Passenger car and an ambulance come to the rectangular crossroad, the ambulance left. Passenger car at speed 39 km/h and ambulance at 41 km/h.

Calculate such a relative speed of the ambulance move to the car.

Result

v =  56.59 km/h

#### Solution:

$v_{1}=39 \ \text{km/h} \ \\ v_{2}=41 \ \text{km/h} \ \\ \ \\ s^2=s_{1}^2 + s_{2}^2 \ \\ (vt)^2=(v_{1} \ t)^2 + (v_{2} \ t)^2 \ \\ \ \\ v^2=v_{1}^2+v_{2}^2 \ \\ \ \\ v=\sqrt{ v_{1}^2+v_{2}^2 }=\sqrt{ 39^2+41^2 } \doteq \sqrt{ 3202 } \doteq 56.5862 \doteq 56.59 \ \text{km/h}$

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