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Trace: streetmath


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. What distance $D$ does the bicycle $B$ 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: It takes \[ T = \frac{10 \,\text{m}}{30 \frac{\,\text{km}}{\,\text{h}} - 20 \frac{\,\text{km}}{\,\text{h}}} = \frac{10 \,\text{m}}{30 \frac{1000 \,\text{m}}{3600 \,\text{s}} - 20 \frac{1000 \,\text{m}}{3600 \,\text{s}}} = \frac{10 \,\text{m}}{10 \frac{1000 \,\text{m}}{3600 \,\text{s}}} = \frac{36 \,\text{s}}{10 \,\text{m}} = 3.6 \,\text{s} \] for $C$ to overtake $B$ if their speed difference is 10 km/h.

Now, using $T$ and the reference speed $v_B$ of the cyclist, it is easy to calculate \[ D = v_B \cdot T. \]

Again, in our running example we have \[ D = 20 \,\frac{\text{km}}{\text{h}} \cdot 3.6 \,\text{s} = 20 \,\frac{1000\,\text{m}}{3600\,\text{s}} \cdot 3.6 \,\text{s} = 20 \,\text{m}. \]

Now it remains to add the relative window size $w$ to $D$ in order to get the distance $C$ travels from the beginning of the overtake until its end.

In total, overtaking a cyclist riding at 20 km/h while driving a car at 30 km/h takes 3.6 s and meanwhile the cyclist travels 20 m. In order to avoid hitting the cyclist, the driver must have the full overview over the road 30 m ahead.