Explanation Given Information: The distance ( r 0 ) of the helicopter is 3000 ft . The angle of elevation ( θ 0 ) of the helicopter is 20 ° . The distance ( r 4 ) of the helicopter after 4 sec is 3320 ft . The angle of elevation ( θ 4 ) of the helicopter after 4 sec is 23.1 ° . The time interval ( Δ t ) is 4 sec. Calculation: Show position of helicopter at r 0 and r 4 as in Figure (1). Calculate the change in radius ( Δ r ) by apply cosine rule for Figure (1): Δ r 2 = r 0 2 + r 4 2 − 2 r 0 r 4 cos ( θ 4 − θ 0 ) Substitute 3000 ft for r 0 , 20 ° for θ 0 , 3320 ft for r 4 and 23.1 ° for θ 4 . Δ r 2 = 3000 2 + 3320 2 − 2 × ( 3000 ) × ( 3320 ) cos ( 23.1 ° − 20 ° ) Δ r 2 = 9 , 000 , 000 + 11 , 022 , 400 − 19 , 890 , 850.47 Δ r = 131 , 549.53 Δ r = 362.7 ft Calculate the average speed ( v ave ) of the helicopter using the relation: v ave = Δ r Δ t Substitute 362

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 11.5, Problem 11.172P

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The average speed (vavg) and the angle of climb β of the helicopter during the 4 sec of time (t) interval.

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Chapter 11.5, Problem 11.171P

Chapter 11.5, Problem 11.173P

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Vector Mechanics for Engineers: Statics and Dynamics

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The average speed ( v a v g ) and the angle of climb β of the helicopter during the 4 sec of time ( t ) interval.

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The average speed (vavg) and the angle of climb β of the helicopter during the 4 sec of time (t) interval.

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