Explanation Given data: y One airplane leaves Chicago at 12:00 P.M. due south at 500 miles/hour. Another airplane travelling at 650 miles/ hour to East arrives Chicago at 2:00 PM. Method/Formula used: Distance travelled by an object; D = speed×time . Calculation: The airplane A s travelling due south at 500 miles/hour leaves Chicago at 12:00 P.M. and the air plane A e travelling due east at 650 miles/ hour arrives Chicago at 2:00 P.M. Let when airplane A e leaves Chicago, the airplane A e was at point A that reaches at point O in two hours. Therefore, AO=650 × 2=1300 miles. The situation is shown in figure here. Further, let in time t the airplane reaches point B and the airplane reaches at point D so that the distance BD between two airplanes is the closest. Therefore, AD= 650 × t ⇒ DO=1300-650 t = 650 ( 2 − t ) and OB= 500 × t AD. The angle ∠ D O B = 90 0 . Therefore, from Pythagoras theorem D B 2 =D O 2 +O B 2 ⇒ D B 2 = { 650(2- t ) } 2 + ( 500 t ) 2 ⇒ DB=50 169(2- t ) 2 + 100 t 2 .... (1) This is the closest distance between two air planes in terms of “ t ”. Let we write DB = C C =50 169(2- t ) 2 + 100 t 2 ...

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Chapter 3.4, Problem 56A

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Chapter 3.4, Problem 55A

Chapter 3.4, Problem 57A

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Chapter 3 Solutions

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The time when two air planes were at the closest and the closest distance also.

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