MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4a Kinematics Motion Variables

What this quiz covers

This quiz focuses on 4a Kinematics Motion Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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 marker's acceleration must have been negative during the transition because velocity decreased while direction stayed positive. (correct answer)
2. The marker's acceleration must have been positive because the marker continued moving in the + direction.
3. The marker's speed increased because the marker covered additional displacement each second.
4. The marker's velocity change is best described in $\text{m/s}^2$ because velocity is an acceleration-like quantity.

Explanation: This question tests understanding of kinematics and motion variables. Acceleration is the rate of change of velocity, and its sign depends on whether velocity is increasing or decreasing in the chosen positive direction. In this treadmill study, the marker's velocity decreases from +1.5 m/s to +0.5 m/s while remaining positive, indicating a decrease in speed without direction change. Choice A is correct because the negative acceleration reflects the velocity decrease in the positive direction. A common misconception is that positive motion implies positive acceleration, as in choice B, but acceleration opposes motion when speed decreases. To check similar problems, calculate the change in velocity and divide by time to find average acceleration. Ensure the sign convention is consistent with the defined positive direction.

Question 2

1. Acceleration is zero because speed is constant.
2. Acceleration can be nonzero because velocity changes when direction changes, even at constant speed. (correct answer)
3. Acceleration must be negative because the cart is turning.
4. Acceleration is measured in $\text{m/s}$ because it reflects changing direction.

Explanation: This question tests understanding of kinematics and motion variables. Acceleration is the rate of change of velocity vector, nonzero if direction changes even at constant speed. The cart's constant speed but changing direction means changing velocity. Choice B is correct because direction change implies acceleration. A common misconception is that constant speed means zero acceleration, as in choice A. Recall vector nature in circular motion. Check if speed constant but path curved.

Question 3

1. The particle's displacement is not changing during the interval, consistent with zero velocity. (correct answer)
2. The particle's acceleration must be nonzero because its position is constant.
3. The particle's speed must be increasing because it is not moving.
4. The particle's velocity is zero, so it must have moved zero total distance over any longer time period.

Explanation: This question tests understanding of kinematics and motion variables. Zero velocity means no change in position over time. The particle's v=0 corresponds to constant position observed. Choice A is correct because zero velocity implies unchanging displacement. A common misconception is that constant position implies nonzero acceleration, as in choice B. Verify with position data matching velocity. Integrate velocity to confirm position constancy.

Question 4

1. Velocity should be approximately constant (no systematic increase or decrease). (correct answer)
2. Velocity should decrease to zero because acceleration is zero.
3. Velocity should increase because the cart is moving in the + direction.
4. Velocity must change sign because acceleration is zero for multiple seconds.

Explanation: This question tests understanding of kinematics and motion variables. In kinematics, acceleration is defined as the rate of change of velocity over time, so if acceleration is zero, the velocity remains constant. In this scenario, the cart is moving in the positive direction with zero acceleration measured over a 3-second window, and instrument uncertainty is negligible. Therefore, the cart's velocity should be approximately constant, showing no systematic increase or decrease, which aligns with choice A. A common misconception is that zero acceleration means the velocity decreases to zero, as in choice B, but this confuses acceleration with velocity; zero acceleration simply means no change in velocity, not that velocity is zero. To check similar problems, verify if acceleration is zero, which implies constant velocity regardless of direction. Additionally, ensure that initial conditions like direction do not imply a change unless acceleration is non-zero.

Question 5

1. Segment II has larger acceleration because the velocities are larger.
2. Segment I has larger acceleration because it starts from rest.
3. Both segments have the same average acceleration because the velocity changes by the same amount in the same time. (correct answer)
4. Both segments have zero acceleration because velocity remains positive throughout.

Explanation: This question tests understanding of kinematics and motion variables. Average acceleration depends on Δv/Δt, same if changes are equal over equal times. Both segments have Δv = +2 m/s in 1 s, so same a. Choice C is correct because accelerations match. A common misconception is that higher velocities mean larger acceleration, as in choice A. Compute a separately for each segment. Ensure time intervals are identical for comparison.

Question 6

1. The cart's speed increases by $3.0\ \text{m/s}$ because acceleration is constant.
2. The cart's velocity becomes more positive by $3.0\ \text{m/s}$ over 3.0 s.
3. The cart's velocity decreases by $3.0\ \text{m/s}$ over 3.0 s. (correct answer)
4. The cart's acceleration changes from $-1.0\ \text{m/s}$ to $-3.0\ \text{m/s}$.

Explanation: This question tests understanding of kinematics and motion variables, specifically how constant acceleration affects velocity over time. When an object has constant acceleration, its velocity changes at a steady rate given by Δv = a × Δt. The cart starts with velocity +2.0 m/s and experiences acceleration -1.0 m/s² for 3.0 s, so the velocity change is (-1.0 m/s²)(3.0 s) = -3.0 m/s. This means the velocity decreases by 3.0 m/s, making the final velocity +2.0 m/s - 3.0 m/s = -1.0 m/s. Choice A incorrectly states that speed increases, confusing the magnitude of velocity change with an increase in speed. To verify constant acceleration problems, always use Δv = a × Δt and pay attention to signs: negative acceleration with positive initial velocity means the object slows down and may reverse direction.

Question 7

1. Average acceleration is positive. (correct answer)
2. Average acceleration is negative because initial velocity was negative.
3. Average acceleration is zero because the speed values are equal.
4. Average acceleration has units of m/s, so its sign cannot be determined.

Explanation: This question tests understanding of kinematics and motion variables, particularly determining acceleration sign from velocity data. Average acceleration equals Δv/Δt = [+0.5 - (-0.5)]/(4 - 0) = (+1.0)/(4) = +0.25 m/s². The velocity change is positive (from negative to positive), making the average acceleration positive. This represents motion that slows down in the negative direction, stops, then speeds up in the positive direction. Choice C incorrectly claims zero acceleration because speeds are equal, ignoring that velocity includes direction and the signs differ. When calculating acceleration, always account for velocity signs: changing from -0.5 to +0.5 m/s is a positive change of 1.0 m/s, not zero.

Question 8

1. Velocity is unchanged from 0–2 s, then decreases by $2\ \text{m/s}$ from 2–3 s. (correct answer)
2. Velocity decreases steadily from 0–2 s because the runner is moving forward.
3. Acceleration is $-2\ \text{m/s}$ during 2–3 s because velocity is in m/s.
4. The runner's displacement becomes negative during 2–3 s because acceleration is negative.

Explanation: This question tests understanding of kinematics and motion variables, particularly the relationship between velocity and acceleration. Acceleration is defined as the time rate of change of velocity (a = Δv/Δt), meaning when acceleration is zero, velocity remains constant. During the first 2 seconds, the runner maintains constant velocity at +6 m/s with zero acceleration, so velocity is unchanged. During the interval from 2-3 s, the runner experiences constant acceleration of -2 m/s², causing the velocity to decrease by (-2 m/s²)(1 s) = -2 m/s, from +6 m/s to +4 m/s. Choice B incorrectly assumes that forward motion requires deceleration, confusing position with velocity. When analyzing motion segments, identify periods of constant velocity (a = 0) versus constant acceleration (a ≠ 0), and apply Δv = a × Δt only during accelerated motion.

Question 9

1. Velocity changes by about $9.8\ \text{m/s}$ in the downward direction. (correct answer)
2. Velocity changes by about $9.8\ \text{m/s}$ upward because gravity is negative.
3. Velocity remains zero because initial velocity was zero.
4. Acceleration changes by $9.8\ \text{m/s}$ because velocity is changing.

Explanation: This question tests understanding of kinematics and motion variables in the context of free fall under gravity. When an object is released from rest, it experiences constant gravitational acceleration g = 9.8 m/s² downward. Using Δv = at with initial velocity v₀ = 0, after 1 second the velocity change is (9.8 m/s²)(1 s) = 9.8 m/s in the downward direction. Since the package was dropped (not thrown), it moves downward with increasing downward velocity. Choice B incorrectly interprets the negative sign convention, failing to recognize that 'downward' is the positive direction for falling objects in this context. When solving free fall problems, establish a clear sign convention: if taking down as positive, then g = +9.8 m/s² and downward velocities are positive.

Question 10

1. Acceleration is in the negative direction because velocity becomes more negative. (correct answer)
2. Acceleration is in the positive direction because speed increased.
3. Acceleration is zero because the sled did not change direction.
4. Acceleration has units of m/s, so its direction is undefined.

Explanation: This question tests understanding of kinematics and motion variables, particularly determining acceleration direction from velocity changes. When velocity changes from -2 m/s to -5 m/s, the change is Δv = -5 - (-2) = -3 m/s. Since Δv is negative, the acceleration must be negative (leftward), making the sled go faster in the negative direction. Both velocity and acceleration point in the same negative direction, so the sled speeds up while moving leftward. Choice B incorrectly claims positive acceleration because speed increased, failing to distinguish between speed (magnitude) and velocity (vector). When velocity becomes more negative, acceleration is negative - this represents speeding up in the negative direction, not slowing down.

Question 11

1. Acceleration is constant and positive because velocity is positive.
2. Acceleration is approximately zero during 5–9 s. (correct answer)
3. Displacement must be zero because acceleration is zero.
4. Speed must be decreasing because time is increasing.

Explanation: This question tests understanding of kinematics and motion variables, specifically the relationship between constant velocity and acceleration. When velocity is constant, there is no change in velocity over time, which means acceleration equals zero by definition (a = Δv/Δt = 0/Δt = 0). During the 5-9 s interval, the chair maintains constant translational velocity at +0.80 m/s, so the acceleration is zero throughout this period. This is true regardless of the velocity's magnitude or sign - constant velocity always implies zero acceleration. Choice A incorrectly assumes that positive velocity requires positive acceleration, confusing the state of motion with changes in motion. Remember that acceleration describes how velocity changes, not the velocity itself: an object can move at high constant speed with zero acceleration.

Question 12

1. Velocity is positive but decreasing in magnitude over time. (correct answer)
2. Velocity is negative because acceleration is negative.
3. Velocity is constant because acceleration is constant.
4. Velocity increases because acceleration is negative.

Explanation: This question tests understanding of kinematics and motion variables in projectile motion with opposing velocity and acceleration. When a ball is thrown upward (positive velocity) with downward gravitational acceleration (-9.8 m/s²), the velocity decreases over time. While moving upward before reaching peak height, velocity remains positive but decreases in magnitude as Δv = at where a is negative. The ball slows down until velocity reaches zero at the peak. Choice B incorrectly claims velocity is negative while moving upward, confusing acceleration direction with velocity direction. During upward motion against gravity, velocity is positive and decreasing - the ball slows down but continues upward until v = 0.

Question 13

1. Acceleration is leftward (negative) during the interval. (correct answer)
2. Acceleration is rightward (positive) because the carriage is moving rightward.
3. Acceleration must be zero because the carriage still moves rightward.
4. Displacement is negative because the carriage slows down.

Explanation: This question tests understanding of kinematics and motion variables, specifically the relationship between velocity and acceleration when an object slows down. When an object moves rightward (positive velocity) but slows down, its velocity decreases, meaning Δv is negative. Since acceleration a = Δv/Δt and Δt is positive, acceleration must be negative (leftward). This demonstrates that acceleration opposes velocity when an object slows down. Choice B incorrectly assumes acceleration must match the direction of motion, confusing velocity direction with acceleration direction. Remember: when an object slows down, acceleration always opposes velocity direction regardless of which way the object moves.

Question 14

1. Average acceleration is negative during the interval. (correct answer)
2. Average acceleration is positive because velocity is positive.
3. Average acceleration is zero because the cart never stops.
4. Average acceleration has units of m/s, matching velocity change.

Explanation: This question tests understanding of kinematics and motion variables, specifically the calculation and interpretation of average acceleration. Average acceleration is defined as the change in velocity divided by the time interval: a_avg = Δv/Δt = (v_f - v_i)/Δt. With initial velocity +3.0 m/s and final velocity +1.0 m/s over 1.0 s, the average acceleration is (+1.0 - 3.0)/(1.0) = -2.0 m/s². The negative sign indicates the acceleration opposes the positive velocity direction, causing the cart to slow down. Choice B incorrectly assumes positive acceleration because velocity remains positive, confusing the sign of velocity with the sign of velocity change. To determine acceleration sign, always calculate Δv first: if Δv and v have opposite signs, the object is slowing down.

Question 15

1. The cart's velocity was zero at least once (it momentarily stopped). (correct answer)
2. The cart's acceleration was zero the entire time.
3. The cart's displacement was zero the entire time.
4. The cart's speed was negative at least once.

Explanation: This question tests understanding of kinematics and motion variables, particularly velocity changes and direction reversal. When velocity changes from +1.0 m/s to -1.0 m/s, the cart must have changed direction during the interval. Since velocity is continuous for physical objects, it must pass through zero when changing sign, meaning the cart momentarily stopped (v = 0) at some instant between t = 0 and t = 2 s. This occurs when positive initial velocity is opposed by negative acceleration. Choice B incorrectly suggests zero acceleration, which would maintain constant velocity rather than allowing the observed change. When velocity changes sign, the object must instantaneously stop at the reversal point - this is a fundamental consequence of velocity continuity.

Question 16

1. Velocity must instantly become negative when the force is applied.
2. Velocity decreases over time while the leftward acceleration acts. (correct answer)
3. Velocity stays constant because mass is constant.
4. Displacement becomes zero because acceleration is nonzero.

Explanation: This question tests understanding of kinematics and motion variables, specifically how forces create acceleration that changes velocity. A constant leftward (negative) force produces constant leftward acceleration according to Newton's second law. With initial velocity +1.5 m/s (rightward) and leftward acceleration, the puck's velocity decreases over time as Δv = a × Δt where a is negative. The puck slows down, may stop momentarily, and could reverse direction if the force acts long enough. Choice A incorrectly suggests instantaneous velocity change, violating the principle that velocity changes continuously under constant acceleration. Remember that forces cause acceleration, which gradually changes velocity - velocity cannot jump discontinuously under finite forces.

Question 17

1. Acceleration is near zero before and after the ramp, and nonzero mainly during the ramp. (correct answer)
2. Acceleration is constant and nonzero for the full 30 s because speed is nonzero.
3. Acceleration is negative during the ramp because the belt speed increases.
4. Acceleration equals the belt speed because both describe motion.

Explanation: This question tests understanding of kinematics and motion variables, specifically identifying acceleration patterns from velocity behavior. During constant-speed phases (before and after the ramp), velocity is constant so acceleration equals zero. During the 1-s ramp period when belt speed increases, velocity changes, requiring nonzero acceleration. The acceleration profile shows near-zero values except during the brief ramp transition where acceleration spikes to accomplish the speed change. Choice C incorrectly assigns negative acceleration to increasing speed, misunderstanding sign conventions. For piecewise motion analysis, acceleration is zero during constant-velocity segments and nonzero only when velocity changes - acceleration exists only during transitions between different steady states.

Question 18

1. The drone's speed will decrease to zero and then increase in the opposite direction (velocity becomes negative). (correct answer)
2. The drone's speed will increase because acceleration magnitude is nonzero.
3. The drone's displacement will stop changing immediately because acceleration is negative.
4. The drone's acceleration must become zero once its velocity reaches zero.

Explanation: This question tests understanding of kinematics and motion variables. Constant negative acceleration will reverse positive velocity over time. Starting at +3 m/s with a=-1 m/s², it slows, stops, then speeds up negatively. Choice A is correct because sustained a reverses direction. A common misconception is that negative a stops at zero without reversal, as in choice D. Use v = u + at to find reversal time. Assume constancy unless stated otherwise.

Question 19

1. Acceleration is positive because the drone's velocity is positive.
2. Acceleration is negative because velocity is decreasing in the positive direction. (correct answer)
3. Acceleration has units of m/s, so its sign cannot be determined.
4. Acceleration is zero because the drone never reverses direction.

Explanation: This question tests understanding of kinematics and motion variables, particularly determining acceleration sign from velocity changes. Acceleration is calculated as a = Δv/Δt = (v_f - v_i)/(t_f - t_i), and its sign indicates whether velocity is increasing (positive) or decreasing (negative) in the positive direction. The drone's velocity changes from +6 m/s to +2 m/s over 2 seconds, giving acceleration a = (2 - 6)/(2 - 0) = -4/2 = -2 m/s², which is negative because velocity decreases while remaining positive. Choice A incorrectly assumes acceleration must match velocity's sign, missing that negative acceleration can slow positive motion. To determine acceleration sign, calculate the velocity change: if final velocity is less than initial velocity (both positive), acceleration is negative.

Question 20

1. The bed's acceleration changes from approximately zero to a nonzero value in the direction of motion. (correct answer)
2. The bed's velocity becomes zero because its speed is increasing.
3. The bed's displacement becomes zero because the speed was constant previously.
4. The bed's acceleration must point opposite the motion whenever speed increases.

Explanation: This question tests understanding of kinematics and motion variables, specifically how acceleration relates to changes in speed. When an object moves at constant speed, its acceleration is zero; when speed increases in the direction of motion, acceleration becomes positive (same direction as velocity). The bed transitions from constant speed (a = 0) to increasing speed while maintaining direction, requiring positive acceleration in the motion direction. The bed's acceleration changes from approximately zero to a nonzero value in the direction of motion. Choice D incorrectly claims acceleration opposes motion when speed increases, confusing the conditions for speeding up versus slowing down. For speed changes: acceleration in motion direction increases speed, acceleration opposite motion direction decreases speed.