Question

A car 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.

Answer 1

Answer:

15.8 ft/s

Explanation:

$\frac{da}{dt}$ = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

$\frac{db}{dt}$ = 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

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s$

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