# Driver

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

Correct result:

v2 =  90 km/h

#### Solution:

$v_{1}=72 \ \text{km/h} \ \\ t=1 + 1/4=\dfrac{ 5 }{ 4 }=1.25 \ \text{h} \ \\ s_{1}=30 \ \text{km} \ \\ s=v_{1} \cdot \ t=72 \cdot \ 1.25=90 \ \text{km} \ \\ s_{2}=s-s_{1}=90-30=60 \ \text{km} \ \\ t_{1}=s_{1} / v_{1}=30 / 72 \doteq \dfrac{ 5 }{ 12 } \doteq 0.4167 \ \text{h} \ \\ t_{2}=t - t_{1} - 10/60=1.25 - 0.4167 - 10/60 \doteq \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \text{h} \ \\ v_{2}=s_{2}/t_{2}=60/0.6667=90 \ \text{km/h}$

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