Explanation Given Info: The travelled distances in respective directions shown by figure. Figure (I) Flight Distances are, A → = 170 km A → is the displacement in east of north B → = 230 km B → is the displacement in south of east Formula to calculate the sum of all the vector displacement is, A → + B → = R → R → is the distance from starting point to end point Formula of standard form of vector A → in 2 direction is, A → = A x → i ^ + A y → j ^ (I) A x → is a vector component in x direction A y → is a vector component in y direction i ^ is the unit vector in x direction j ^ is the unit vector in y direction From figure, Component of vector A → in positive x direction is, A x = A → sin θ Substitute 170 km A → and 68 ° for θ A x = 170 sin 68 ° = 157.62 km Component of vector A → in positive y direction is, A y = A → cos θ Substitute 170 km A → and 68 ° for θ A y = 170 cos 60 ° = 63.68 km Substitute 157 .62 km for A x → and 63 .68 km for A y → in equation (I) to find A → A → = 157.62 i ^ + 63.68 j ^ Formula of standard form of vector B → in 2 direction is, B → = B x → i ^ + B y → j ^ (II) B x → is a vector component in x direction B y → is a vector component in y direction Component of vector B → in positive x direction is, B x = B → cos θ Substitute 230 km B → and 36 ° for θ B x = 230 cos 36 ° = 186 km Component of vector B → in positive y direction is, B y = B → sin θ Substitute − 230 km B → and 36 ° for θ B y = − 230 sin 36 ° = − 135

Sears And Zemansky's University Physics With Modern Physics

13th Edition

ISBN: 9780321897961

Author: YOUNG, Hugh D./

Publisher: Pearson College Div

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Chapter 1, Problem 68E

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The distance to fly and the direction of flight to reach the final point of destination.

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Chapter 1, Problem 67E

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Sears And Zemansky's University Physics With Modern Physics

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