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. And what distance $D$ do both vehicles travel while the overtake takes place?

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