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?
-19.2 m/s
To solve this problem, we can use the concept of related rates from calculus. We need to find the rate at which the distance between the two planes is changing with respect to time.
Let's denote the distance traveled by the slower plane as x (in miles) and the distance traveled by the faster plane as y (in miles). We are given that the slower plane is 5 miles farther east than the faster plane, so we have the following relationship:
x = y + 5
We also know that the planes are flying eastwards on parallel courses 12 miles apart. Therefore, the distance between the two planes is given by the difference in their distances traveled:
d = y - x
To find the rate at which the distance between the two planes is changing, we need to differentiate this equation with respect to time (t) using the chain rule:
d/dt = d/dy(dy/dt) - d/dx(dx/dt)
Since the planes are flying eastwards, the distances x and y are increasing with time. Therefore, dx/dt and dy/dt are positive.
Given that the slower plane is flying at 240 m.p.h and the faster plane is flying at 300 m.p.h, we have:
dx/dt = 240 m.p.h
dy/dt = 300 m.p.h
Now we can substitute these values into the equation:
d/dt = (d/dy)(dy/dt) - (d/dx)(dx/dt)
= (1)(300) - (-1)(240) (since d/dy = 1 and d/dx = -1, as x = y + 5)
= 300 + 240
= 540 m.p.h
Therefore, the distance between the two planes is changing at a rate of 540 miles per hour when the slower plane is 5 miles farther east than the faster plane.