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
[tex]z^2=x^2+y^2[/tex]
[tex]z^2=0.4^2+0.3^2=0.25\\z=\sqrt{0.25}=0.5km[/tex]
Taking derivative of [tex]z^2=x^2+y^2[/tex]
[tex]2z\frac{dz}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}[/tex]
[tex]\frac{dx}{dt}=90km/hr, \frac{dy}{dt}=95km/hr[/tex]
[tex]2(0.5)\frac{dz}{dt}= 2(0.4)X90+2X0.3X95\\\frac{dz}{dt}=72+57=129[/tex]
The distance z, between the cars is changing at a rate of 129 km/hr.
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 = [tex]\frac{Distance}{Speed}[/tex]
= [tex]\frac{0.3}{95}[/tex]
= 0.00316 h
The time for car B to get to the intersection = [tex]\frac{Distance}{Speed}[/tex]
= [tex]\frac{0.4}{90}[/tex]
= 0.00444 h
Total time = [tex]\sqrt{0.00316^{2} + 0.00444^{2} }[/tex]
= 0.00545 h
Total distance between the two cars at that moment = [tex]\sqrt{0.3^{2} + 0.4^{2} }[/tex]
= 0.5 Km
The rate at which the distance between the cars is changing = [tex]\frac{0.5}{0.00545}[/tex]
= 91.743 Km/h
