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streetmath [2020/03/08 11:50] – mario | streetmath [2020/03/08 17:20] (current) – mario | ||
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==== Overtaking ==== | ==== Overtaking ==== | ||
- | Let C (as in ' | + | Let $C$ (as in ' |
Questions: | Questions: | ||
- | - 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? | + | - Given the speed difference between |
- | - And what distance $D$ do both vehicles | + | - What distance $D$ does the bicycle $B$ travel while the overtake takes place? |
=== Mathematical Model === | === Mathematical Model === | ||
- | Assume C and B are points and the road is a straight line. Given quantities: | + | Assume |
* $v_C$: speed of the car | * $v_C$: speed of the car | ||
* $v_B$: speed of the bicycle | * $v_B$: speed of the bicycle | ||
* $w$: | * $w$: | ||
- | 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. | + | <WRAP center round info 60%> |
+ | 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 | ||
+ | </ | ||
- | 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 | + | 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 |
\[ T = \frac{w}{\Delta_v}.\] | \[ T = \frac{w}{\Delta_v}.\] | ||
- | Example: | + | <WRAP center round info 60%> |
+ | Example: | ||
\[ T | \[ T | ||
- | = \frac{10 \text{m}}{30 \frac{\text{km}}{\text{h} - 20 \frac{\text{km}}{\text{h}} | + | = \frac{10 |
- | = \frac{10 \text{m}}{30 | + | = \frac{10 |
- | = \frac{10 \text{m}}{10 | + | = \frac{10 |
- | = 10 \text{m} \cdot \frac{3600 \text{s}}{1000 \text{m}} | + | = \frac{36 \, |
- | = 3.6 \text{s}.\] | + | = 3.6 \, |
+ | \] | ||
+ | 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 | ||
+ | \[ | ||
+ | D = v_B \cdot T. | ||
+ | \] | ||
+ | |||
+ | <WRAP center round info 60%> | ||
+ | Again, in our running example we have | ||
+ | \[ | ||
+ | D = 20 \,\frac{\text{km}}{\text{h}} \cdot 3.6 \,\text{s} | ||
+ | = 20 \, | ||
+ | = 20 \,\text{m}. | ||
+ | \] | ||
+ | </ | ||
+ | |||
+ | 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. | ||
+ | |||
+ | <WRAP center round info 60%> | ||
+ | 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. | ||
+ | </ |