DokuWiki - fricklers.org

Trace: streetmath

Streetmath

This is an old revision of the document!


Overtaking

Let $C$ (as in 'car') and $B$ (as in 'bicycle') be two vehicles moving along a road. Let $C$ be behind $B$, but driving at a higher speed than $B$, i.e. $C$ will overtake $B$ (on the opposite lane keeping a safety distance of at least 1.50 m). Questions:

  1. Given the speed difference between $C$ and $B$ and the total speed of $B$, what time $T$ does it take for $C$ to overtake $B$?
  2. And what distance $D$ do both vehicles travel while the overtake takes place?

Mathematical Model

Assume $C$ and $B$ are points and the road is a straight line. Given quantities:

  • $v_C$: speed of the car
  • $v_B$: speed of the bicycle
  • $w$: relative window size

For example, the driver travels at $v_C$ = 30 km/h, the cyclist at $v_B$ = 20 km/h, and the relative window size is $w$ = 10 m, that is $C$ switches the lane 5 m behind $B$, passes $B$, and switches the lane again when $C$ is 5 meters in front of $B$.

Consider the speed difference $\Delta_v = v_C - v_B$ of $C$ and $B$. Regarding question 1, take $B$ as the reference point. Then $C$ approaches $B$ with speed $\Delta_v$ and needs to travel a distance of $w$ in order to safely overtake $B$. We obtain \[ T = \frac{w}{\Delta_v}.\]

Example: \[ T = \frac{10 \,\text{m}}{30 \frac{\,\text{km}}{\,\text{h}} - 20 \frac{\,\text{km}}{\,\text{h}}} = \frac{10 \,\text{m}}{30 \frac{1000 \,\text{m}}{3600 \,\text{s}} - 20 \frac{1000 \,\text{m}}{3600 \,\text{s}}} = \frac{10 \,\text{m}}{10 \frac{1000 \,\text{m}}{3600 \,\text{s}}} = \frac{36 \,\text{s}}{10 \,\text{m}} = 3.6 \,\text{s} \]