Highway section length

A passenger car drove a highway section at a constant speed. At a speed of 20 km/h higher, the ride would take 6 minutes less. At a speed of 20 km/h lower, it would take 9 minutes more. Calculate the length of the highway section.

Final Answer:

x =  60 km

Step-by-step explanation:

$$\begin{gathered} \Delta =20\text{km/h} \\ {t}_1=6\text{min}\to \text{h}=6:60\text{h}=0.1\text{h} \\ {t}_2=9\text{min}\to \text{h}=9:60\text{h}=0.15\text{h} \\ x=v\cdot t \\ x=(v+\Delta )\cdot (t-t_1) \\ x=(v-\Delta )\cdot (t+t_2) \\ v\cdot t=(v+\Delta )\cdot (t-t_1) \\ v\cdot t=(v-\Delta )\cdot (t+t_2) \\ v\cdot t=vt+\Delta t-t_{1}v-t_1\Delta \\ v\cdot t=vt-\Delta t+t_{2}v-t_2\Delta \\ 0=\Delta \cdot t-t_1\cdot v-t_1\cdot \Delta \\ 0=-\Delta \cdot t+t_2\cdot v-t_2\cdot \Delta \\ 0=20\cdot t-0.1\cdot v-0.1\cdot 20 \\ 0=-20\cdot t+0.15\cdot v-0.15\cdot 20 \\ 20t-0.1v=2 \\ 20t-0.15v=-3 \\ Row2-Row1\to Row2 \\ -0.05v=-5 \\ v=\frac{-5}{-0.05}=100 \\ t=\frac{2+0.1v}{20}=\frac{2+0.1\cdot 100}{20}=0.6 \\ t=\frac{3}{5}=0.6 \\ v=100 \\ x=v\cdot t=100\cdot 0.6=60\text{km} \\ \text{ Verifying Solution: } \\ {v}_2=v+\Delta =100+20=120\text{km/h} \\ {v}_3=v-\Delta =100-20=80\text{km/h} \\ {s}_1=v\cdot t=100\cdot 0.6=60\text{km} \\ {s}_2=v_2\cdot (t-t_1)=120\cdot (0.6-0.1)=60\text{km} \\ {s}_3=v_3\cdot (t+t_2)=80\cdot (0.6+0.15)=60\text{km} \end{gathered}$$

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