Question

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 95 km/h while car B is 0.4 km from the intersection and traveling at 90 km/h. How fast is the distance between the cars changing at that moment? km/h

Answer 1

Answer:

The rate at which the distance changes at that moment is 91.743  Km/h

Explanation:

The time for car a to get to the intersection = $\frac{Distance}{Speed}$

                 = $\frac{0.3}{95}$

                = 0.00316 h

The time for car B to get to the intersection = $\frac{Distance}{Speed}$

                = $\frac{0.4}{90}$

                = 0.00444 h

Total time = $\sqrt{0.00316^{2} + 0.00444^{2} }$

                = 0.00545 h

Total distance between the two cars at that moment = $\sqrt{0.3^{2} + 0.4^{2} }$

                = 0.5 Km

The rate at which the distance between the cars is changing = $\frac{0.5}{0.00545}$

                =  91.743  Km/h

Answer 2

Answer:

129 km/hr

Explanation:

Distance of Car A North of the Intersection, y=0.3km

Distance of Car B West of the Intersection, x=0.4 km

The distance z,  between A and B is determined by the Pythagoras theorem

$z^2=x^2+y^2$

$z^2=0.4^2+0.3^2=0.25\\z=\sqrt{0.25}=0.5km$

Taking derivative of $z^2=x^2+y^2$

$2z\frac{dz}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}$

$\frac{dx}{dt}=90km/hr, \frac{dy}{dt}=95km/hr$

$2(0.5)\frac{dz}{dt}= 2(0.4)X90+2X0.3X95\\\frac{dz}{dt}=72+57=129$

The distance z, between the cars is changing at a rate of 129 km/hr.