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two cars start moving from the same point. one travels south at 36 mi/h and the other travels west at 27 mi/h. at what rate (in mi/h) is the distance between the cars increasing four hours later?

User Quetcy
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To find the rate at which the distance between the cars is increasing, we can use the concept of relative velocity. Let's consider the southward car as car A and the westward car as car B.

The rate at which the distance between the cars is increasing is equal to the magnitude of the relative velocity between the cars.

We can use the Pythagorean theorem to find the distance between the cars at any given time. Let's denote this distance as D(t).

D(t) = √((Distance traveled by car A)^2 + (Distance traveled by car B)^2)

After 4 hours, the distance traveled by car A is (36 mi/h) * (4 h) = 144 miles.

The distance traveled by car B is (27 mi/h) * (4 h) = 108 miles.

Substituting these values into the equation for D(t):

D(t) = √((144)^2 + (108)^2)

D(t) = √(20736 + 11664)

D(t) = √32400

D(t) = 180 miles

Now, let's differentiate the equation D(t) with respect to time t to find the rate of change of D(t) with respect to time:

d(D(t))/dt = (d/dt)√((144)^2 + (108)^2)

To simplify the differentiation, we can consider (144)^2 + (108)^2 as a constant value, say C.

d(D(t))/dt = (d/dt)√C

Using the chain rule, we have:

d(D(t))/dt = (1/2) * C^(-1/2) * (dC/dt)

Since C = (144)^2 + (108)^2, the derivative of C with respect to time is zero because it is a constant.

d(D(t))/dt = (1/2) * C^(-1/2) * (dC/dt) = (1/2) * C^(-1/2) * 0 = 0

Therefore, the rate at which the distance between the cars is increasing four hours later is 0 mi/h.

User Eugene Myasyshchev
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