A skateboarder moves in 1D; positive is uphill. A velocity–time graph is a straight line that goes from at to at .
What is the skateboarder’s acceleration during this interval?
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AP Physics 1 Quiz
Practice Displacement Velocity And Acceleration in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A skateboarder moves in 1D; positive is uphill. A velocity–time graph is a straight line that goes from v=0m/s at t=0s to v=6m/s at t=3s.
What is the skateboarder’s acceleration during this interval?
This quiz focuses on Displacement Velocity And Acceleration, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A skateboarder moves in 1D; positive is uphill. A velocity–time graph is a straight line that goes from v=0m/s at t=0s to v=6m/s at t=3s.
What is the skateboarder’s acceleration during this interval?
Explanation: This question examines determining acceleration from a velocity-time graph in one-dimensional motion. Acceleration is the slope of the velocity-time graph, given by the change in velocity over the change in time. From v=0 m/s at t=0 s to v=6 m/s at t=3 s, Δv = 6 m/s - 0 m/s = 6 m/s and Δt = 3 s - 0 s = 3 s, so a = 6 m/s / 3 s = 2 m/s², corresponding to choice C. The positive acceleration means the velocity is increasing uphill, consistent with the direction. Choice A, 18 m/s², is erroneous, likely from multiplying velocity and time instead of dividing Δv by Δt. A transferable approach is to always calculate the slope of v-t graphs for acceleration, verifying units match m/s².
A cart moves in 1D; positive is to the right. It travels from x=0m to x=5m, then back to x=2m.
What is the cart’s total distance traveled?
Explanation: This question tests distinguishing between displacement and total distance traveled in one-dimensional motion. Total distance is the sum of absolute path lengths, not the net change. The cart moves from 0 m to 5 m (distance 5 m), then back to 2 m (distance 3 m), totaling 5 m + 3 m = 8 m, which is choice D. This accounts for the actual ground covered in both directions. Choice C, 7 m, is wrong, perhaps from adding initial and final positions without considering the reversal. To solve path-dependent problems, break the motion into segments and sum the absolute distances, unlike net displacement which is just final minus initial position.
A cyclist moves along a straight road; positive is east. A velocity–time graph is a straight line decreasing from v=+6m/s at t=0s to v=0m/s at t=3s.
What is the cyclist’s displacement from t=0 to t=3s?
Explanation: This question evaluates calculating displacement from a velocity-time graph with changing velocity. Displacement is the area under the v-t curve, here a triangle decreasing from 6 m/s to 0 m/s over 3 s. The area is (1/2) * base * height = (1/2) * 3 s * 6 m/s = 9 m, or using average velocity: v_avg = (6 + 0)/2 = 3 m/s, then Δx = 3 m/s * 3 s = 9 m, matching choice B. The positive displacement indicates net eastward movement as velocity slows to stop. Choice A, 18 m, is incorrect, likely from multiplying maximum velocity by time without averaging or using the triangle area formula. For decelerating motion on graphs, use geometric area or average velocity times time as a consistent method to find displacement.
A cart moves along a straight track; positive is to the right. Its position is recorded: at t=0s, x=0m; at t=2s, x=6m; at t=5s, x=3m. Assume the cart moves smoothly between these times.
What is the cart’s average velocity from t=2s to t=5s?
Explanation: This question assesses the skill of calculating average velocity from position-time data in one-dimensional motion. Average velocity is defined as the change in position divided by the change in time, so for the interval from t=2 s to t=5 s, Δx = 3 m - 6 m = -3 m and Δt = 5 s - 2 s = 3 s. Thus, the average velocity v_avg = Δx / Δt = -3 m / 3 s = -1.0 m/s, which corresponds to choice B. The negative sign indicates the cart is moving to the left, toward the negative direction, even though it initially moved right. Choice A, +1.0 m/s, is incorrect because it ignores the direction of the displacement, treating it as positive when the cart actually moves leftward. To solve similar problems, always use the formula v_avg = (x_final - x_initial) / (t_final - t_initial) and pay attention to signs for direction.
A ball rolls in 1D; positive is to the right. A velocity–time graph is a horizontal line at v=2m/s from t=0 to t=5s.
What is the ball’s acceleration during 0≤t≤5s?
Explanation: This question assesses finding acceleration from a velocity-time graph in uniform motion. Acceleration is the slope of the v-t graph, and a horizontal line indicates constant velocity. The graph at v=2 m/s from t=0 to t=5 s has Δv=0 over Δt=5 s, so a=0 m/s², which is choice D. This means no change in velocity, consistent with uniform motion to the right. Choice A, 2 m/s², is wrong, possibly from confusing velocity value with acceleration or misreading the graph. A reliable strategy is to compute the slope of v-t plots for acceleration; zero slope always means zero acceleration in kinematics.
A cart moves along a line; positive is to the right. A position–time graph has a horizontal segment (constant x) from t=4s to t=6s.
During t=4s to t=6s, what is the cart’s velocity?
Explanation: This question assesses interpreting velocity from a position-time graph in linear motion. A horizontal segment on a position-time graph indicates constant position, meaning no change in x over time. Therefore, the velocity, which is the slope of this segment, is zero, as Δx = 0 over Δt = 2 s, giving v = 0 m/s, which is choice C. This implies the cart is at rest during that interval, regardless of prior motion. Choice A, positive and constant, is incorrect, as it might stem from confusing constant position with constant positive velocity, ignoring the zero slope. For similar graphs, identify velocity by examining the slope: flat lines mean zero velocity, a key strategy in kinematics analysis.
A robot moves along the x-axis; positive is to the right. Its velocity is v(t)=−3m/s for 0≤t≤4s.
Which statement about the robot’s motion is correct?
Explanation: This question tests understanding of velocity and speed in constant-velocity motion. Velocity is a vector with magnitude and direction, while speed is the magnitude alone. Given v(t) = -3 m/s, the robot moves left (negative direction) at a constant speed of 3 m/s, as stated in choice B. This describes steady motion without changing speed or direction during the interval. Choice A is incorrect because speed cannot be negative; it mistakenly equates speed with velocity. To analyze motion, distinguish between velocity (directional) and speed (scalar), using signs to determine direction in one-dimensional problems.
A toy car moves along the x-axis; positive is to the right. Its position as a function of time is shown by a straight line on an x vs. t graph passing through (t=1s,x=2m) and (t=5s,x=10m).
What is the car’s velocity during this interval?
Explanation: This question tests the skill of finding velocity from a position-time graph in straight-line motion. Velocity is the slope of the position-time graph, calculated as the change in position over the change in time. Using the points (1 s, 2 m) and (5 s, 10 m), Δx = 10 m - 2 m = 8 m and Δt = 5 s - 1 s = 4 s, so v = 8 m / 4 s = 2 m/s, which is choice A. The positive value indicates constant motion to the right throughout the interval. Choice B, 8 m/s, is incorrect, possibly from miscalculating Δx as the final position without subtracting the initial or using incorrect points. Remember, for linear graphs, compute slope as rise over run, ensuring to use Δx/Δt for velocity in kinematics problems.
A car moves on a straight road; positive is north. At t=0s the car is at x=5m, and at t=8s it is at x=−3m.
What is the car’s displacement over the 8s interval?
Explanation: This question evaluates calculating displacement in one-dimensional motion using position data. Displacement is the change in position, Δx = x_final - x_initial, independent of the path taken. Here, x_initial = 5 m at t=0 s and x_final = -3 m at t=8 s, so Δx = -3 m - 5 m = -8 m, matching choice B. The negative value indicates a net movement southward over the interval. Choice A, 8 m, is wrong because it represents the magnitude (distance) rather than the directed displacement. Always use Δx = x_f - x_i for displacement problems, remembering it includes direction unlike total distance traveled.
A runner moves on a straight path; positive is east. A velocity–time graph shows v=+4m/s constant from t=0 to t=3s, then v=−2m/s constant from t=3 to t=7s.
What is the runner’s displacement from t=0 to t=7s?
Explanation: This question evaluates the ability to determine displacement from a velocity-time graph in one-dimensional kinematics. Displacement is the area under the velocity-time graph, calculated separately for each segment. From t=0 to t=3 s, the area is (4 m/s) * (3 s) = +12 m; from t=3 to t=7 s, it is (-2 m/s) * (4 s) = -8 m, yielding a total displacement of +12 m - 8 m = +4 m, matching choice B. The positive net displacement indicates an overall eastward movement despite the westward segment. Choice C, -4 m, is wrong as it might result from mistakenly adding the areas without considering signs or reversing the directions. A useful strategy for these problems is to break the graph into geometric shapes and sum their signed areas to find net displacement.
A robot moves along a straight line, with +x to the right. Its velocity changes from v=+5m/s at t=1s to v=−1m/s at t=4s.
What is the robot’s average acceleration from t=1s to t=4s?
Explanation: This question tests determining average acceleration from velocity changes in straight-line motion. Average acceleration is change in velocity divided by change in time. From v=+5 m/s at t=1 s to v=-1 m/s at t=4 s, Δv = -1 - 5 = -6 m/s, Δt=3 s, so a_avg = -6/3 = -2 m/s². The negative value indicates deceleration or direction change toward the left. Choice D, +6 m/s², is incorrect by using absolute Δv and ignoring signs. Always calculate a_avg as Δv/Δt, including signs to capture direction.
A cart moves along a straight track, with +x to the right. A motion sensor records the cart’s position as a function of time: x=0m at t=0s, x=4m at t=2s, and x=4m at t=6s. The cart then moves to x=1m at t=8s. Assume the motion between listed times is continuous.
What is the cart’s average velocity from t=6s to t=8s?
Explanation: This question assesses the skill of calculating average velocity from position-time data in one-dimensional kinematics. Average velocity is defined as the change in position divided by the change in time over the specified interval. From t=6 s to t=8 s, the position changes from x=4 m to x=1 m, so Δx = 1 m - 4 m = -3 m and Δt = 8 s - 6 s = 2 s, yielding v_avg = -3 m / 2 s = -1.5 m/s. This negative value indicates the direction of the average velocity is to the left, consistent with the +x to the right convention. Choice C, +3 m/s, is incorrect because it uses the magnitude of the displacement without considering the direction, ignoring the sign of Δx. To find average velocity over any interval, always compute Δx/Δt, ensuring to account for the direction in the coordinate system.
A cart moves along a straight line, with +x to the right. It travels from x=0m to x=5m, then back to x=2m. All motion is along the same line.
What is the cart’s displacement for the entire trip?
Explanation: This question examines the distinction between displacement and distance in one-dimensional kinematics. Displacement is the net change in position, final minus initial, regardless of path. The cart goes from x=0 m to 5 m then back to 2 m, so Δx = 2 m - 0 m = +2 m. This positive value shows net motion to the right. Choice A, -4 m, might confuse total distance with displacement or misapply signs. Remember, displacement is vectorial; compute it as x_final - x_initial for transferable accuracy.