Driver

The driver of the supply car reckoned that at the average speed of 72 km/h arrived at the warehouse for 1 1/4 hours. After 30 km, however, he unintentionally drove to the gas station and had ten minutes delay. At what average speed would you have to go the rest of the way to catch up with delay?

Correct answer:

v₂ =  90 km/h

Step-by-step explanation:

$$\begin{gathered} v_1=72\text{km/h} \\ t=1\frac{1}{4}=1+\frac{1}{4}=\frac{1\cdot 4+1}{4}=\frac{4+1}{4}=\frac{5}{4}=1.25\text{h} \\ {s}_1=30\text{km} \\ s=v_1\cdot t=72\cdot \frac{5}{4}=\frac{72\cdot 5}{4}=\frac{360}{4}=90\text{km} \\ {s}_2=s-s_1=90-30=60\text{km} \\ {t}_1=s_1/v_1=30/72=\frac{5}{12}\doteq 0.4167\text{h} \\ {t}_2=t-t_1-10/60=\frac{5}{4}-\frac{5}{12}-10/60=\frac{2}{3}\doteq 0.6667\text{h} \\ {v}_2=s_2/t_2=60/\frac{2}{3}=60:\frac{2}{3}=60\cdot \frac{3}{2}=\frac{60\cdot 3}{2}=\frac{180}{2}=90\text{km/h} \end{gathered}$$

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