📝 Displacement vs distance traveled integration (26 MCQs)

📖 Explanation: Displacement is the integral of velocity: $\int_0^4 (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_0^4 = (64 - 96 + 36) - 0 = 4$ m. The particle ends 4 meters from its starting position. It is important to remember that displacement is the net change in position and can be positive, negative, or zero.

📖 Explanation: Distance traveled requires integrating the absolute value of velocity: $\int_0^4 |3t^2 - 12t + 9| dt$. The velocity is zero at $t=1$ and $t=3$. On [0,1], v≥0; on [1,3], v≤0; on [3,4], v≥0. Distance = $\int_0^1 v dt - \int_1^3 v dt + \int_3^4 v dt = 4 - (-4) + 4 = 12$ m. The particle changes direction, so distance traveled is greater than displacement.

📖 Explanation: The student computed displacement (net change in position), not distance traveled. Distance traveled is $\int_0^\pi |\cos t| dt$. Since $\cos t$ is positive on [0, π/2] and negative on [π/2, π], the correct distance is $\int_0^{π/2} \cos t dt - \int_{π/2}^π \cos t dt = 1 - (-1) = 2$. The student's error is a classic mistake: confusing net displacement with total path length.

📖 Explanation: The velocity function is $v(t) = t + 2$. Displacement = $\int_0^4 (t+2) dt = [t^2/2 + 2t]_0^4 = 8 + 8 = 16$. This can also be found as the area under the velocity curve, which is a trapezoid with bases 2 and 6 and height 4: area = $(1/2)(2+6)(4) = 16$. Since the velocity is always positive, displacement equals distance traveled.

📖 Explanation: $v(t) = \int 6t dt = 3t^2 + C$. With $v(0) = 2$, $C = 2$, so $v(t) = 3t^2 + 2$, which is always positive. Distance = displacement = $\int_0^2 (3t^2 + 2) dt = [t^3 + 2t]_0^2 = 8 + 4 = 12$. The acceleration is always positive, so velocity increases from 2 to 14, never changing sign, so distance equals displacement.

📖 Explanation: $s(t) = \int_0^t (3u^2 - 6u) du = [u^3 - 3u^2]_0^t = t^3 - 3t^2$. At $t=3$, $s(3) = 27 - 27 = 0$. The particle returns to the starting point. This is a straightforward Easy of the relationship between velocity and position.

📖 Explanation: The area under the velocity curve is the area of a semicircle with radius 2: area = $\frac{1}{2}\pi(2)^2 = 2\pi$. Since the velocity is always nonnegative, displacement equals the area under the curve. The graph is a semicircle, so the integral is simply the geometric area.

📖 Explanation: Since the velocity is always nonnegative (the semicircle is above the t-axis), the distance traveled equals the displacement. Both are the area under the curve, which is $2\pi$. This is a simple case where distance and displacement are equal because the particle never changes direction.

📖 Explanation: Average velocity = $\frac{1}{4-1} \int_1^4 (2t - 3) dt = \frac{1}{3} [t^2 - 3t]_1^4 = \frac{1}{3} [(16 - 12) - (1 - 3)] = \frac{1}{3} [4 - (-2)] = \frac{1}{3} [6] = 2$. The average velocity is the displacement divided by the time interval, which is also the average value of the velocity function.

📖 Explanation: v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3). Velocity is positive on (0,1), negative on (1,3), positive on (3,4). Distance = $\int_0^1 v(t) dt - \int_1^3 v(t) dt + \int_3^4 v(t) dt$. $s(1) = 1 - 6 + 9 + 2 = 6$. $s(3) = 27 - 54 + 27 + 2 = 2$. $s(4) = 64 - 96 + 36 + 2 = 6$. Displacement from 0 to 1 is $s(1)-s(0) = 6-2 = 4$. From 1 to 3 is $s(3)-s(1) = 2-6 = -4$. From 3 to 4 is $s(4)-s(3) = 6-2 = 4$. Total distance = $4 + 4 + 4 = 12$. This method uses the position function directly to find distances between turning points.

📖 Explanation: $\int_0^\pi \cos(2t) dt = [\sin(2t)/2]_0^\pi = 0$. The integral of a cosine over a full period is zero. The particle returns to its starting position after one full period of the velocity function.

📖 Explanation: $\cos(2t)$ changes sign at $t = \pi/4, 3\pi/4$. Distance = $\int_0^{\pi/4} \cos(2t) dt - \int_{\pi/4}^{3\pi/4} \cos(2t) dt + \int_{3\pi/4}^{\pi} \cos(2t) dt$. Each of the three integrals has magnitude $1/2$. Actually, $\int_0^{\pi/4} \cos(2t) dt = [\sin(2t)/2]_0^{\pi/4} = 1/2$. The integral from $\pi/4$ to $3\pi/4$ is $-1$, so its absolute value is 1. The integral from $3\pi/4$ to $\pi$ is $1/2$. Total distance = $1/2 + 1 + 1/2 = 2$. The particle moves back and forth, so distance is greater than displacement.

📖 Explanation: The student is correct. Distance is $\int |v| dt$, and displacement is $\int v dt$. Since $|v| \ge v$, the integral of $|v|$ is always greater than or equal to the integral of $v$. They are equal only when $v$ does not change sign (i.e., the particle never reverses direction). This is a fundamental inequality in calculus.

📖 Explanation: $\int_0^3 2t dt = [t^2]_0^3 = 9$. Since the velocity is always positive, distance equals displacement. This is a straightforward integration problem.

📖 Explanation: The velocity is always nonnegative (it's a square), so it never changes sign. Distance = displacement = $\int_0^2 3(t-1)^2 dt = [(t-1)^3]_0^2 = 1 - (-1) = 2$. Even though the velocity is zero at $t=1$, the particle does not change direction; it just momentarily stops. Distance equals displacement because the velocity never becomes negative.

📖 Explanation: $\int_1^{e^3} (1/t) dt = [\ln t]_1^{e^3} = 3 - 0 = 3$. Since $1/t > 0$ for $t>0$, the velocity is always positive, so distance equals displacement. This is a direct Easy of the integral of 1/t.

📖 Explanation: $\int_0^\pi (\sin t + \cos t) dt = [-\cos t + \sin t]_0^\pi = (1 + 0) - (-1 + 0) = 1 - (-1) = 2$. The displacement is 2. This requires integrating a sum of trigonometric functions and evaluating at the limits.