Question

Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 101010 meters per second and the second car's velocity is 666 meters per second. At a certain instant, the first car is 444 meters from the intersection and the second car is 333 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)

Answer 1

Answer:

D(L)/dt  = 407,6 m/s

Step-by-step explanation:

Let call A the intersection point.

As the cars are driving from perpendicular directions, they form with a coordinates x and y, a right  triangle, and distance between them is the hypotenuse (L), then

L² = x² + y²

Taking derivatives with respect to time we have:

2*L* D(L)/dt  =  2*x *D(x)/dt  + 2*y* D(y)/dt      (1)

In this equation we know: At a certain time

x  =  444 m       and  D(x)/dt  =  10 m/s

y  =  333 m       and  D(y)/dt  = 666 m/s

And   L = √(x)²  + (y)²    ⇒  L  = √ (444)² +( 333)² ⇒ L = √197136 + 110889

L = √308025

L = 555 m

Thn plugin these values in euatn (1) we get

2* 555 * D(L)/dt (m) =  2* 444* 10 + 2*333*666 (m*m/s)

D(L)/dt  = ( 4440 + 221778)/555  (m/s)

D(L)/dt  = 407,6 m/s