Question

Starting from the relative position shown, aircraft B is to rendezvous with the refueling tanker A. If B is to arrive in close proximity to A in a 2.9- minute time interval, what absolute velocity vector should B acquire and maintain? The velocity of the tanker A is 308 mi/hr along the constant-altitude path shown. 10170 2330 B Answer: V (i i+ j) mi/hr i

Answer 1

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Answer 2

SOLUTION :

Let aircraft have absolute velocity Vx and Vy in horizontal and vertical direction respectively to rendezvous with tanker air craft A in 2.9 min. , which is moving horizontally at V = 308 mi/h.

Aircraft speeds are usually in nautical miles per hour.

1 nautical miles = 6076.12 feet 

Required relative velocity of B to A in horizontal direction

= distance / time 

= 10170 / (2.9 / 60) feet/hour

=  210413.79 feet/hr

= 210413.79 / 6076.12 mi/hour

= 34.63 mi/hour

So,

Absolute Vx = 308 + 34.63 = 342.63

Relative velocity required for B to A in vertical direction 

= distance / time

= 2330 / (2.9/60) feet/hour

= 48206.90 feet/hour

= 48206.90/6076.12 mi/hour

= 7.93 miles per hour 

B has zero velocity in vertical direction.

So, 

Absolute Vy = 7.83 + 0 = 7.93 mi/hour.

Hence absolute Velocity of B to meet A :

= V(B) = (342.63 I + 7.93 j) mi/hour (ANSWER).

After rendezvous, B should maintain the same speed as that of A 

= 308 mi/hour in horizontal direction. (ANSWER).