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AP Physics 1 Quiz

AP Physics 1 Quiz: Scalars And Vectors In One Dimension

Practice Scalars And Vectors In One Dimension in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Along a line, up is positive and down is negative. An elevator moves from y=+2 my=+2\,\text{m}y=+2m to y=−4 my=-4\,\text{m}y=−4m in 3 s3\,\text{s}3s. Which statement about average velocity and average speed is correct?

Select an answer to continue

What this quiz covers

This quiz focuses on Scalars And Vectors In One Dimension, 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.

All questions

Question 1

Along a line, up is positive and down is negative. An elevator moves from y=+2 my=+2\,\text{m}y=+2m to y=−4 my=-4\,\text{m}y=−4m in 3 s3\,\text{s}3s. Which statement about average velocity and average speed is correct?

  1. Average velocity is +2 m/s+2\,\text{m/s}+2m/s because the magnitude of displacement is 6 m6\,\text{m}6m.
  2. Average speed is −2 m/s-2\,\text{m/s}−2m/s because motion is downward.
  3. Average velocity is −2 m/s-2\,\text{m/s}−2m/s if the distance traveled equals 6 m6\,\text{m}6m.
  4. Average velocity is −2 m/s-2\,\text{m/s}−2m/s because displacement is −6 m-6\,\text{m}−6m. (correct answer)

Explanation: This question assesses the skill of distinguishing between scalars and vectors in one-dimensional motion. Scalars are magnitude-only, like average speed, which is total distance over time and always positive. Vectors include magnitude and direction, such as average velocity, which is displacement over time and can be negative. The core difference is that scalars do not consider direction, remaining positive, while vectors use signs for direction. One distractor, choice B, incorrectly assigns a negative sign to average speed, treating it as if it includes direction like a vector. A transferable strategy is to use displacement for vector calculations and distance for scalars to compute velocities and speeds accurately.

Question 2

A puck slides on a straight line where forward is positive. During an interval, its displacement is 0 m0\text{ m}0 m. Which statement correctly distinguishes scalar from vector reasoning?

  1. The puck’s distance traveled must be 0 m0\text{ m}0 m since displacement is zero.
  2. The puck’s average speed must be 0 m/s0\text{ m/s}0 m/s because displacement is zero.
  3. The puck could have traveled a nonzero distance even though its displacement is zero. (correct answer)
  4. The puck’s velocity was zero at all times because displacement is a scalar.

Explanation: This question tests understanding of scalars and vectors in one dimension. Displacement is a vector showing net position change, while distance is a scalar showing total path length. Zero displacement means the puck returned to its starting position, but it could have traveled any distance to get there (forward then back). Distance traveled can be nonzero even when displacement is zero, and average speed can be positive while average velocity is zero. Choice A incorrectly assumes zero displacement means no motion occurred, confusing the vector and scalar concepts. When displacement is zero, remember that the object returned to its starting point but may have taken any path to do so, accumulating positive distance along the way.

Question 3

On a line, right is positive. A cart’s velocity changes from vi=+4 m/sv_i=+4\text{ m/s}vi​=+4 m/s to vf=−1 m/sv_f=-1\text{ m/s}vf​=−1 m/s in 1 s1\text{ s}1 s. Which quantity is a vector?

  1. Speed
  2. Time
  3. Acceleration (correct answer)
  4. Mass

Explanation: This question tests understanding of scalars and vectors in one dimension. Acceleration is a vector quantity that represents the rate of change of velocity, including direction. The acceleration is calculated as (vf - vi)/t = (-1 - 4)/1 = -5 m/s², which is negative indicating leftward acceleration. Speed and mass are scalar quantities with only magnitude, while time is also a scalar. Among the choices, only acceleration has both magnitude and direction, making it the vector quantity. When identifying vectors versus scalars, remember that vectors have direction (can be positive or negative in 1D), while scalars only have magnitude.

Question 4

A ball moves along a line where up is positive and down is negative. During a time interval, its displacement is +4 m+4\text{ m}+4 m. Which statement must be true?

  1. The distance traveled is +4 m+4\text{ m}+4 m.
  2. The average velocity is positive for that interval. (correct answer)
  3. The speed is positive because displacement is positive.
  4. The ball never moved downward during the interval.

Explanation: This question tests understanding of scalars and vectors in one dimension. Displacement is a vector quantity showing net change in position with direction, while average velocity is displacement divided by time. Since displacement is +4 m (positive), the average velocity must also be positive because it equals positive displacement divided by positive time. Distance traveled could be 4 m or more (if the ball changed direction), and speed is always positive regardless of displacement direction. Choice D is incorrect because the ball could have moved down then up more, still yielding positive net displacement. To analyze motion with given displacement, remember that average velocity has the same sign as displacement, but the actual path taken could be more complex.

Question 5

On a straight track, east is positive and west is negative. A runner’s velocity changes from +3 m/s+3\,\text{m/s}+3m/s to −3 m/s-3\,\text{m/s}−3m/s over some time interval. Which statement correctly compares speed and velocity?

  1. The speed changes from +3 m/s+3\,\text{m/s}+3m/s to −3 m/s-3\,\text{m/s}−3m/s.
  2. The velocity changes direction, but the speed can remain 3 m/s3\,\text{m/s}3m/s. (correct answer)
  3. The speed must be zero whenever velocity is negative.
  4. Speed and velocity are identical, so both must change sign.

Explanation: This question assesses the skill of distinguishing between scalars and vectors in one-dimensional motion. Scalars have magnitude only, such as speed, which remains positive regardless of direction changes. Vectors have both magnitude and direction, like velocity, which changes sign when direction reverses. The distinction is that scalars disregard direction, staying the same even if direction flips, while vectors reflect directional changes. One distractor, choice A, incorrectly applies a sign change to speed, treating it like a vector. A transferable strategy is to check if the quantity changes when only direction reverses; if it stays the same, it's a scalar like speed.

Question 6

On a line, right is positive and left is negative. A ball has velocity −2 m/s-2\,\text{m/s}−2m/s for 4 s4\,\text{s}4s. Which statement about the ball’s change in position is correct?

  1. The ball’s displacement is +8 m+8\,\text{m}+8m because speed is 2 m/s2\,\text{m/s}2m/s.
  2. The ball’s displacement is −8 m-8\,\text{m}−8m because velocity includes direction. (correct answer)
  3. The ball’s distance traveled is −8 m-8\,\text{m}−8m because it moved left.
  4. The ball’s displacement is 0 m0\,\text{m}0m because velocity is constant.

Explanation: This question assesses the skill of distinguishing between scalars and vectors in one-dimensional motion. Scalars are magnitude-only, like distance traveled, which is positive and ignores direction. Vectors include magnitude and direction, such as displacement, calculated with signs from velocity and time. Scalars focus on absolute change, while vectors incorporate directional information. One distractor, choice C, incorrectly makes distance negative, treating it like a vector. A transferable strategy is to use the sign from velocity in displacement calculations, but take absolute value for distance.

Question 7

On a one-dimensional track, right is positive and left is negative. A runner’s velocity is v=−3 m/sv=-3\,\text{m/s}v=−3m/s for 2 s. Which statement is correct?

  1. The runner’s speed is −3 m/s-3\,\text{m/s}−3m/s because velocity is negative.
  2. The runner’s displacement is −6 m-6\,\text{m}−6m, a vector quantity. (correct answer)
  3. The runner’s displacement is +6 m+6\,\text{m}+6m because speed is positive.
  4. The runner’s distance traveled is −6 m-6\,\text{m}−6m because motion is left.

Explanation: This question tests understanding of scalars and vectors in one dimension. Scalars have only magnitude (always positive), while vectors have both magnitude and direction (can be positive or negative). With velocity v = -3 m/s for 2 s, the runner moves left (negative direction). Displacement = velocity × time = (-3 m/s)(2 s) = -6 m (a vector pointing left). Speed is the magnitude of velocity: |−3| = 3 m/s (scalar, always positive), and distance traveled = speed × time = 6 m (scalar). Choice B correctly identifies displacement as -6 m, a vector quantity. Remember that speed and distance are scalars (never negative), while velocity and displacement are vectors (use signs for direction).

Question 8

Along a straight hallway, north is positive and south is negative. A student walks from x=0x=0x=0 to x=+5 mx=+5\,\text{m}x=+5m, then to x=+2 mx=+2\,\text{m}x=+2m. Which quantity must be negative?

  1. The displacement for the entire trip
  2. The total distance traveled
  3. The displacement during the second part of the walk (correct answer)
  4. The average speed for the entire trip

Explanation: This question tests understanding of scalars and vectors in one dimension. Scalars have only magnitude (always positive), while vectors have both magnitude and direction (can be positive or negative). The student walks from x = 0 to x = +5 m (displacement₁ = +5 m), then from x = +5 m to x = +2 m (displacement₂ = +2 - 5 = -3 m). Total displacement = +2 m (vector), total distance = 5 + 3 = 8 m (scalar), and average speed = 8 m/time (scalar, positive). Only the displacement during the second part is negative (-3 m). To identify negative quantities, look for vectors where motion opposes the positive direction.

Question 9

Along a straight path, right is positive. A block moves with constant acceleration a=−1 m/s2a=-1\,\text{m/s}^2a=−1m/s2 for 3 s. Which statement best describes aaa?

  1. Acceleration is a scalar, so −1 m/s2-1\,\text{m/s}^2−1m/s2 means it is slowing down.
  2. Acceleration is a vector pointing left (negative direction) with magnitude 1 m/s21\,\text{m/s}^21m/s2. (correct answer)
  3. Acceleration cannot be negative because time is always positive.
  4. Acceleration equals speed, so the block’s speed is −1 m/s-1\,\text{m/s}−1m/s.

Explanation: This question tests understanding of scalars and vectors in one dimension. Scalars have only magnitude (always positive), while vectors have both magnitude and direction (can be positive or negative). Acceleration is a vector quantity, so a = -1 m/s² means the acceleration has magnitude 1 m/s² and points in the negative direction (left). The negative sign indicates direction, not that acceleration is "slowing down" (which depends on velocity direction too). Choice B correctly identifies acceleration as a vector pointing left with magnitude 1 m/s². When working with vector quantities, always interpret the sign as indicating direction relative to the chosen coordinate system.

Question 10

Along a line, define east as positive. A car’s displacement is −30 m-30\,\text{m}−30m over 10 s10\,\text{s}10s. Which statement about average speed and average velocity is correct?

  1. Average velocity is 3 m/s3\,\text{m/s}3m/s and average speed is −3 m/s-3\,\text{m/s}−3m/s.
  2. Average velocity is −3 m/s-3\,\text{m/s}−3m/s and average speed is 3 m/s3\,\text{m/s}3m/s. (correct answer)
  3. Average velocity equals average speed because both use total time.
  4. Average speed must be 0 m/s0\,\text{m/s}0m/s since displacement is negative.

Explanation: This question tests understanding of scalars and vectors in one dimension. Average velocity is a vector quantity calculated as displacement divided by time, while average speed is a scalar quantity calculated as total distance divided by time. With displacement of -30 m over 10 s, the average velocity is -30 m ÷ 10 s = -3 m/s (westward). Since we don't know the exact path taken, we must assume the car traveled directly, making distance = |-30| = 30 m, and average speed = 30 m ÷ 10 s = 3 m/s. Choice D incorrectly claims average speed is zero when displacement is negative, but speed is always non-negative. Remember that average velocity includes direction (can be negative), while average speed is always positive or zero.

Question 11

Along a straight road, east is positive and west is negative. A car’s average velocity over a trip is vˉ=0 m/s\bar v=0\,\text{m/s}vˉ=0m/s. Which statement must be true?

  1. The car’s average speed is 0 m/s0\,\text{m/s}0m/s.
  2. The car did not move during the trip.
  3. The car’s net displacement is 0 m0\,\text{m}0m. (correct answer)
  4. The car traveled equal distances east and west.

Explanation: This question tests understanding of scalars and vectors in one dimension. Scalars have only magnitude (always positive), while vectors have both magnitude and direction (can be positive or negative). Average velocity = displacement/time, so if v̄ = 0 m/s, then displacement must equal 0 m (the car returned to its starting position). However, the car could have traveled any distance—for example, 10 km east then 10 km west gives zero displacement but 20 km total distance. Choice C correctly identifies that net displacement must be 0 m. To analyze average velocity, focus on displacement (a vector) not distance traveled (a scalar).

Question 12

Along a straight road, east is positive and west is negative. A car’s average velocity over a trip is 0 m/s0\text{ m/s}0 m/s. Which statement must be true?

  1. The car’s average speed is 0 m/s0\text{ m/s}0 m/s.
  2. The car never moved during the trip.
  3. The car’s displacement over the trip is 0 m0\text{ m}0 m. (correct answer)
  4. The car’s distance traveled over the trip is 0 m0\text{ m}0 m.

Explanation: This question tests understanding of scalars versus vectors in one dimension. Average velocity is a vector quantity calculated as total displacement divided by total time, while average speed is a scalar calculated as total distance divided by total time. If average velocity equals 0 m/s, this means the total displacement must be 0 m - the car returned to its starting position. However, the car could have traveled any distance (forward then backward) as long as it ended where it started, so distance traveled is not necessarily zero. Choice A incorrectly assumes zero average velocity means zero average speed, but the car could have moved significantly. When average velocity is zero, displacement must be zero, but distance traveled can be any non-negative value.

Question 13

On a straight line, define up as positive and down as negative. An elevator’s velocity changes from −2 m/s-2\text{ m/s}−2 m/s to +2 m/s+2\text{ m/s}+2 m/s over 4 s. Which statement about acceleration is correct?

  1. The average acceleration is 0 m/s20\text{ m/s}^20 m/s2 because the speeds are equal.
  2. The average acceleration is +1 m/s2+1\text{ m/s}^2+1 m/s2. (correct answer)
  3. The average acceleration is −1 m/s2-1\text{ m/s}^2−1 m/s2 because the initial velocity was negative.
  4. The average acceleration is +4 m/s2+4\text{ m/s}^2+4 m/s2 because the velocity changed by 4 m/s4\text{ m/s}4 m/s.

Explanation: This question tests understanding of scalars and vectors in one dimension, specifically acceleration as a vector quantity. Acceleration is defined as the change in velocity divided by time, where both initial and final velocities must include their signs to properly account for direction. The elevator's velocity changes from -2 m/s (downward) to +2 m/s (upward), so the change in velocity = final - initial = +2 m/s - (-2 m/s) = +4 m/s. Average acceleration = (+4 m/s)/4 s = +1 m/s² (positive, indicating upward acceleration). Choice A incorrectly assumes zero acceleration because the speeds (magnitudes) are equal, ignoring that the velocities have opposite directions. When calculating acceleration, always use signed velocities to account for direction, not just speed magnitudes.

Question 14

A cart moves along a straight track where right is positive and left is negative. It starts at x=0 mx=0\text{ m}x=0 m, travels to +6 m+6\text{ m}+6 m, then travels to +2 m+2\text{ m}+2 m. Which statement correctly distinguishes a scalar from a vector for this motion?

  1. The cart’s distance traveled is +2 m+2\text{ m}+2 m because it ends at +2 m+2\text{ m}+2 m.
  2. The cart’s displacement is +2 m+2\text{ m}+2 m and its distance traveled is 10 m10\text{ m}10 m. (correct answer)
  3. The cart’s displacement is 10 m10\text{ m}10 m because displacement is always positive.
  4. The cart’s distance traveled is +10 m+10\text{ m}+10 m since distance must include direction.

Explanation: This question tests understanding of scalars versus vectors in one dimension. Scalars are quantities that have only magnitude (size), while vectors have both magnitude and direction. Distance traveled is a scalar that represents the total length of the path taken, regardless of direction, so the cart travels |6-0| + |2-6| = 6 + 4 = 10 m total. Displacement is a vector that represents the change in position from start to finish, including direction, so the cart's displacement is +2 m - 0 m = +2 m. Choice A incorrectly confuses final position with distance traveled. When distinguishing scalars from vectors, remember that scalars never have negative values or directional signs, while vectors in one dimension use positive/negative signs to indicate direction.

Question 15

A ball moves along a line where upward is positive and downward is negative. It is thrown upward with velocity +8 m/s+8\text{ m/s}+8 m/s and later moves downward with velocity −8 m/s-8\text{ m/s}−8 m/s. Which statement correctly compares speed and velocity?

  1. The speeds are equal, but the velocities are opposite. (correct answer)
  2. The velocities are equal because their magnitudes match.
  3. The speeds are opposite because one direction is negative.
  4. Both speed and velocity are negative on the way down.

Explanation: This question tests understanding of scalars versus vectors in one dimension. Speed is a scalar that represents only the magnitude of velocity, while velocity is a vector that includes both magnitude and direction. The ball moving upward at +8 m/s has a speed of 8 m/s, and when moving downward at -8 m/s, it still has a speed of 8 m/s (the magnitude). The velocities are opposite in sign (+8 m/s versus -8 m/s), indicating opposite directions, even though the speeds are equal. Choice B incorrectly claims the velocities are equal, ignoring that opposite signs mean opposite directions. When comparing motion in opposite directions, speeds will be equal if magnitudes match, but velocities will have opposite signs.

Question 16

On a straight hallway, take east as positive and west as negative. A student walks from x=0 mx=0\text{ m}x=0 m to x=+6 mx=+6\text{ m}x=+6 m in 3 s, then to x=+2 mx=+2\text{ m}x=+2 m in the next 2 s. Which statement must be true about the student’s motion?

  1. The student’s average speed is 0.8 m/s0.8\text{ m/s}0.8 m/s.
  2. The student’s average velocity is +0.4 m/s+0.4\text{ m/s}+0.4 m/s. (correct answer)
  3. The student’s displacement is 8 m8\text{ m}8 m.
  4. The student’s average speed equals the magnitude of the average velocity.

Explanation: This question tests understanding of scalars and vectors in one dimension, specifically the difference between average speed and average velocity. Average speed is a scalar quantity that only considers the total distance traveled divided by time, while average velocity is a vector quantity that considers displacement (change in position) divided by time, including direction. The student travels from 0 m to +6 m (distance = 6 m) then to +2 m (distance = 4 m), for a total distance of 10 m in 5 s, giving average speed = 10/5 = 2 m/s. The displacement is final position minus initial position: +2 m - 0 m = +2 m, so average velocity = +2 m/5 s = +0.4 m/s. Choice A incorrectly calculates average speed as 0.8 m/s instead of 2 m/s. When working with motion problems, always distinguish between scalar quantities (distance, speed) that have only magnitude and vector quantities (displacement, velocity) that have both magnitude and direction.

Question 17

Along a straight track, right is positive and left is negative. During one interval, a cart’s velocity changes from −1 m/s-1\text{ m/s}−1 m/s to +3 m/s+3\text{ m/s}+3 m/s. Which statement is correct?

  1. The cart’s change in velocity is Δv=+4 m/s\Delta v=+4\text{ m/s}Δv=+4 m/s. (correct answer)
  2. The cart’s change in speed is +4 m/s+4\text{ m/s}+4 m/s.
  3. The cart’s average speed must be +1 m/s+1\text{ m/s}+1 m/s.
  4. Because velocity became positive, the cart’s speed became negative.

Explanation: This question tests understanding of scalars versus vectors in one dimension. Change in velocity is a vector quantity calculated by subtracting initial from final velocity, while change in speed involves comparing magnitudes only. The change in velocity is Δv = vf - vi = (+3 m/s) - (-1 m/s) = +4 m/s, correctly accounting for both magnitudes and directions. The initial speed was |-1| = 1 m/s and final speed is |+3| = 3 m/s, so the change in speed is 3 - 1 = 2 m/s (not 4 m/s). Choice B incorrectly applies the velocity calculation to speed, ignoring that speed uses only magnitudes. When calculating changes in vector quantities, use signed values throughout; for scalar changes, work with magnitudes only.

Question 18

A runner moves along a line where north is positive and south is negative. During one interval, the runner’s velocity is −3 m/s-3\,\text{m/s}−3m/s. Which quantity must be negative in this situation?

  1. Speed
  2. Distance traveled
  3. Velocity (correct answer)
  4. Time

Explanation: This question assesses the understanding of scalars and vectors in one dimension. Scalars are quantities that describe only magnitude, such as speed or time, and do not incorporate direction. Vectors include both magnitude and direction, like velocity or displacement, allowing them to take negative values in one-dimensional scenarios to denote opposite directions. Thus, while scalars are always non-negative, vectors can be negative to reflect directional changes along a defined axis. Choice A is a distractor because speed is a scalar and cannot be negative, even if the motion is in the negative direction; it only represents the magnitude of velocity. A transferable strategy is to identify vectors by checking if the quantity must include a directional sign to fully describe the physical situation.

Question 19

A toy car moves on a straight line where right is positive and left is negative. Over 4 s4\,\text{s}4s it has displacement +12 m+12\,\text{m}+12m. Which statement correctly compares average speed and average velocity?

  1. Average velocity must equal +3 m/s+3\,\text{m/s}+3m/s, while average speed must be 3 m/s3\,\text{m/s}3m/s or greater. (correct answer)
  2. Average speed must equal +3 m/s+3\,\text{m/s}+3m/s, while average velocity must be 3 m/s3\,\text{m/s}3m/s or greater.
  3. Average speed and average velocity must be equal because the displacement is positive.
  4. Average velocity is always positive, but average speed can be negative if the car moved left.

Explanation: This question assesses the understanding of scalars and vectors in one dimension. Scalars, such as average speed, only quantify magnitude and are calculated using total distance traveled, which is always positive. Vectors, like average velocity, incorporate both magnitude and direction by using displacement, which can be positive or negative based on net position change. Therefore, average speed is always greater than or equal to the magnitude of average velocity, as distance accounts for all path lengths while displacement considers only the straight-line net change. Choice D is a distractor because it wrongly states average speed can be negative, but speed is a scalar and cannot have a negative value regardless of direction. A transferable strategy is to compare calculations: use total distance for scalars like speed and net displacement for vectors like velocity to distinguish them.

Question 20

A robot moves along the xxx-axis where positive is to the right. It reports an average speed of 2 m/s2\text{ m/s}2 m/s for 3 s3\text{ s}3 s and an average velocity of −2 m/s-2\text{ m/s}−2 m/s. Which statement is correct?

  1. The robot’s displacement is +6 m+6\text{ m}+6 m because speed is positive.
  2. The robot’s displacement is −6 m-6\text{ m}−6 m, while its distance traveled is 6 m6\text{ m}6 m. (correct answer)
  3. The robot’s distance traveled is −6 m-6\text{ m}−6 m because velocity is negative.
  4. Average speed and average velocity must have the same sign.

Explanation: This question tests understanding of scalars and vectors in one dimension. Average speed is a scalar (always positive) representing total distance over time, while average velocity is a vector including direction. With average speed 2 m/s and average velocity -2 m/s for 3 s, the robot traveled distance = 2 × 3 = 6 m total, but displacement = -2 × 3 = -6 m (6 m left). This means the robot moved entirely leftward, covering 6 m distance with -6 m displacement. Choice A incorrectly uses speed to determine displacement direction, while C wrongly assigns negative to distance. To distinguish scalar and vector motion quantities, remember that speed and distance are always positive, while velocity and displacement include direction through their sign.