Two automobiles start from a point A at the same time. One travels west at 60km/hr and the other travels north 35 km/hr. how fast is the distance between them increasing 3hr later?

at time t hours, the distance d is

d^2 = (60t)^2 + (35t)^2
when t=3, then, d=208.39

Since

2d dd/dt = 7200t + 2450t = 9650t
at t=3,

dd/dt = 28950/(2*208.39) = 69.46 km/hr

very goog

To find how fast the distance between the two automobiles is increasing, we can use the Pythagorean theorem.

Let's denote the distance between the two automobiles as "d" at any given time t.

According to the Pythagorean theorem, we have:

d^2 = (distance traveled by the westbound automobile)^2 + (distance traveled by the northbound automobile)^2

The distance traveled by the westbound automobile is given by:

distance = speed x time
distance = 60 km/hr x t

The distance traveled by the northbound automobile is given by:

distance = speed x time
distance = 35 km/hr x t

Substituting these values into the equation:

d^2 = (60t)^2 + (35t)^2

Expanding the equation:

d^2 = (3600t^2) + (1225t^2)
d^2 = 4825t^2

To find how fast the distance is increasing, we need to find d/dt (the derivative of d with respect to t).

Taking the derivative of both sides of the equation:

2d(d/dt) = 2*4825t
(d/dt) = 4825t / d

Since we are interested in finding the rate of change 3 hours later, we can substitute t = 3 into the equation:

(d/dt) = 4825(3) / d

Now, we need to find the distance "d" at t = 3 hours. Using the Pythagorean theorem, we plug in t = 3 into the original equation:

d^2 = (60(3))^2 + (35(3))^2
d^2 = 10800 + 3675
d^2 = 14475
d = √14475
d ≈ 120.3 km

Substituting d ≈ 120.3 km and t = 3 back into the derivative equation:

(d/dt) = 4825(3) / 120.3
(d/dt) ≈ 120.34 km/hr

Therefore, the distance between the two automobiles is increasing at a rate of approximately 120.34 km/hr 3 hours later.

To find the rate at which the distance between the two automobiles is increasing, we can use the concept of related rates. We can break down the problem into two components: the horizontal component (westward) and the vertical component (northward).

Let's denote the horizontal distance traveled by the westward automobile as x (in km) and the vertical distance traveled by the northward automobile as y (in km). We need to find the rate of change of the distance between them, which is the hypotenuse of the right triangle formed by their positions.

Using the Pythagorean theorem, we have:

distance^2 = x^2 + y^2

To find how the distance is changing with respect to time, we need to take the derivative of both sides of the equation and solve for the rate of change of the distance (d(distance)/dt).

Taking the derivative with respect to time (t) of both sides, we get:

2 * distance * (d(distance)/dt) = 2x * (dx/dt) + 2y * (dy/dt)

Since dx/dt represents the westward speed of the first automobile (60 km/hr) and dy/dt represents the northward speed of the second automobile (35 km/hr), we can substitute these values into the equation. Let's assume t = 3 hours:

2 * distance * (d(distance)/dt) = 2 * x * 60 + 2 * y * 35

Now we have an equation with the unknown values of x and y. To find these values, we can use the fact that both automobiles started at the same time. Since they traveled for 3 hours, we can calculate their distances separately:

x = 60 km/hr * 3 hr = 180 km (westward distance)
y = 35 km/hr * 3 hr = 105 km (northward distance)

Substituting these values into the equation, we have:

2 * distance * (d(distance)/dt) = 2 * 180 * 60 + 2 * 105 * 35

Calculate the right side of the equation to find the total rate of change of the distance:

2 * distance * (d(distance)/dt) = 21600 + 7350

2 * distance * (d(distance)/dt) = 28950

Divide both sides of the equation by 2 * distance to solve for (d(distance)/dt):

d(distance)/dt = 28950 / (2 * distance)

Now, substitute the initial value of distance to find the rate at which the distance is increasing after 3 hours. Unfortunately, the initial distance is not provided in the question. Please provide the initial distance value to complete the calculation.