Differences

This shows you the differences between two versions of the page.

--- streetmath [2020/03/08 11:33] – mario
+++ streetmath [2020/03/08 17:20] (current) – mario
@@ Line 1: / Line 1: @@
 ==== 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). Given the speed difference between C and B and the total speed of B, how long does it take for C to overtake B? And what distance do both vehicles travel while the overtake takes place?
+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:
+  - Given the speed difference between $C$ and $B$ and the total speed of $B$, what time $T$ does it take for $C$ to overtake $B$?
+  - 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:
+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.
+<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 $B$, passes $B$, and switches the lane again when $C$ is 5 meters in front of $B$.
+</WRAP>
+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
+\[ T = \frac{w}{\Delta_v}.\]
+<WRAP center round info 60%>
+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.
+</WRAP>
+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
+\[
+D = 20 \,\frac{\text{km}}{\text{h}} \cdot 3.6 \,\text{s}
+= 20 \,\frac{1000\,\text{m}}{3600\,\text{s}} \cdot 3.6 \,\text{s}
+= 20 \,\text{m}.
+\]
+</WRAP>
+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.
+</WRAP>