Two commercial airplanes are flying at an altitude of 40,000 ft along straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 427 knots (nautical miles per hour; a nautical mile is 2000 yd or 6000 ft.) Plane B is approaching the intersection at 439 knots.
At what rate is the distance between the planes decreasing when Plane A is 5 nautical miles from the intersection point and Plane B is 5 nautical miles from the intersection point?
To find the rate at which the distance between the planes is decreasing, we can use the concept of relative motion. We will consider the distance between the planes as the hypotenuse of a right triangle, with the distances of Plane A and Plane B from the intersection point being the legs of the triangle.
Let's label the distance of Plane A from the intersection point as x and the distance of Plane B from the intersection point as y.
According to the problem, the planes are approaching the intersection point at speeds of 427 knots and 439 knots, respectively.
Let's find the rates of change for both x and y.
Rate of change for x (distance of Plane A from the intersection point):
dx/dt = -427 knots (since Plane A is approaching the intersection point)
Rate of change for y (distance of Plane B from the intersection point):
dy/dt = -439 knots (since Plane B is approaching the intersection point)
Now, let's find the rate of change for the distance between the planes (d):
d = sqrt(x^2 + y^2) (using Pythagoras' theorem)
Differentiating both sides with respect to time (t), we get:
dd/dt = (1/2) * (2x(dx/dt) + 2y(dy/dt))
= x(dx/dt) + y(dy/dt)
Substituting the given values, we have:
dd/dt = (5 * (-427) + 5 * (-439))
Simplifying this expression, we find:
dd/dt = -2135 - 2195
dd/dt = -4330 knots
Therefore, the rate at which the distance between the planes is decreasing is -4330 knots.
To find the rate at which the distance between the two planes is decreasing, we need to use the concept of relative velocity.
Let's consider a coordinate system where the intersection point is the origin (0,0). At any given time, Plane A will be at a position (x, 0) and Plane B will be at a position (0, y). We want to find the rate of change of the distance between the two planes, which we'll call D.
Since Plane A is approaching the intersection point at a speed of 427 knots, its position can be described by the equation x = 5 - 427t, where t is the time in hours. Similarly, the position of Plane B can be described by the equation y = 5 - 439t.
Now, let's find the expression for the distance D between the two planes using the Pythagorean theorem:
D^2 = (x - 0)^2 + (0 - y)^2
D^2 = x^2 + y^2
Substituting the expressions for x and y, we get:
D^2 = (5 - 427t)^2 + (5 - 439t)^2
To find the rate at which the distance D is changing, we differentiate this expression with respect to time:
2D * dD/dt = 2(5 - 427t)(-427) + 2(5 - 439t)(-439)
Simplifying further:
2D * dD/dt = -854(5 - 427t) - 878(5 - 439t)
2D * dD/dt = -854(5) + 854(427t) - 878(5) + 878(439t)
2D * dD/dt = -4270 + 364558t - 4390 + 384142t
2D * dD/dt = 750700t - 8660
Now, we need to find the time t when Plane A is 5 nautical miles away from the intersection. We know that the distance x is given by x = 5 - 427t. Setting this equal to 5, we get:
5 - 427t = 5
-427t = 0
t = 0
So, when Plane A is 5 nautical miles away from the intersection, t = 0.
Substituting t = 0 into the expression for the rate of change of distance:
2D * dD/dt = 750700(0) - 8660
2D * dD/dt = -8660
dD/dt = -8660 / (2D)
Since we want to find the rate at which the distance between the planes is decreasing, we take the negative value:
Rate of change of distance = - dD/dt = 8660 / (2D)
Now, we need to find the value of D when Plane A is 5 nautical miles away from the intersection. Substituting x = 5 into the expression for D:
D^2 = (5 - 427t)^2 + (5 - 439t)^2
D^2 = (5 - 427(0))^2 + (5 - 439(0))^2
D^2 = (5 - 0)^2 + (5 - 0)^2
D^2 = 25 + 25
D^2 = 50
D = √50
D ≈ 7.07 nautical miles
Now, we can substitute this value of D into the expression for the rate of change of distance:
Rate of change of distance = 8660 / (2D)
Rate of change of distance = 8660 / (2 * 7.07)
Rate of change of distance ≈ 614.65 knots
Therefore, the rate at which the distance between the planes is decreasing when Plane A is 5 nautical miles from the intersection point and Plane B is 5 nautical miles from the intersection point is approximately 614.65 knots.
distanceapart^2=distaneAfromcollision^2+distanceBfromcollision^2
Take the derivative..
s ds/dt=2xdx/dt + 2y dy/dt where x, y are the distances from the intersection point, and s is sqrt(x^2+y^2)
you are given dx/dt, dy/dt, solve for ds/dt