AP Physics 1 Quiz: Representing Motion

What this quiz covers

This quiz focuses on Representing Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

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.

Question 1

1. The ball’s velocity is constant and negative. (correct answer)
2. The ball’s acceleration is constant and negative.
3. The ball is slowing down because its position is decreasing.
4. The ball must cross $x=0$ during the interval.

Explanation: This question assesses the skill of representing motion using position-time graphs in AP Physics 1. To extract motion information, the slope of the position-time graph indicates velocity. A straight line with constant negative slope means constant negative velocity, as position decreases at a steady rate. From t=0 to t=2 s, this linear graph with negative slope supports unchanging speed in the negative direction. A common distractor is choice B, which assumes negative acceleration from the negative slope, but constant slope means zero acceleration. A transferable strategy is to distinguish between the sign of the slope for velocity direction and changes in slope for acceleration in position-time graphs.

Question 2

1. The robot moves in the negative direction and its speed decreases. (correct answer)
2. The robot moves in the negative direction and its speed increases.
3. The robot’s velocity is positive because the curve is concave up.
4. The robot’s acceleration is zero because the graph is smooth.

Explanation: This question assesses interpreting motion from position-time graphs with decreasing concave up curves. On a position-time graph, the slope represents velocity, and a decreasing curve that is concave up means the slope starts negative but becomes less negative over time. The robot moves in the negative direction (negative slope) but its speed decreases because the magnitude of the negative slope decreases. The concave up shape indicates positive acceleration, which reduces the magnitude of negative velocity. Choice B incorrectly suggests increasing speed, but concave up while decreasing means the negative velocity magnitude decreases (speed decreases).

Question 3

1. The cart moves in the positive direction while slowing down. (correct answer)
2. The cart moves in the negative direction while slowing down.
3. The cart’s velocity is constant because the graph increases.
4. The cart’s acceleration is zero because the graph never decreases.

Explanation: This question assesses interpreting motion from position-time graphs that are concave down while increasing. On a position-time graph, increasing position indicates positive velocity (positive direction motion), while concave down curvature means the slope becomes less positive over time—velocity decreases. The cart moves in the positive direction while slowing down because positive acceleration decreases or becomes negative. The cart continues forward but at a decreasing rate. Choice B incorrectly suggests negative direction motion, but increasing position indicates positive direction movement.

Question 4

1. The cart’s velocity is constant during that section. (correct answer)
2. The cart’s acceleration is constant and nonzero during that section.
3. The cart is at rest during that section because the graph is above $x=0$.
4. The cart moves backward during that section because $x$ is increasing.

Explanation: This question tests recognizing constant velocity from linear sections of position-time graphs. On a position-time graph, the slope represents velocity, and a straight line section indicates constant slope, meaning constant velocity during that interval. When the overall increasing graph contains a linear section, the cart moves with constant velocity during that specific section, even though it may have different velocities before or after. The straight line portion shows uniform motion. Choice C incorrectly assumes rest from position above zero, but position values don't determine motion state—slope does.

Question 5

1. The ball’s acceleration is constant and positive. (correct answer)
2. The ball’s speed is constant because the graph is linear.
3. The ball is at rest for the entire interval because it crosses $v=0$.
4. The ball’s position is constant because the graph crosses the axis.

Explanation: This question assesses recognizing constant acceleration from velocity-time graphs crossing zero. On a velocity-time graph, the slope represents acceleration, and a straight line indicates constant slope, meaning constant acceleration. Since the line slopes upward and crosses from negative to positive velocity, the ball has constant positive acceleration throughout, causing it to change direction when velocity passes through zero. The straight line confirms that acceleration remains constant even as velocity changes sign. Choice C incorrectly assumes the ball is at rest because it crosses v=0, but crossing zero velocity is just an instant of direction change, not a permanent rest state.

Question 6

1. The cart has zero acceleration. (correct answer)
2. The cart has increasing acceleration.
3. The cart is at rest because the line is flat.
4. The cart’s position is constant because velocity is constant.

Explanation: This question assesses interpreting zero acceleration from horizontal velocity-time graphs. On a velocity-time graph, the slope represents acceleration, and a horizontal line has zero slope, indicating zero acceleration. Though the line is above v=0, meaning the cart moves with constant positive velocity, the horizontal nature confirms that acceleration is zero—velocity doesn't change over time. The cart moves at constant speed in the positive direction. Choice B incorrectly suggests increasing acceleration, but a horizontal line indicates zero acceleration, not increasing acceleration.

Question 7

1. The car moves in the negative direction at constant velocity. (correct answer)
2. The car accelerates negatively because $v$ is negative.
3. The car’s position must be negative because $v$ is negative.
4. The car is slowing down because the graph is below zero.

Explanation: This question assesses interpreting constant velocity from velocity-time graphs below zero. On a velocity-time graph, the vertical axis represents velocity, and a horizontal line indicates constant velocity over time. Since the line is horizontal at a negative value, the car moves with constant negative velocity—constant speed in the negative direction. The car's acceleration is zero because velocity remains unchanged (horizontal line has zero slope). Choice B incorrectly assumes negative acceleration from negative velocity, but acceleration depends on the slope of the v-t graph, not the velocity value.

Question 8

1. The cart’s acceleration is positive but decreasing in magnitude. (correct answer)
2. The cart’s velocity is constant because it stays positive.
3. The cart is slowing down because the curve bends downward.
4. The cart’s position is constant because the graph is above zero.

Explanation: This question assesses interpreting motion from velocity-time graphs with concave down curvature above zero. On a velocity-time graph, velocity values above v=0 indicate positive direction motion, while concave down curvature means the slope (acceleration) becomes less positive over time. The cart has positive but decreasing acceleration—it speeds up but at a decreasing rate. The cart moves in the positive direction with acceleration that diminishes toward zero. Choice C incorrectly concludes the cart is slowing down, but positive velocity with positive (though decreasing) acceleration means the cart still speeds up, just less rapidly over time.

Question 9

1. The object’s velocity is zero. (correct answer)
2. The object’s acceleration is constant and nonzero.
3. The object’s speed increases because time increases.
4. The object moves in the negative direction because the slope is zero.

Explanation: This question tests recognizing zero velocity from position-time graphs with zero slope. On a position-time graph, the slope represents velocity, and a straight line with zero slope (horizontal) indicates zero velocity. When the slope is zero, the object remains at rest at a constant position throughout the time interval. The object's acceleration is also zero since velocity remains constant at zero. Choice B incorrectly suggests nonzero acceleration, but zero slope indicates both zero velocity and zero acceleration.

Question 10

1. The car’s velocity is zero during the interval shown. (correct answer)
2. The car moves with constant speed of $5\,\text{m/s}$.
3. The car accelerates because $x$ is not zero.
4. The car moves in the negative direction because $x$ is positive.

Explanation: This question tests recognizing zero velocity from horizontal position-time graphs. On a position-time graph, the slope represents velocity, and a horizontal line has zero slope, indicating zero velocity. When the graph is horizontal at x=5m, the car remains at a constant position of 5m, meaning its velocity is zero during the entire interval shown. The car is at rest at that specific location. Choice B incorrectly interprets the constant position value as constant speed, but zero slope means zero velocity, not constant speed.

Question 11

1. The object moves in the negative direction and speeds up. (correct answer)
2. The object moves in the positive direction and speeds up.
3. The object’s velocity is constant because the graph is smooth.
4. The object is at rest because the curve is concave up.

Explanation: This question tests interpreting motion from position-time graphs that are concave up while decreasing. On a position-time graph, decreasing values indicate negative velocity (negative direction motion), while concave up curvature means the slope becomes less negative over time. The object moves in the negative direction and speeds up because the magnitude of negative velocity increases (slope becomes more negative). The concave up shape while decreasing indicates the object accelerates in the negative direction. Choice B incorrectly suggests positive direction motion, but decreasing position values clearly indicate negative direction movement.

Question 12

1. The sprinter’s acceleration increases over time. (correct answer)
2. The sprinter’s velocity is constant because $v$ stays positive.
3. The sprinter’s acceleration is zero because the graph is smooth.
4. The sprinter slows down because the graph is curved.

Explanation: This question assesses interpreting acceleration from velocity-time graphs with increasing slope. On a velocity-time graph, the slope represents acceleration, and a curve that gets steeper with time indicates that acceleration increases over time. The sprinter experiences increasing acceleration, meaning the rate at which velocity increases is itself increasing. This produces motion where speed increases at an accelerating rate. Choice B incorrectly concludes constant velocity from positive v values, but constant velocity would appear as a horizontal line, while this curved upward graph shows changing (increasing) acceleration.

Question 13

1. The cart’s velocity is increasing over time. (correct answer)
2. The cart’s velocity is constant because $x$ increases steadily.
3. The cart’s acceleration is zero because the graph never decreases.
4. The cart moves backward because the graph is not a straight line.

Explanation: This question tests recognizing increasing velocity from position-time graphs with concave up curvature. On a position-time graph, the slope represents velocity, and a curve that is concave up while increasing means the slope becomes more positive over time—velocity increases. The cart's speed increases because the rate of position change (velocity) increases over time. The concave up shape indicates positive acceleration. Choice B incorrectly concludes constant velocity from steadily increasing position, but constant velocity would produce a straight line, not a concave up curve.

Question 14

1. The marble’s acceleration is positive because $x$ increases.
2. The marble’s velocity decreases over time. (correct answer)
3. The marble’s velocity is constant because the graph is continuous.
4. The marble is moving backward because the curve is not a line.

Explanation: This question assesses interpreting motion from position-time graphs with decreasing slope. On a position-time graph, the slope represents velocity, and when the slope becomes less positive over time, velocity decreases. Though position continues to increase (positive slope), the rate of increase slows down, indicating the marble's velocity decreases over time. The curve shows positive but decreasing velocity, meaning the marble slows down while moving in the positive direction. Choice A incorrectly assumes positive acceleration from increasing position, but acceleration depends on changes in velocity (slope), not position values.

Question 15

1. The cart has constant negative velocity. (correct answer)
2. The cart has constant negative acceleration.
3. The cart’s speed increases because the line goes downward.
4. The cart is at rest because the graph is a straight line.

Explanation: This question tests interpreting constant negative velocity from decreasing linear position-time graphs. On a position-time graph, the slope represents velocity, and a straight line indicates constant slope, meaning constant velocity. Since the graph decreases linearly, the cart has constant negative velocity—it moves with unchanging speed in the negative direction. The linear nature confirms zero acceleration since velocity doesn't change. Choice B incorrectly identifies negative acceleration, but constant negative velocity means zero acceleration.

Question 16

1. The cart speeds up in the positive direction. (correct answer)
2. The cart slows down because the slope is positive.
3. The cart’s velocity is constant because the graph is a line.
4. The cart moves in the negative direction because acceleration is positive.

Explanation: This question assesses interpreting motion from velocity-time graphs with positive velocity and positive slope. On a velocity-time graph, positive velocity indicates motion in the positive direction, and positive slope indicates positive acceleration. When both velocity and acceleration are positive, the object moves in the positive direction and speeds up—both quantities reinforce each other. The cart accelerates in the positive direction, increasing its positive velocity. Choice C incorrectly concludes constant velocity from a straight line, but on a v-t graph, a straight line indicates constant acceleration, not constant velocity.