Peleton

In a cycling race, the peloton travels at an average speed of 36 km/h. One competitor stopped for 5 minutes to repair a puncture. By how many kilometres per hour did his speed exceed the speed of the peloton when he caught up with it in 20 minutes? How long would it take him to catch up if the peloton immediately increased its speed to 40 km/h after he stopped?

Final Answer:

x =  9 km/h
t₃ =  40 min

Step-by-step explanation:

$$\begin{gathered} v_1=36\text{km/h} \\ {t}_1=5\text{min}\to \text{h}=5:60\text{h}=0.08333\text{h} \\ {t}_2=20\text{min}\to \text{h}=20:60\text{h}=0.33333\text{h} \\ {s}_1=s_2 \\ {v}_1\cdot (t_1+t_2)=(v_1+x)\cdot t_2 \\ x=v_1\cdot \frac{t_1+t_2}{t_2}-v_1=36\cdot \frac{0.0833+0.3333}{0.3333}-36=9\text{km/h} \end{gathered}$$

$$\begin{gathered} v_2=v_1+x=36+9=45\text{km/h} \\ {v}_3=40\text{km/h} \\ {s}_3=s_4 \\ {v}_3\cdot (t_1+y)=v_2\cdot y \\ 40\cdot (0.083333333333333+y)=45\cdot y \\ 5y=3.333333 \\ y=\frac{3.33333333}{5}=0.66666667 \\ y=\frac{2}{3}\doteq 0.666667 \\ {t}_3=y\to \text{min}=y\cdot 60\text{min}=0.6667\cdot 60\text{min}=40\text{min}=40\text{min} \end{gathered}$$

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