dry asphalt use $\mu=0.70$. Compute the maximum safe speed in both cases and evaluate the claim. Ignore air resistance and treat the car as a point mass.

Use a browser-based simulation such as PhET Collision Lab or an online stopping-distance model. Test the claim: “Doubling speed doubles stopping distance.” Set a fixed mass and friction, run the model at $30$, $60$, and $90$ km/h, and record the total stopping distance including a fixed reaction time if the simulation allows it. Plot $s$ against $v$ and compute $s/v^2$; write a short evaluation of the claim, naming at least two assumptions. Do not attempt measurements on a real road; all testing stays in the simulation.

To make the comparison between the claimed stopping distance and the computed total stopping distance visible, separating reaction distance from braking distance.

A scaled stacked bar or number line for $v=50$ km/h: reaction segment $s_r\approx11.1$ m, braking segment $s_b\approx14.8$ m, total $25.9$ m; a vertical marker at the claimed $13$ m shows that the claim gives too short a distance.

A car of mass $1500\ \mathrm{kg}$ hits a crash-test barrier at $50\ \mathrm{km/h}=13.9\ \mathrm{m/s}$. Its front crumple zone deforms by $0.80\ \mathrm{m}$ before the passenger cell stops. The average stopping force is about $F = m v^2/(2s) = 1500\cdot 13.9^2/(2\cdot0.80) \approx 1.8\times10^5\ \mathrm{N}$. The safety claim is: 'this crumple zone keeps the average force below $2.0\times10^5\ \mathrm{N}$'.

The scenario tests the claim with the impulse-momentum theorem: $F\Delta t = \Delta p$, and for uniform deceleration $\Delta t = 2s/v$, so $F = m v^2/(2s)$. Comparing the calculated force with the claim's threshold is the evaluation.

A compact car of mass $1200\ \mathrm{kg}$ travels at $15\ \mathrm{m/s}$ and rear-ends a stationary SUV of mass $1800\ \mathrm{kg}$; the bumpers lock. Before impact the system momentum is $p = 1200\cdot15 = 1.8\times10^4\ \mathrm{kg\cdot m/s}$. After impact the common speed is $v = p/(1200+1800) = 1.8\times10^4/3000 = 6.0\ \mathrm{m/s}$. The claim is: 'the SUV is safer because it is heavier'.

The scenario uses conservation of momentum in a perfectly inelastic collision. The compact car's velocity change is $\Delta v = 15-6.0 = 9.0\ \mathrm{m/s}$, while the SUV's is $\Delta v = 6.0-0 = 6.0\ \mathrm{m/s}$; this shows that greater mass alone does not remove the larger velocity change of the other participant.

In a collision a pedestrian is thrown horizontally from a height $h = 1.2\ \mathrm{m}$ and lands at a distance $d = 6.0\ \mathrm{m}$ from the impact point. The fall time is $t = \sqrt{2h/g} = \sqrt{2\cdot1.2/9.81} \approx 0.495\ \mathrm{s}$. The horizontal throw speed is $v_x = d/t = 6.0/0.495 \approx 12.1\ \mathrm{m/s} = 43.6\ \mathrm{km/h}$. The driver claims: 'I was travelling at less than $30\ \mathrm{km/h}$'.

The scenario applies projectile motion: vertical fall and horizontal flight are independent, so the measured range gives an estimate of the initial speed. This estimate directly tests the numerical speed claim.