Acar leaves an intersection traveling west. its position 4 sec later is 21 ft from the intersection. at the same time, another car leaves the same intersection heading north so that its position 4 sec later is 28 ft from the intersection. if the speeds of the cars at that instant of time are 9 ft/sec and 13 ft/sec, respectively, find the rate at which the distance between the two cars is changing.

15.8 ft/s

Explanation:

= Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

= Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

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