AP PHYSICS: Vector Addition

Q: An ant travels long distances each day in search of food. On one such occasion, the ant travels 3.0 meters at an angle of 30⁰ north of east. Afterward, the ant travels 2.0 meters to the north. Finally, the ant finds food after traveling 2.0 meters at an angle of 60⁰ south of west. What is the ant’s final position relative to its starting point?

A: The ant’s path of travel is as follows:

A total distance of 7.0 meters was traveled prior to arrival at the final destination. This location was identified using tip-to-tail vector addition. What if, however, we want a quantitative displacement value relative to the starting point? After breaking each vector into x-and-y components, we subsequently add the components to determine the net x/y displacement values. We subsequently use the Pythagorean Theorem to determine the magnitude of displacement from the starting point:

( 3.0 m )( cos 30⁰ ) + ( 2.0 m )( cos 90⁰ ) + ( 2.0 m )( cos 240⁰ ) = Net x-component

2.6 m + 0 m + ( -1.0 m ) = 1.6 m E

( 3.0 m )( sin 30⁰ ) + ( 2.0 m )( sin 90⁰ ) + ( 2.0 m )( sin 240⁰ ) = Net y-component

1.5 m + 2.0 m + ( -1.7 m ) = 1.8 m N

Finalmente,

a² + b² = c²

( 1.6 m E )² + ( 1.8 m N )² = ?

2.6 m² E² + 3.2 m² N² = ?

Note: Even though the east and west magnitudes were included in the derivation up to this point, they were not necessary. Taking note of the directions of travel makes it easier to visualize the appropriate orientation of the final vector within the x/y plane.

? = √( 5.8 m ) = 2.4 m

We now use the tangent θ function to determine which quadrant the final vector is located:

tan θ = opp/adj

tan θ = ( 1.8 m / 1.6 m )

tan θ = ( 1.1 )

θ = 48⁰

The ant’s net displacement is 2.4 m at an angle of 48⁰ relative to the x-axis.