Question

An aircraft we will call the "receiver" is approaching a second aircraft in mid-air, which we will call the "tanker", for refueling. The receiver is following the same path as that of the tanker. The tanker is moving at a constant velocity of 500 feet per second, with its position at time t minutes given by s[t]=500t.

At time t = 0 seconds, the receiver is 2000 feet behind the tanker and has velocity 1000 feet per second. At this time, the receiver begins to decelerate at a constant rate, so that its position at time t (along the path of the tanker) is given by:

r[t]=-2000+1000t+1/2 at^2

where the distance is measured in feet, and a is the acceleration (which is negative). This latter function is carefully calculated by the crew so that in a short amount of time the receiver will catch up to the tanker, and at that instant they will match speeds. The receiver will then link up with the tanker and begin refueling.

Observing that both the position and the velocity of the receiver will equal those of the tanker at the time refueling begins, set up a system of equations for a and t (where a is the acceleration of the receiver and t is the time when the two aircraft meet). Solve your system to determine the following:

(a). How long will it take the receiver to catch up to the tanker?

(b). What is the deceleration (negative acceleration) of the receiver?

(c). Draw motion curves (position, velocity, and acceleration curves) for the receiver and the tanker and write a paragraph describing their motions in detail.

(d). Create a graph of the receiver's position and the tanker's position on the same axes. Also, define a function giving the distance between the receiver and tanker, and graph that function as well. Describe what these graphs illustrate.

Answer 1

A,B) Thus it takes 8 seconds to catch up and -62.5 feet per second^2 to deacclerate.

C. For the receiver the position vs time graph is inverted parabolic till t = 8 seconds and then it increases linearly with slope 500 feets per second as both tanker and receiver travel at same velocity of 500 feets per second.

The velocity of receiver decreases from 1000 feets per seconds for 8 seconds and when it reaches near the tanker it then travels with it at the speed of 500 feets per seconds.

The acceleration of the receiver is constant at -62.5 feet per second^2 upto t = 8 seconds and then it becomes zero at it catches up with the tanker.

While the tanker position increases with time linearly with a slop 500 feets per seconds. Thus the velocity is constant and accleration is zero all the time.

D. The position of receiver increases parabolically till t = 8 seconds and then it becomes a asymptote to the position curve of the tanker as both moves with same velocity after that. While the difference in the positions between the two increases from negative to 0, parabolically till t = 8 seconds and the remains constant after that.