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AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 1: KinematicsVectors & MotionVectors & Motion in Two Dimensions

Vectors & Motion in Two Dimensions (College Board AP® Physics 1: Algebra-Based)

Study Guide

Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Analyzing motion in two dimensions

  • The same kinematic equations can be used to analyze motion in two dimensions

  • Two-dimensional motion can be separated into one-dimensional components

    • The kinematic equations can be used on each component

Diagram of a two-dimensional vector A with components Ax and Ay. The vector forms a right triangle with these components on an x-y coordinate plane. Labels identify 1D and 2D vectors.
Two-dimensional vectors can be split into component vectors

Worked Example

A carrier jet is an aircraft that takes off from a large boat called an aircraft carrier. The jet is catapulted from the aircraft carrier to reach the velocity required for takeoff on such a short runway.

At the moment of takeoff, a jet has a horizontal acceleration of 25 space straight m divided by straight s squared and a vertical acceleration of 11 space straight m divided by straight s squared. The jet's initial velocity is 64 space straight m divided by straight s in the horizontal direction and 19 space straight m divided by straight s in the vertical direction.

Calculate the resultant final velocity of the jet after 7 space straight s.

Answer:

Step 1: Analyze the scenario

  • To find the resultant final velocity, the component final velocities must first be calculated

Step 2: List the known quantities

  • Taking forward and upward to be positive

  • Horizontal acceleration, a subscript x space equals space 25 space straight m divided by straight s squared

  • Horizontal initial velocity, v subscript x space 0 end subscript space equals space 64 space straight m divided by straight s

  • Vertical acceleration, a subscript y space equals space 11 space straight m divided by straight s squared

  • Vertical initial velocity, v subscript y space 0 end subscript space equals space 19 space straight m divided by straight s

  • Time interval, t space equals space 7 space straight s

Step 3: Select a suitable kinematic equation to calculate the final velocity components

  • Displacement is not required

v subscript x space equals space v subscript x space 0 end subscript space plus space a subscript x t

v subscript x space equals space 64 space plus space open parentheses 25 space times space 7 close parentheses space equals space 239 space straight m divided by straight s

v subscript y space equals space 19 space plus space open parentheses 11 space times space 7 close parentheses space equals space 96 space straight m divided by straight s

Step 4: Calculate the magnitude of the resultant final velocity

v subscript x space plus space y end subscript space equals space square root of 239 squared space plus space 96 squared end root space equals space 258 space straight m divided by straight s

Step 5: Calculate the direction of the resultant final velocity

v subscript x plus y end subscript space equals space tan to the power of negative 1 end exponent open parentheses v subscript y over v subscript x close parentheses

v subscript x space plus space y end subscript space equals space tan to the power of negative 1 end exponent space open parentheses 96 over 239 close parentheses space equals space 22 degree from the horizontal

Step 6: State the resultant final velocity

v subscript x space plus space y end subscript space equals space 258 space straight m divided by straight s space at space 22 degree space to space the space horizontal

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    Author
    Reviewer
    Leander Oates

    Author: Leander Oates

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.