Two aeroplanes fly eastwards on parallel courses 12 miles apart. One flies at 240 m.p.h. and the other at 300 m.p.h. How fast is the distance between them changing when the slower plane is 5 miles farther east than the faster plane?
Well, if you've ever watched a turtle race, you know that the slower one is always playing catch-up! In this case, let's call the distance between the faster plane and the slower plane "d" (in miles). So, when the slower plane is 5 miles farther east than the faster plane, we have d = 5 miles.
Now, let's take a look at their speeds. The faster plane is flying at 300 m.p.h., while the slower one is going at 240 m.p.h. That means the distance between them is changing at a rate of (300 - 240) m.p.h. = 60 m.p.h.
So, the distance between them is changing at a rate of 60 miles per hour. And since d = 5 miles, we can say that the distance between them is changing at a rate of 60 miles per hour when the slower plane is at a distance of 5 miles from the faster plane.
Hope that puts a smile on your face!
To solve this problem, we can use the concept of related rates. Let's denote the distance between the two planes as "d" and the time as "t".
We are given that the planes are flying eastwards on parallel courses, 12 miles apart. This means that the distance between them, d, remains constant.
We also know that one plane flies at 240 mph and the other at 300 mph. Let's denote the slower plane's distance as "x" and the faster plane's distance as "y". It's given that when the slower plane is 5 miles farther east than the faster plane, x = y + 5.
We need to find the rate of change of the distance between the planes, dd/dt, when the slower plane is 5 miles farther east than the faster plane. In other words, we need to find d(dt).
To solve for dd/dt, we can use the Pythagorean theorem. According to the theorem, d² = x² + y².
Differentiating both sides of this equation with respect to time t, we get:
2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)
Since the planes are traveling eastwards, we have dx/dt = 240 mph and dy/dt = 300 mph (negative because the planes are moving closer to each other).
Substituting the values, and using x = y + 5, we get:
2d(dd/dt) = 2(y + 5)(240) + 2y(-300)
Simplifying this equation, we have:
2d(dd/dt) = (480y + 2400) - (600y)
2d(dd/dt) = 480y - 600y + 2400
2d(dd/dt) = -120y + 2400
Now, we know that d is constant, so dd/dt = 0.
Finally, solving for y, when x = y + 5, we get y = -2.5.
Substituting this value in the equation, we have:
2d(0) = -120(-2.5) + 2400
0 = 300 + 2400
0 = 2700
Since this equation has no solution, there seems to be an error in the problem statement or calculations. Please double-check the information provided or let me know if there's anything else I can assist you with.
To solve this problem, we can use the concept of related rates. We want to find the rate at which the distance between the planes is changing with respect to time. Let's denote the distance between the planes as "d" and the time as "t."
Given information:
- The planes are flying eastwards on parallel courses.
- The initial distance between the planes is 12 miles.
- The speed of the slower plane is 240 mph.
- The speed of the faster plane is 300 mph.
- The slower plane is 5 miles farther east than the faster plane.
Let's break down the problem and find an equation that relates the variables:
1. The rate of change of distance between the planes can be represented as "dd/dt" (the derivative of 'd' with respect to 't').
2. The rate at which the slower plane is moving eastward can be represented as "dr/dt"
3. We want to find the value of "dd/dt" when the slower plane is 5 miles farther east than the faster plane.
To find the equation relating these variables, we need to consider that the relative motion between the two planes causes the distance between them to change. The rate at which this change occurs can be calculated as follows:
dd/dt = dr/dt - df/dt
Where:
- dd/dt is the rate of change of distance 'd' with respect to time 't'
- dr/dt is the rate at which the slower plane is moving eastward
- df/dt is the rate at which the faster plane is moving eastward
From the problem statement, we know that the rate at which the slower plane is moving eastward (dr/dt) is 240 mph. Given that the faster plane's speed (df/dt) is 300 mph, we can express the rate of change of the distance between the two planes as:
dd/dt = 240 mph - 300 mph
Simplifying further:
dd/dt = -60 mph
It means the distance between the planes is decreasing at a rate of 60 mph.
Hence, the rate at which the distance between the planes is changing when the slower plane is 5 miles farther east than the faster plane is -60 mph.
I'll assume you meant that the faster plane is 5 miles farther east...
say the slow one is flying along the x-axis, and the faster one along the line y=12. Then at time t hours, the distance z between them is
z^2 = (300t - 240t)^2 + 12^2
= 3600t^2 + 144
300t-240t = 5 when t=1/12
at that time, z=13
2z dz/dt = 7200t
dz/dt = 20 mi/hr