Question 6 of 6: Automatic Object-Detection & Collision-Avoidance — Reaction-Time and Safe-Speed Analysis

Question 6: Automatic Object-Detection & Collision-Avoidance — Reaction-Time and Safe-Speed Analysis (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Detection range (normal conditions), d_detect = 50 m; reference speed, v_ref = 60 km/h; braking-only stopping distance at v_ref (dry road), d_brake,dry = 30 m; braking-only stopping distance at v_ref (slippery road), d_brake,wet = 50 m; foggy-day detection range, d_fog = 30 m.

Find. (1) Longest allowable reaction time t_react at 60 km/h with the normal 50 m detection range. (2) Maximum safe speed in fog (30 m detection, same t_react). (3) Maximum safe speed on a slippery road (50 m detection restored, worse braking).

Approach. The total distance needed to stop safely is reaction distance (v · t_react, travelled at constant speed before the brake engages) plus pure braking distance; the given 30 m figure is the braking-only distance at 60 km/h (brake "fully engaged," i.e. reaction time already excluded), so it is first used to back out the vehicle's braking deceleration, which then carries forward unchanged into parts (2) and (3) except where the road condition itself is stated to change it.

1. Back out the braking deceleration from the given dry-road stopping distance. Using v² = 2ad with v_ref = 60 km/h = 16.67 m/s and d_brake,dry = 30 m: $$a = \frac{v_{\text{ref}}^2}{2\,d_{\text{brake,dry}}} = \frac{16.67^2}{2\times30} = 4.63\ \text{m/s}^2$$
2. Part (1): solve for the longest safe reaction time at 60 km/h. The total available stopping room is the 50 m detection range; the reaction distance is whatever is left after the 30 m pure-braking distance is subtracted. $$v_{\text{ref}}\cdot t_{\text{react}} + d_{\text{brake,dry}} \le d_{\text{detect}} \quad\Longrightarrow\quad t_{\text{react}} = \frac{d_{\text{detect}} - d_{\text{brake,dry}}}{v_{\text{ref}}} = \frac{50-30}{16.67} = \boxed{1.2\ \text{s}}$$ Any reaction time up to 1.2 s (detection processing, decision logic, and physical brake actuation combined) still allows the vehicle to stop within the 50 m detected range.
3. Part (2): solve for the maximum safe speed in fog (detection reduced to 30 m, same 1.2 s reaction time, dry-road deceleration unchanged). Both the reaction distance and the braking distance now scale with the unknown speed v: $$v\cdot t_{\text{react}} + \frac{v^2}{2a} = d_{\text{fog}} \quad\Longrightarrow\quad \frac{v^2}{2a} + t_{\text{react}}\,v - d_{\text{fog}} = 0$$ Substituting t_react = 1.2 s, a = 4.63 m/s², d_fog = 30 m and solving the quadratic for the positive root: $$v_{\text{fog}} = \frac{-t_{\text{react}} + \sqrt{t_{\text{react}}^2 + 4\cdot\frac{1}{2a}\cdot d_{\text{fog}}}}{2\cdot\frac{1}{2a}} = 12.01\ \text{m/s} = \boxed{43.2\ \text{km/h}}$$
4. Part (3): solve for the maximum safe speed on a slippery road (detection restored to the normal 50 m, but stopping now takes 50 m — not 30 m — from 60 km/h). First re-derive the slippery-road deceleration the same way as Step 1: $$a_{\text{wet}} = \frac{v_{\text{ref}}^2}{2\,d_{\text{brake,wet}}} = \frac{16.67^2}{2\times50} = 2.78\ \text{m/s}^2$$ Then solve the same quadratic form with the normal 50 m detection range restored: $$v_{\text{wet}} = \frac{-t_{\text{react}} + \sqrt{t_{\text{react}}^2 + 4\cdot\frac{1}{2a_{\text{wet}}}\cdot d_{\text{detect}}}}{2\cdot\frac{1}{2a_{\text{wet}}}} = 13.66\ \text{m/s} = \boxed{49.2\ \text{km/h}}$$ The slippery-road case permits a slightly higher safe speed than the foggy case (49.2 vs. 43.2 km/h) even though braking itself is far worse (2.78 vs. 4.63 m/s² deceleration), because the detection range is much better (50 m vs. only 30 m in fog) — sight distance dominates over braking performance in this comparison.