📝 Average velocity calculus (25 MCQs)

📖 Explanation: The average velocity is the change in position divided by the change in time. Compute $s(4) = 4^3 - 6(4)^2 + 9(4) = 64 - 96 + 36 = 4$ and $s(1) = 1 - 6 + 9 = 4$. The displacement is $4 - 4 = 0$, so the average velocity is $0/(4-1) = 0$ m/s. Even though the particle moved during this interval, it returned to its starting point.

📖 Explanation: Average velocity is total displacement divided by total time, not total distance. The displacement vector is $30\hat{j} + 40\hat{i}$, with magnitude $\sqrt{30^2 + 40^2} = 50$ km. Total time is 1.5 hours. So average velocity magnitude is $50/1.5 \approx 33.3$ km/h. Wait, I need to recalculate: The question asks for the magnitude of average velocity. $50/1.5 = 33.3$ km/h. However, option C is 28 km/h, which is total distance (70 km) / total time (1.5 h). Option A is the displacement magnitude divided by something else. Let's correct: Total time = 0.5 + 1 = 1.5 h. Displacement magnitude = 50 km. Average velocity = 50/1.5 = 33.3 km/h. None of the options match exactly. Let's check the options again. Option B is 46.7 km/h which is 70/1.5. Option D is 35.6 km/h. I'll recalculate: 50/1.5 = 33.33. There's no 33.3. Let's look at the options: A: 50, B: 46.7, C: 28, D: 35.6. The closest is none. Let me re-evaluate the problem. Perhaps the car travels 30 km north in 0.5 h, then 40 km east in 1 h. Displacement magnitude is 50 km. Total time = 1.5 h. Average velocity = 33.33 km/h. Since it's not listed, maybe I made an error. Let's check if the trip is 30 km north, then 40 km east. The displacement is 50 km. 50/1.5 = 33.33. The options don't match. I'll choose the closest, which is D (35.6). But 35.6 is not 33.33. Let's re-calculate the displacement: $\sqrt{30^2 + 40^2} = 50$. Total time = 1.5. 50/1.5 = 33.33. The options are all wrong. I'll change the options to include 33.3. Let me adjust the options: A: 50 km/h, B: 46.7 km/h, C: 33.3 km/h, D: 28 km/h. Correct is C. So the answer is C.

📖 Explanation: The average velocity is defined as total displacement divided by total time. The average of the initial and final velocities is only equal to the average velocity if the acceleration is constant. In the case where an object returns to its starting point, the displacement is zero, so the average velocity is zero. However, the average of the initial and final velocities could be non-zero if the speeds are different (e.g., moving right then left with different speeds). This misconception arises from confusing average velocity with average of velocities.

📖 Explanation: Average acceleration is the change in velocity divided by the change in time. We compute $v(4) = 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9$ and $v(1) = 3 - 12 + 9 = 0$. Change in velocity is $9 - 0 = 9$. Change in time is $4 - 1 = 3$. So average acceleration is $9/3 = 3$ m/s². This is conceptually related to average velocity, but applied to acceleration.

📖 Explanation: Average velocity is zero when the displacement is zero, i.e., $s(b) = s(a)$. We check each option: For [0,4], $s(0) = 5$, $s(4) = 2(16) - 32 + 5 = 32 - 32 + 5 = 5$. So displacement is zero, average velocity is zero. For [1,3], $s(1) = 2 - 8 + 5 = -1$, $s(3) = 18 - 24 + 5 = -1$. Also zero. So both A and B are correct. Let me adjust the options to have only one correct. I'll change option B to [1,2]. $s(1) = -1$, $s(2) = 8 - 16 + 5 = -3$. Not zero. So correct is A.

📖 Explanation: Average speed is total distance divided by total time: $400/50 = 8$ m/s. Average velocity is total displacement divided by total time. After completing a lap, the displacement is zero because the runner returns to the starting point. Therefore, average velocity is $0/50 = 0$ m/s. This highlights the distinction between distance (scalar) and displacement (vector), and between speed and velocity.

📖 Explanation: For the round trip, the total displacement is zero because the train starts and ends at the same point (station A). Therefore, the average velocity, which is total displacement divided by total time, is $0 / \text{total time} = 0$ km/h. This is a classic trap; students often calculate the average speed (which would be $2(60)(40)/(60+40) = 48$ km/h) but forget that velocity is a vector quantity that depends on displacement, not distance.

📖 Explanation: The slope of a position-time graph represents the velocity. A straight line indicates constant velocity. A negative slope indicates that the position is decreasing with time, meaning the particle is moving in the negative direction with constant velocity. This is a direct interpretation of the graph, testing the understanding that velocity is the slope of the position-time graph.

📖 Explanation: We compute $s(\pi) = \sin(\pi) = 0$ and $s(0) = \sin(0) = 0$. The displacement is $0 - 0 = 0$. The time interval is $\pi - 0 = \pi$. Therefore, the average velocity is $0/\pi = 0$. Even though the particle moved (it went up to 1 and back down to 0), its net displacement is zero, so its average velocity is zero. This reinforces the idea that average velocity depends only on the initial and final positions.

📖 Explanation: Average velocity is defined as $(s(t_2) - s(t_1))/(t_2 - t_1)$. If this value is negative, it means the numerator is negative (since $t_2 - t_1 > 0$), so $s(t_2) < s(t_1)$. This means the final position is less than the initial position, indicating the particle moved in the negative direction overall. It does not necessarily mean the particle moved a greater distance in the negative direction (that would be about total distance, not displacement), nor does it directly relate to speed or acceleration.

📖 Explanation: Average velocity when acceleration is constant is the average of the initial and final velocities: $(10 + 30)/2 = 20$ m/s. This is a standard formula for constant acceleration. However, it's important to note that this only works for constant acceleration; otherwise, it's not valid. This question tests the recall of that specific formula.

📖 Explanation: Average velocity over $[a,b]$ is $(s(b) - s(a))/(b-a)$. For the average velocity to be negative, we need $s(b) < s(a)$. This means the function is decreasing on average. $s(t) = (t-1)(t-3)$. The vertex is at $t=2$. The function decreases on $(-\infty, 2)$ and increases on $(2, \infty)$. So for any interval within $(-\infty, 2)$, the average velocity will be negative. The interval $(1,3)$ is not entirely within the decreasing part; it includes increasing part as well. So the correct answer is A.

📖 Explanation: Total distance = 200 + 100 + 300 = 600 m. Total time = 4 + 2 + 6 = 12 s. Since the motion is in the same direction, displacement equals distance. Average velocity = total displacement / total time = 600 / 12 = 50 m/s. This tests the ability to calculate average velocity from multiple segments of motion, emphasizing that it's the total displacement over total time, not the average of the segment velocities (which would be $(200/4 + 100/2 + 300/6)/3 = (50+50+50)/3 = 50$ m/s in this case, but that's a coincidence).

📖 Explanation: Average velocity is zero if and only if total displacement is zero. Displacement is zero if and only if the final position equals the initial position. This does not require the particle to be at rest, nor does it require constant speed or equal distances in opposite directions (it could move in a complex path and return to the start). The only condition is $s(t_2) = s(t_1)$. This is a fundamental definition.

📖 Explanation: We can find the displacement by integrating velocity: $s(5) - s(1) = \int_1^5 (2t - 4) dt = [t^2 - 4t]_1^5 = (25 - 20) - (1 - 4) = 5 - (-3) = 8$. Time interval is 4 seconds. Average velocity = 8/4 = 2 m/s. Alternatively, average of velocities: $(v(1)+v(5))/2 = (-2+6)/2 = 2$. This tests the relationship between velocity and position through integration.

📖 Explanation: Compute $s(3) = 3(9) - 18 + 2 = 27 - 18 + 2 = 11$. $s(1) = 3 - 6 + 2 = -1$. Displacement = 11 - (-1) = 12. Time = 3 - 1 = 2. Average velocity = 12/2 = 6 m/s. This is a straightforward calculation, testing the basic definition.

📖 Explanation: Constant speed means the magnitude of velocity is constant, but the direction could change (e.g., circular motion). Average velocity is a vector quantity defined as total displacement divided by time. If the direction changes, the displacement is less than the total distance, so the magnitude of average velocity is less than the average speed. Even in straight-line motion, if the direction reverses, the average velocity could be less than the speed. The correct statement is that average speed equals constant speed, not average velocity. This addresses a common misconception.

📖 Explanation: For a parabola $s(t) = at^2 + bt + c$, the vertex is at $t = -b/(2a)$. If the interval is symmetric about the vertex, say $[t_0 - h, t_0 + h]$, then $s(t_0 - h) = s(t_0 + h)$ because the parabola is symmetric. Therefore, the displacement is zero, and the average velocity is zero. This tests the interpretation of the graph and the symmetry of quadratic functions.

📖 Explanation: $s(4) = 16 - 16 = 0$. $s(0) = 0$. Displacement is 0. Average velocity is 0. This particle moves up and then down, returning to the origin. The average velocity is zero, highlighting that even with motion, the net displacement can be zero.

📖 Explanation: The Mean Value Theorem for derivatives states that if a function is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c$ in $(a,b)$ such that f&#039;(c) = (f(b) - f(a))/(b-a). Here, f&#039;(c) is the instantaneous velocity at $c$, and the right-hand side is the average velocity. So the Mean Value Theorem guarantees the existence of such a point. This is a higher-order question linking calculus theorems to physical concepts.

📖 Explanation: Let the total time be $2T$. Distance in first half = $40T$. Distance in second half = $60T$. Total distance = $100T$. Total time = $2T$. Average velocity = $100T / 2T = 50$ km/h. This is different from the case where the car travels equal distances at different speeds. This tests the understanding of the difference between averaging over time vs. averaging over distance.

📖 Explanation: $s(\pi) - s(0) = \int_0^\pi \sin(t) dt = [-\cos(t)]_0^\pi = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2$. Wait, that's displacement = 2. Time = $\pi$. So average velocity = $2/\pi$. Let me recalculate: $\int_0^\pi \sin(t) dt = 2$. So average velocity = $2/\pi$. Option B is correct. I'll adjust. The correct answer is B.

📖 Explanation: Displacement in first 10 seconds = $5 \times 10 = 50$ m. Displacement in next 10 seconds = $-5 \times 10 = -50$ m. Total displacement = 0. Average velocity = 0. This shows that even with motion in both directions, the net displacement can be zero, leading to zero average velocity.

📖 Explanation: We integrate to find displacement: $s(3) - s(0) = \int_0^3 (3t^2 - 6t) dt = [t^3 - 3t^2]_0^3 = (27 - 27) - 0 = 0$. Average velocity = 0/3 = 0. This demonstrates that even with a changing velocity, the net displacement can be zero.

📖 Explanation: Time for quarter circle = $\frac{(1/4)(2\pi r)}{v} = \frac{\pi r}{2v} = \frac{5\pi}{4}$ s. Displacement is the chord of the quarter circle, which is $r\sqrt{2} = 5\sqrt{2}$ m. Average velocity magnitude = displacement / time = $\frac{5\sqrt{2}}{5\pi/4} = \frac{4\sqrt{2}}{\pi}$ m/s. This is a Easy problem that requires understanding that displacement is the straight-line distance between start and end points, not the arc length, and calculating the time from the speed.