Displacement, Velocity, and Acceleration - AP Physics C Mechanics

  1. What is the formula for final velocity squared in uniformly accelerated motion?
    vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a \Delta x. Derived from kinematic equations eliminating time.
  2. Define instantaneous velocity.
    Instantaneous velocity is the velocity at a specific point in time. The limit of average velocity as time interval approaches zero.
  3. What does a flat line on a velocity-time graph indicate?
    Constant velocity. Zero slope means no change in velocity over time.
  4. What does it mean if displacement is negative?
    The final position is less than the initial position. Object moved backward from its starting position.
  5. What is the initial velocity if v=20 m/sv = 20 \text{ m/s}, a=2 m/s2a = 2 \text{ m/s}^2, t=5 st = 5 \text{ s}?
    vi=10 m/sv_i = 10 \text{ m/s}. Using v=vi+atv = v_i + at and solving for viv_i.
  6. What is the effect of doubling time on displacement in uniform acceleration?
    Displacement quadruples. Since Δx∝t2\Delta x \propto t^2 for constant acceleration.
  7. Calculate the time for displacement Δx=50 m\Delta x = 50 \text{ m} with v=10 m/sv = 10 \text{ m/s}.
    t=5 st = 5 \text{ s}. Using t=Δxvt = \frac{\Delta x}{v} for constant velocity.
  8. Find velocity when displacement is 20 m20 \text{ m} and time is 4 s4 \text{ s}.
    v=5 m/sv = 5 \text{ m/s}. Using v=ΔxΔtv = \frac{\Delta x}{\Delta t} formula.
  9. Calculate the displacement given v=5 m/sv = 5 \text{ m/s} and t=10 st = 10 \text{ s}.
    Δx=50 m\Delta x = 50 \text{ m}. Using Δx=vt\Delta x = vt with constant velocity.
  10. What is the area under a velocity-time graph?
    The area represents displacement. Velocity times time gives change in position.
  11. How does velocity change when acceleration is opposite to velocity?
    Velocity decreases. Negative acceleration opposes motion direction.
  12. What does the area under an acceleration-time graph represent?
    Change in velocity. Acceleration times time gives velocity change.
  13. State the formula for velocity.
    v=ΔxΔtv = \frac{\Delta x}{\Delta t}. Change in position divided by time interval.
  14. What is the definition of displacement?
    Displacement is the change in position of an object. Vector quantity representing the straight-line distance between start and end points.
  15. How is acceleration related to velocity?
    Acceleration is the rate of change of velocity. Describes how quickly velocity changes over time.
  16. What is the formula for displacement from initial and final velocity and time?
    Δx=(vi+vf)2⋅t\Delta x = \frac{(v_i + v_f)}{2} \cdot t. Average velocity times time equals displacement.
  17. Calculate acceleration for vi=0 m/sv_i = 0 \text{ m/s}, vf=10 m/sv_f = 10 \text{ m/s}, and t=5 st = 5 \text{ s}.
    a=2 m/s2a = 2 \text{ m/s}^2. Using a=ΔvΔt=10−05a = \frac{\Delta v}{\Delta t} = \frac{10-0}{5}.
  18. Describe the motion when acceleration is zero.
    The object moves with constant velocity. No acceleration means velocity doesn't change.
  19. Calculate the acceleration given Δv=15 m/s\Delta v = 15 \text{ m/s} and Δt=3 s\Delta t = 3 \text{ s}.
    a=5 m/s2a = 5 \text{ m/s}^2. Using a=ΔvΔta = \frac{\Delta v}{\Delta t} formula directly.
  20. What does the slope of a position-time graph represent?
    The slope represents velocity. Rise over run gives rate of position change with time.
  21. What is the difference between speed and velocity?
    Velocity includes direction; speed does not. Velocity is a vector; speed is a scalar quantity.
  22. What does the slope of a velocity-time graph represent?
    The slope represents acceleration. Rise over run gives rate of velocity change with time.
  23. What is the definition of uniform acceleration?
    Acceleration that does not change over time. Constant rate of velocity change over time.
  24. What is the acceleration due to gravity on Earth?
    9.8 m/s29.8 \text{ m/s}^2. Standard gravitational acceleration value near Earth's surface.
  25. Find the time taken if vi=0v_i = 0, a=9.8 m/s2a = 9.8 \text{ m/s}^2, vf=19.6 m/sv_f = 19.6 \text{ m/s}.
    t=2 st = 2 \text{ s}. Using vf=vi+atv_f = v_i + at and solving for tt.
  26. Calculate displacement for vi=3 m/sv_i = 3 \text{ m/s}, a=2 m/s2a = 2 \text{ m/s}^2, t=4 st = 4 \text{ s}.
    Δx=28 m\Delta x = 28 \text{ m}. Using Δx=vit+12at2\Delta x = v_i t + \frac{1}{2}at^2 formula.
  27. State the formula for displacement in uniformly accelerated motion.
    Δx=vit+12at2\Delta x = v_i t + \frac{1}{2} a t^2. Kinematic equation for position with constant acceleration.
  28. What is the effect on velocity when acceleration is zero?
    Velocity remains constant. No acceleration means no velocity change occurs.
  29. Determine acceleration if vi=4 m/sv_i = 4 \text{ m/s}, vf=12 m/sv_f = 12 \text{ m/s}, t=2 st = 2 \text{ s}.
    a=4 m/s2a = 4 \text{ m/s}^2. Using a=vf−vita = \frac{v_f - v_i}{t} calculation.
  30. Define average velocity.
    Average velocity is the total displacement divided by total time. Total displacement over total time elapsed.
  31. What is the formula for average acceleration?
    a=ΔvΔta = \frac{\Delta v}{\Delta t}. Change in velocity divided by time interval.
  32. Find final velocity if vi=0v_i = 0, a=9.8 m/s2a = 9.8 \text{ m/s}^2, Δx=19.6 m\Delta x = 19.6 \text{ m}.
    vf=19.6 m/sv_f = 19.6 \text{ m/s}. Using vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a\Delta x equation.
  33. What is the formula for average velocity in terms of initial and final velocity?
    vavg=vi+vf2v_{avg} = \frac{v_i + v_f}{2}. Valid only for constant acceleration motion.
  34. Find final velocity if vi=5 m/sv_i = 5 \text{ m/s}, a=3 m/s2a = 3 \text{ m/s}^2, t=4 st = 4 \text{ s}.
    vf=17 m/sv_f = 17 \text{ m/s}. Using vf=vi+atv_f = v_i + at kinematic equation.

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