Question

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

Answer 1

At 4:00 PM, the distance between the ships is changing at a rate of 6160 km/h.

Step-by-step explanation:

To find the rate at which the distance between the ships is changing, we can use the concept of relative velocity. At 4:00 PM, ship A has been sailing east for 4 hours, covering a distance of 40 km/h * 4 h = 160 km. Ship B has been sailing north for 4 hours, covering a distance of 20 km/h * 4 h = 80 km.

To find the distance between the ships at 4:00 PM, we can use the Pythagorean theorem. The distance between the ships is the hypotenuse of a right triangle with legs of 170 km and 80 km. Using the theorem, we find that the distance between the ships is √(170^2 + 80^2) ≈ 186.45 km.

To find the rate at which the distance is changing, we can use the derivative of the distance formula. Let D be the distance between the ships. Then, D^2 = (170 + 40t)^2 + (80 + 20t)^2, where t is the time in hours. Taking the derivative of D^2 with respect to t and solving for dD/dt, we get: dD/dt = (170 + 40t)(40) + (80 + 20t)(20).

Substituting t = 4 into the expression, we find that dD/dt = (170 + 40(4))(40) + (80 + 20(4))(20) = 6160 km/h. Therefore, the distance between the ships is changing at a rate of 6160 km/h at 4:00 PM.

Answer 2

40 times 4hours=160 km east for ship A Ship B goes 80 km north.
170-160=10 10^2=100 80^2=6400 6400+100=6500 square root 6500 and you get the distance they are away from each other at four o'clock which is about 80.622 km away.