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.

This question tests understanding of kinematics and motion variables, specifically the calculation of average acceleration from velocity data. Average acceleration equals the change in velocity divided by the time interval: a_avg = Δv/Δt = (v_f - v_i)/(t_f - t_i). With velocities of +4.0 m/s at both t = 10 s and t = 12 s, the velocity change is 0 m/s over 2 s, giving a_avg = 0/2 = 0 m/s². Zero acceleration indicates constant velocity motion. Choice B incorrectly equates the velocity value with acceleration, confusing different kinematic quantities with different units. To find average acceleration, always calculate the velocity difference first - if velocities are equal, acceleration is zero regardless of the velocity magnitude.

This question tests understanding of kinematics and motion variables, specifically the fundamental relationship between constant acceleration and velocity. When acceleration is constant and nonzero, velocity must change at a constant rate given by Δv = a × Δt. Throughout the 1-second interval, velocity increases or decreases linearly depending on acceleration's sign, creating a straight-line velocity-time graph with nonzero slope. Choice B incorrectly claims velocity must be constant when acceleration is constant, confusing zero acceleration with nonzero constant acceleration. Remember that constant nonzero acceleration means uniform velocity change - velocity varies linearly with time, changing by the same amount in each equal time interval.

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.

This question tests understanding of kinematics and motion variables, specifically sign conventions for velocity and acceleration during deceleration. Moving leftward means negative velocity (v < 0) since rightward is positive. Slowing down means acceleration opposes velocity, so with negative velocity, acceleration must be positive (a > 0) to reduce the speed. This gives opposite signs for v and a, which always indicates slowing down. Choice C incorrectly suggests both negative, which would mean speeding up leftward rather than slowing down. When an object slows down, velocity and acceleration must have opposite signs - this is a fundamental rule for interpreting motion regardless of direction.

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.

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.

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.

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.

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.