A car leaves an intersection traveling west. Its position $4 \mathrm { sec }$ later is $26 \mathrm { ft }$ from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is $26 \mathrm { ft }$ from the intersection. If the speeds of the cars at that instant of time are $12 \mathrm { ft } / \mathrm { sec }$ and 10 $\mathrm { ft } / \mathrm { sec }$, respectively, find the rate at which the distance between the two cars is changing. Round to the nearest tenth if necessary.
Select the correct answer.
A) $\quad 15.6 \mathrm { ft } / \mathrm { sec }$
B) $3.7 \mathrm { ft } / \mathrm { sec }$
C) $3.1 \mathrm { ft } / \mathrm { sec }$
D) $36.8 \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The correct answer is A.The distance between the two cars is the hypotenuse of a right triangle with legs 26 ft and 26 ft. The rate at which the distance is changing is the rate at which the hypotenuse is changing. The rate at which the hypotenuse is changing is the square root of the sum of the squares of the rates at which the legs are changing. The rate at which the westbound car is moving away from the intersection is 12 ft/sec, and the rate at which the northbound car is moving away from the intersection is 10 ft/sec. Therefore, the rate at which the distance between the two cars is changing is the square root of (12^2 + 10^2) ft/sec = 15.6 ft/sec.

A Car Leaves an Intersection Traveling West 4sec4 \mathrm { sec }4sec

A Car Leaves an Intersection Traveling West $4 \mathrm { sec }$