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.

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.

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.

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.

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.

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.

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.