at 12 noon ship A is 65 km due north of a second ship B. Ship A sails south at a rate of 14km/hr, and ship B sails west at a rate of 16km/hr. How fast are the two ships approaching each other 1.5 hours later at 1:30pm?
The answer is -4.629...
Aprox -4.623
To find out how fast the two ships are approaching each other at 1:30 pm, we need to determine their positions at that time. To do that, we first need to find the positions of the ships at 12 noon.
At 12 noon, ship A is 65 km due north of ship B. This means that the vertical distance between them is 65 km.
Ship A sails south at a rate of 14 km/hr for 1.5 hours, which means it travels a distance of 14 km/hr × 1.5 hr = 21 km.
Since ship A is sailing south, the vertical distance between the two ships decreases by 21 km, so the vertical distance between them at 1:30 pm is 65 km - 21 km = 44 km.
Ship B sails west at a rate of 16 km/hr for 1.5 hours, which means it travels a distance of 16 km/hr × 1.5 hr = 24 km.
Since ship B is sailing west, the horizontal distance between the two ships decreases by 24 km, so the horizontal distance between them at 1:30 pm is 0 km - 24 km = -24 km. The negative sign means that ship B has moved to the west of its starting position, which is expected.
Now that we know the horizontal and vertical distances between the two ships at 1:30 pm, we can use the Pythagorean theorem to find the distance between them. The Pythagorean theorem states that for a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
The horizontal distance is -24 km and the vertical distance is 44 km. Using the Pythagorean theorem, we can calculate the distance between the two ships at 1:30 pm:
Distance^2 = (-24 km)^2 + 44 km^2
Distance^2 = 576 km^2 + 1936 km^2
Distance^2 = 2512 km^2
Distance = √(2512 km^2)
Distance ≈ 50.12 km
Now we have the distance between the two ships at 1:30 pm. To find out how fast they are approaching each other, we need to find the rate at which this distance is changing. We can do this by taking the derivative of the distance equation with respect to time.
d(Distance)/dt = d(√(576 km^2 + 1936 km^2))/dt
To calculate this derivative, we need to know the rates at which the horizontal and vertical distances are changing with time. The rate of change of the horizontal distance is the velocity of ship B, which is given as 16 km/hr. The rate of change of the vertical distance is the velocity of ship A, which is -14 km/hr (negative because it is moving south).
Now, we can substitute these values into the derivative equation:
d(Distance)/dt = (16 km/hr)(-24 km) / √(576 km^2 + 1936 km^2) + (-14 km/hr)(44 km) / √(576 km^2 + 1936 km^2)
Simplifying this equation gives us:
d(Distance)/dt ≈ -384 km²/hr - 616 km²/hr / √(576 km² + 1936 km²)
d(Distance)/dt ≈ -1000 km²/hr / √(576 km² + 1936 km²)
Therefore, the two ships are approaching each other at a rate of approximately -1000 km²/hr / √(576 km² + 1936 km²). Note that the negative sign indicates that they are getting closer to each other.