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November 26, 2014

November 26, 2014

Posted by **Anonymous** on Tuesday, August 2, 2011 at 2:23am.

The prototype is 100 meters south of an intersection, while the diesel car is 100 meters east of the intersection. Both vehicles start moving at the same time. The prototype moves north, toward the intersection, and the diesel car moves east, away from the intersection. If the prototype is traveling at a velocity of 3 meters per second and the diesel car is traveling at 2 meters per second, what is the rate of change of the distance between the cars after four seconds? Round off your answer to two decimal places.

- Math -
**drwls**, Tuesday, August 2, 2011 at 7:14amThe Y-coordinate of car 1, relative to the intersection, are

Y = -100 + 3t (Its X value is 0)

The X-coordinate of car 2, relative to the intersection, is

X = 100 + 2t (Its Y value is zero)

The distance between the two cars is

R = sqrt(X^2 + Y^2) = sqrt(2*10^4 -600t + 9t^2 +400t +4t^2) = sqrt(2*10^4 +13t^2 -200t)

The rate of change of separation distance is

dR/dt = (1/2)*(26t -200)/sqrt(2*10^4 +13t^2 -200t)

Compute the value of dR/dt at t = 4s.

(The denominator is the distance apart, which is 139.13 m at t = 4 s)

- Math -
**honey**, Tuesday, March 19, 2013 at 12:01pmarea of parallelogram

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