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streetmath [2020/03/08 12:00] mariostreetmath [2020/03/08 17:20] (current) mario
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 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 $C$ and $B$ and the total speed of $B$, what time $T$ does it take for $C$ to overtake $B$? 
-  - And what distance $D$ do both vehicles travel while the overtake takes place?+  - What distance $D$ does the bicycle $B$ travel while the overtake takes place?
  
 === Mathematical Model === === Mathematical Model ===
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   * $w$:   relative window size   * $w$:   relative window size
  
 +<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 $B$.  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 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
 \[ T = \frac{w}{\Delta_v}.\] \[ T = \frac{w}{\Delta_v}.\]
  
-Example:+<WRAP center round info 60%> 
 +Example: It takes
 \[ T  \[ T 
 = \frac{10 \,\text{m}}{30 \frac{\,\text{km}}{\,\text{h}} - 20 \frac{\,\text{km}}{\,\text{h}}} = \frac{10 \,\text{m}}{30 \frac{\,\text{km}}{\,\text{h}} - 20 \frac{\,\text{km}}{\,\text{h}}}
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 = 3.6 \,\text{s} = 3.6 \,\text{s}
 \] \]
 +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>