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:

$v_{1}=39 \ \text{km/h} \ \\ v_{2}=41 \ \text{km/h} \ \\ \ \\ s^2=s_{1}^2 + s_{2}^2 \ \\ (vt)^2=(v_{1} \ t)^2 + (v_{2} \ t)^2 \ \\ \ \\ v^2=v_{1}^2+v_{2}^2 \ \\ \ \\ v=\sqrt{ v_{1}^2+v_{2}^2 }=\sqrt{ 39^2+41^2 } \doteq \sqrt{ 3202 } \doteq 56.5862 \doteq 56.59 \ \text{km/h}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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dont understand Tips to related online calculators
Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.
Do you want to convert velocity (speed) units?
Pythagorean theorem is the base for the right triangle calculator.

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