Question 11.SP.13: Airplane B, which is travelling at a constant 560 km/h, is p...
Airplane B, which is travelling at a constant 560 km/h, is pursuing airplane A, which is travelling northeast at a constant 800 km/hr. At time t = 0, airplane A is 640 km east of airplane B. Determine (a) the direction of the course airplane B should follow (measured from the east) to intercept plane A, (b) the rate at which the distance between the airplanes is decreasing, (c) how long it takes for airplane B to catch airplane A.
STRATEGY: To find when B intercepts A, you just need to find out
when the two planes are at the same location. The rate at which the distance is decreasing is the magnitude of v_{B/A} , so you can use the relative velocity equation.
MODELING and ANALYSIS: Choose x to be east, y to be north, and
place the origin of your coordinate system at B (Fig. 1).
Positions of the Planes: You know that each plane has a constant speed, so you can write a position vector for each plane. Thus,
a. Direction of B. Plane B will catch up when they are at the same location, that is, r_{A}= r_{B}. You can equate components in the j direction to find
v_{A} \sin 45^{\circ} t_{1}=v_{B} \sin \theta t_{1}After you substitute in values,
b. Rate. The rate at which the distance is decreasing is the magnitude
of v_{B/A}, so
c. Time for B to catch up with A. To find the time, you equate the i components of each position vector, giving
\left(v_{A} \cos 45^{\circ}\right) t_{1}+640 \mathrm{~km}=\left(v_{B} \cos \theta\right) t_{1}Solve this for t_{1}. Thus,
REFLECT and THINK: The relative velocity is only in the horizontal (eastern) direction. This makes sense, because the vertical (northern) components have to be equal in order for the two planes to intersect.
