Car A is traveling west at 80 𝑘𝑚/ℎ and car B is traveling north at 100𝑘𝑚/ℎ. Both are headed for the intersection of two roads. At what rate are the cars approaching each other when car A is 3 𝑘𝑚 and car B is 4 𝑘𝑚 from the intersection?
To find the rate at which the cars are approaching each other, we can use the Pythagorean theorem.
Let x represent the distance between the two cars at any given time, and let y represent the distance car A has traveled and z represent the distance car B has traveled.
Using the Pythagorean theorem, we have:
x^2 = y^2 + z^2
Differentiating both sides with respect to time t, we get:
2x(dx/dt) = 2y(dy/dt) + 2z(dz/dt)
Given that car A is traveling at 80𝑘𝑚/ℎ (dy/dt) and car B is traveling at 100𝑘𝑚/ℎ (dz/dt), we have:
dy/dt = -80 and dz/dt = -100
Substitute y = 3 km, z = 4 km, dy/dt = -80, dz/dt = -100 into the differentiated equation, we get:
2x(dx/dt) = 2(3)(-80) + 2(4)(-100)
2x(dx/dt) = -480 - 800
2x(dx/dt) = -1280
dx/dt = -1280 / (2x)
dx/dt = -640 / x
When we have y = 3 km, z = 4 km, we can find x:
x = sqrt(3^2 + 4^2)
x = 5 km
Now, when x = 5 km, we can find dx/dt:
dx/dt = -640 / 5
dx/dt = -128 km/h
Therefore, when car A is 3 km and car B is 4 km from the intersection, the cars are approaching each other at a rate of 128 km/h.