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Kinematics in 2D quiz

컨트롤 버튼이 '내비게이션' 모드로 변경되었습니다.
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  • What is the main strategy for solving kinematics problems in two dimensions?
    Break the problem into two one-dimensional motions (x and y directions) and solve them separately using the same equations as in 1D motion.
  • How do you decompose a two-dimensional vector into its components?
    Use trigonometric functions: the x-component is the magnitude times cos(theta), and the y-component is the magnitude times sin(theta).
  • What are the five key variables you list for each direction in a 2D kinematics problem?
    The five variables are displacement (Δx or Δy), initial velocity (v₀x or v₀y), final velocity (vfx or vfy), acceleration (ax or ay), and time (t).
  • If a hockey puck’s initial velocity is 8 m/s east, what are its initial velocity components?
    The x-component (east) is 8 m/s, and the y-component (north) is 0 m/s.
  • How do you calculate the x and y components of an acceleration vector at 3 m/s², 37° northeast?
    Ax = 3 × cos(37°) = 2.4 m/s², Ay = 3 × sin(37°) = 1.8 m/s².
  • Which kinematic equation is used to find displacement when final velocity is unknown?
    Use Δx = v₀x·t + 0.5·ax·t² (and similarly for Δy).
  • What is the formula for the magnitude of total displacement in two dimensions?
    The magnitude is √(Δx² + Δy²), using the Pythagorean theorem.
  • How do you find the direction (angle) of the displacement vector?
    Use θ = tan⁻¹(Δy/Δx).
  • What is the displacement in the x direction after 5 seconds for the hockey puck problem?
    Δx = 70 meters.
  • What is the displacement in the y direction after 5 seconds for the hockey puck problem?
    Δy = 22.5 meters.
  • What is the magnitude of the hockey puck’s displacement after 5 seconds?
    The magnitude is 73.5 meters.
  • At what angle (relative to east) does the hockey puck move after 5 seconds?
    The angle is 17.8 degrees north of east.
  • Why is the magnitude of the total displacement greater than either component?
    Because it is the hypotenuse of the right triangle formed by the x and y displacements.
  • Why can you use the same kinematic equations in 2D as in 1D?
    Because each direction (x and y) is treated independently, as separate one-dimensional motions.
  • What is the first step when solving a 2D kinematics problem?
    Draw a diagram and decompose all vectors into their x and y components.