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3 min readโขjune 18, 2024
๐ Welcome to theย AP Physics Unit 1 FRQ (Kinematics) Answers.ย Have your responses handy as you go through the rubrics to see how you did!
โฑ Remember, the AP Physics 1 exam has 5 free-response questions, and you will be given 90 minutes to complete the FRQ section. (This means you shouldย give yourself ~18 minutes to go through each practice FRQ.)
https://www.youtube.com/watch?v=fhOqbAF1Uis
A student of massย m stands on a platform scale in an elevator in a tall building. The positive direction for all vector quantities is upward. The vertical position of an elevator as a function of time is shown below. The velocity at 6 seconds is -30 m/s.
1pt: For a straight line starting at 20 m/s at t=2, and ending at 0 m/s at t=4
1pt: For a straight line starting at 0 m/s at t=4, and ending at -30 m/s at t=6 ๐ Additional Resources
Replay:ย Velocity vs. Time Graphs
1pt: For a description or equation for acceleration either as the slope of theย v vs.ย t graph, or a = โv / โt
1pt: For the work to determine the acceleration, a = (0-20) / (4-2)
1pt: For the correct answer with units, a = -10 m/s^2 ๐ Additional Resources
Study Guide:ย Position, Velocity, and Acceleration
An inspector provides the student with the following graph showing the accelerationย a of the elevator as a function of timeย t.
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3 min readโขjune 18, 2024
๐ Welcome to theย AP Physics Unit 1 FRQ (Kinematics) Answers.ย Have your responses handy as you go through the rubrics to see how you did!
โฑ Remember, the AP Physics 1 exam has 5 free-response questions, and you will be given 90 minutes to complete the FRQ section. (This means you shouldย give yourself ~18 minutes to go through each practice FRQ.)
https://www.youtube.com/watch?v=fhOqbAF1Uis
A student of massย m stands on a platform scale in an elevator in a tall building. The positive direction for all vector quantities is upward. The vertical position of an elevator as a function of time is shown below. The velocity at 6 seconds is -30 m/s.
1pt: For a straight line starting at 20 m/s at t=2, and ending at 0 m/s at t=4
1pt: For a straight line starting at 0 m/s at t=4, and ending at -30 m/s at t=6 ๐ Additional Resources
Replay:ย Velocity vs. Time Graphs
1pt: For a description or equation for acceleration either as the slope of theย v vs.ย t graph, or a = โv / โt
1pt: For the work to determine the acceleration, a = (0-20) / (4-2)
1pt: For the correct answer with units, a = -10 m/s^2 ๐ Additional Resources
Study Guide:ย Position, Velocity, and Acceleration
An inspector provides the student with the following graph showing the accelerationย a of the elevator as a function of timeย t.
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