Position, Velocity, Acceleration via Integrals - Calculus 1

1. A particle has acceleration $a(t)$, initial velocity $v(0)$, and initial position $s(0)$. Which of the following expressions represents the displacement of the particle from $t=0$ to $t=3$?
   Answer: $\int_0^3 \left( v(0) + \int_0^t a(\tau) d\tau \right) dt$ Displacement is the change in position, $s(3) - s(0)$, which is given by the definite integral of the velocity function, $\int_0^3 v(t) dt$. The velocity function $v(t)$ can be expressed in terms of the initial velocity and acceleration: $v(t) = v(0) + \int_0^t a(\tau) d\tau$. Substituting this expression for $v(t)$ into the displacement integral gives $\int_0^3 \left( v(0) + \int_0^t a(\tau) d\tau \right) dt$.
2. A particle moves along a line with acceleration $a(t) = 2e^t$. Its velocity at $t=0$ is $v(0) = -2$. What is the total distance traveled by the particle from $t=0$ to $t=\ln(3)$?
   Answer: $8\ln(2) - 4\ln(3)$ First, find the velocity function: $v(t) = \int 2e^t dt = 2e^t + C$. Using $v(0)=-2$, we get $2e^0 + C = -2 \implies 2+C=-2 \implies C=-4$. So, $v(t) = 2e^t - 4$. Total distance requires integrating $|v(t)|$. Find when $v(t)=0$: $2e^t - 4 = 0 \implies e^t = 2 \implies t = \ln(2)$. This is within the interval $[0, \ln(3)]$. For $t \in [0, \ln(2))$, $v(t) < 0$. For $t \in (\ln(2), \ln(3)]$, $v(t) > 0$. We split the integral: Total distance = $\int_0^{\ln(2)} -(2e^t - 4) dt + \int_{\ln(2)}^{\ln(3)} (2e^t - 4) dt$. First integral: $[4t - 2e^t]_0^{\ln(2)} = (4\ln(2) - 2e^{\ln(2)}) - (0 - 2e^0) = 4\ln(2) - 4 - (-2) = 4\ln(2) - 2$. Second integral: $[2e^t - 4t]_{\ln(2)}^{\ln(3)} = (2e^{\ln(3)} - 4\ln(3)) - (2e^{\ln(2)} - 4\ln(2)) = (6 - 4\ln(3)) - (4 - 4\ln(2)) = 2 - 4\ln(3) + 4\ln(2)$. Total distance = $(4\ln(2) - 2) + (2 - 4\ln(3) + 4\ln(2)) = 8\ln(2) - 4\ln(3)$.
3. A particle's velocity is given by $v(t) = \frac{6}{(t+2)^2}$ for $t \ge 0$. What is the total displacement of the particle from $t=0$ to $t=\infty$?
   Answer: $3$ The total displacement is the value of the improper integral of the velocity function from $t=0$ to $t=\infty$. Displacement $= \int_{0}^{\infty} \frac{6}{(t+2)^2} dt = \lim_{b \to \infty} \int_{0}^{b} 6(t+2)^{-2} dt$. The antiderivative of $6(t+2)^{-2}$ is $\frac{6(t+2)^{-1}}{-1} = -6(t+2)^{-1} = \frac{-6}{t+2}$. Now, we evaluate the definite integral: $\lim_{b \to \infty} [\frac{-6}{t+2}]_{0}^{b} = \lim_{b \to \infty} (\frac{-6}{b+2} - \frac{-6}{0+2})$. As $b \to \infty$, the term $\frac{-6}{b+2}$ approaches 0. The second term is $-\frac{-6}{2} = 3$. So, the total displacement is $0 - (-3) = 3$. A: This is the constant in the denominator of the velocity function, not the displacement. C: This is the numerator of the velocity function, not the displacement. D: This would be the case if the integral diverged, but because the power of $(t+2)$ is greater than 1, the integral converges.
4. A particle's acceleration is given by $a(t) = 12t$. At time $t=1$, its velocity is $v(1)=8$ and its position is $s(1)=-2$. What is the position of the particle at $t=2$?
   Answer: 14 First, find $v(t)$ by integrating $a(t)$: $v(t) = \int 12t dt = 6t^2 + C_1$. Use $v(1)=8$ to find $C_1$: $6(1)^2 + C_1 = 8 \implies C_1 = 2$. So, $v(t) = 6t^2 + 2$. Next, find $s(t)$ by integrating $v(t)$: $s(t) = \int (6t^2 + 2) dt = 2t^3 + 2t + C_2$. Use $s(1)=-2$ to find $C_2$: $2(1)^3 + 2(1) + C_2 = -2 \implies 4 + C_2 = -2 \implies C_2 = -6$. So, $s(t) = 2t^3 + 2t - 6$. Finally, find the position at $t=2$: $s(2) = 2(2)^3 + 2(2) - 6 = 16 + 4 - 6 = 14$.
5. A particle moves with velocity $v(t) = \cos(t)$. For which of the following values of $c > 0$ is the particle's average velocity on the interval $[0, c]$ equal to $\frac{2}{\pi}$?
   Answer: $\frac{\pi}{2}$ The average velocity on $[0, c]$ is given by the formula for the average value of a function: $\frac{1}{c-0} \int_{0}^{c} v(t) dt$. In this case, it is $\frac{1}{c} \int_{0}^{c} \cos(t) dt = \frac{1}{c} [\sin(t)]_{0}^{c} = \frac{1}{c}(\sin(c) - \sin(0)) = \frac{\sin(c)}{c}$. We need to find which of the given options for $c$ satisfies $\frac{\sin(c)}{c} = \frac{2}{\pi}$. We can test the choices: A) For $c=\frac{\pi}{4}$: $\frac{\sin(\pi/4)}{\pi/4} = \frac{\sqrt{2}/2}{\pi/4} = \frac{2\sqrt{2}}{\pi} \neq \frac{2}{\pi}$. B) For $c=\frac{\pi}{3}$: $\frac{\sin(\pi/3)}{\pi/3} = \frac{\sqrt{3}/2}{\pi/3} = \frac{3\sqrt{3}}{2\pi} \neq \frac{2}{\pi}$. C) For $c=\frac{\pi}{2}$: $\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}$. This is correct. D) For $c=\pi$: $\frac{\sin(\pi)}{\pi} = \frac{0}{\pi} = 0 \neq \frac{2}{\pi}$.
6. A particle's position is given by $s(t) = \int_{0}^{t} (x^2-2x)e^{-x} dx$. For $t>0$, when is the particle's acceleration zero?
   Answer: $t = 2 - \sqrt{2}$ The position is given as an integral, so we can find the velocity and acceleration using the Fundamental Theorem of Calculus. Velocity $v(t) = s'(t) = (t^2 - 2t)e^{-t}$. Acceleration $a(t) = v'(t)$. We use the product rule to differentiate $v(t)$. Let $u = t^2 - 2t$ and $w = e^{-t}$. Then $u' = 2t - 2$ and $w' = -e^{-t}$. $a(t) = u'w + uw' = (2t - 2)e^{-t} + (t^2 - 2t)(-e^{-t})$. $a(t) = e^{-t} [ (2t - 2) - (t^2 - 2t) ] = e^{-t} [ 2t - 2 - t^2 + 2t ] = e^{-t} [-t^2 + 4t - 2]$. We set $a(t)=0$. Since $e^{-t}$ is never zero, we only need to solve $-t^2 + 4t - 2 = 0$, or $t^2 - 4t + 2 = 0$. Using the quadratic formula: $t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)} = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$. We have two possible times: $t = 2 - \sqrt{2} \approx 2 - 1.414 = 0.586$ and $t = 2 + \sqrt{2} \approx 3.414$. Both are greater than 0. The question asks 'when is the acceleration zero?' and provides $2-\sqrt{2}$ as an option. Without further constraints, this is a valid answer. B: This is where the velocity is zero, $t(t-2)=0$. C: This is the other time when acceleration is zero. D: This is where the derivative of the polynomial part of acceleration is zero, i.e., the vertex of the parabola $-t^2+4t-2$.
7. The acceleration of a particle is $a(t) = (t+1)^{-3}$ for $t \ge 0$. The particle starts from rest. What is the particle's velocity at $t=1$?
   Answer: $\frac{3}{8}$ First, find the velocity $v(t)$ by integrating the acceleration $a(t)$. 'Starts from rest' means $v(0)=0$. $v(t) = \int (t+1)^{-3} dt = \frac{(t+1)^{-2}}{-2} + C = -\frac{1}{2(t+1)^2} + C$. Now use the initial condition $v(0)=0$ to find $C$. $0 = -\frac{1}{2(0+1)^2} + C \Rightarrow 0 = -\frac{1}{2} + C \Rightarrow C = \frac{1}{2}$. So the velocity function is $v(t) = \frac{1}{2} - \frac{1}{2(t+1)^2}$. Finally, evaluate the velocity at $t=1$. $v(1) = \frac{1}{2} - \frac{1}{2(1+1)^2} = \frac{1}{2} - \frac{1}{2(4)} = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$. B: This is the value of the constant of integration, $C$, not the velocity at $t=1$. C: This would be the result of a sign error when finding $C$, leading to $C = -1/2$. D: This is the value of the integral part of $v(1)$ if the constant $C$ is forgotten.
8. A particle moves with acceleration $a(t) = \sin(t) - \cos(t)$. If $v(0) = 0$, what is the particle's maximum velocity on the interval $0 \le t \le 2\pi$?
   Answer: $1+\sqrt{2}$ First, find the velocity function $v(t)$ by integrating $a(t)$. $v(t) = \int (\sin(t) - \cos(t)) dt = -\cos(t) - \sin(t) + C$. Using $v(0) = 0$, we have $0 = -\cos(0) - \sin(0) + C \Rightarrow 0 = -1 - 0 + C \Rightarrow C = 1$. So, $v(t) = 1 - \cos(t) - \sin(t)$. To find the maximum velocity, we must find the critical points of $v(t)$ by setting its derivative, $v'(t) = a(t)$, to zero. $a(t) = \sin(t) - \cos(t) = 0 \Rightarrow \sin(t) = \cos(t) \Rightarrow \tan(t) = 1$. On the interval $[0, 2\pi]$, this occurs at $t = \pi/4$ and $t = 5\pi/4$. We evaluate $v(t)$ at the critical points and the endpoints $t=0$ and $t=2\pi$: $v(0) = 1 - \cos(0) - \sin(0) = 1 - 1 - 0 = 0$. $v(\pi/4) = 1 - \cos(\pi/4) - \sin(\pi/4) = 1 - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 1 - \sqrt{2}$. (This is a local minimum). $v(5\pi/4) = 1 - \cos(5\pi/4) - \sin(5\pi/4) = 1 - (-\frac{\sqrt{2}}{2}) - (-\frac{\sqrt{2}}{2}) = 1 + \sqrt{2}$. (This is a local maximum). $v(2\pi) = 1 - \cos(2\pi) - \sin(2\pi) = 1 - 1 - 0 = 0$. Comparing the values $0$, $1 - \sqrt{2}$, and $1 + \sqrt{2}$, the maximum velocity is $1 + \sqrt{2}$. B: This is the value of the constant of integration, $C$. C: This would be the constant of integration if the initial condition were $v(0)=1$. D: This is the value of the local minimum velocity, not the maximum.
9. The acceleration of a particle is given by $a(t) = 6t$. The velocity of the particle is 12 at $t=2$. The particle's initial position is $s(0) = 5$. What is the total distance traveled by the particle from $t=0$ to $t=2$?
   Answer: $8$ First, find the velocity function $v(t)$ by integrating $a(t)$. $v(t) = \int 6t dt = 3t^2 + C$. Use the condition $v(2)=12$ to find $C$: $12 = 3(2^2) + C \Rightarrow 12 = 12 + C \Rightarrow C = 0$. So, $v(t) = 3t^2$. To find the total distance traveled on the interval $[0, 2]$, we need to integrate the speed, $|v(t)|$. On the interval $[0, 2]$, $v(t) = 3t^2$ is always non-negative, so $|v(t)| = v(t)$. Therefore, the total distance is equal to the displacement. Total Distance $= \int_{0}^{2} 3t^2 dt = [t^3]_{0}^{2} = 2^3 - 0^3 = 8$. The initial position $s(0)=5$ is extra information not needed for calculating total distance. B: This is the velocity at $t=2$, not the total distance. C: This is the final position of the particle, $s(2) = s(0) + \text{displacement} = 5 + 8 = 13$. D: This could result from an arithmetic error, for example, adding the initial position, velocity, and displacement in some way.
10. A particle moves with velocity $v(t) = 3\sqrt{t}$ for $t \ge 0$. Its position at $t=1$ is $s(1) = -1$. What is the particle's position at $t=4$?
    Answer: $13$ We can find the position $s(4)$ using the Fundamental Theorem of Calculus: $s(4) = s(1) + \int_{1}^{4} v(t) dt$. First, calculate the definite integral: $\int_{1}^{4} 3\sqrt{t} dt = \int_{1}^{4} 3t^{1/2} dt = [3 \cdot \frac{t^{3/2}}{3/2}]_{1}^{4} = [2t^{3/2}]_{1}^{4}$. Evaluate at the limits: $2(4)^{3/2} - 2(1)^{3/2} = 2(8) - 2(1) = 16 - 2 = 14$. This is the displacement from $t=1$ to $t=4$. Now, find the final position: $s(4) = s(1) + 14 = -1 + 14 = 13$. A: This is the result of calculating the displacement from $t=1$ to $t=4$ as $13$ and then subtracting $s(1)$ instead of adding it. C: This is the displacement from $t=1$ to $t=4$, not the final position. D: This is the final position if the displacement were calculated from $t=0$ to $t=4$, which is $s(4) = s(0) + 16$. Assuming $s(0)=s(1)=-1$ yields $s(4)=15$, a common mistake.
11. The acceleration of an object is given by $a(t) = 12t$. The object is at rest at $t=1$. What is its displacement from $t=0$ to $t=2$?
    Answer: $4$ First, find the velocity function $v(t)$ by integrating $a(t)$. $v(t) = \int 12t dt = 6t^2 + C$. The condition 'at rest at t=1' means $v(1)=0$. So, $0 = 6(1)^2 + C \Rightarrow C = -6$. Thus, $v(t) = 6t^2 - 6$. Displacement from $t=0$ to $t=2$ is the definite integral of velocity: $\int_{0}^{2} (6t^2 - 6) dt = [2t^3 - 6t]_{0}^{2}$. Evaluating at the limits: $(2(2)^3 - 6(2)) - (2(0)^3 - 6(0)) = (16 - 12) - 0 = 4$. B: This is the total distance traveled, $\int_{0}^{2} |6t^2 - 6| dt$, a common error when asked for displacement. C: This is the displacement that would be calculated if the constant of integration $C$ was ignored (i.e., assuming $v(0)=0$). D: This is the displacement that would be calculated if a sign error was made in finding the constant of integration (i.e., $C=6$).
12. A particle moves along a line with acceleration $a(t) = -\frac{1}{(t+1)^2}$ for $t \ge 0$. If the particle has an initial velocity of $v(0)=2$, what is its velocity as $t \to \infty$?
    Answer: $1$ To find the velocity function $v(t)$, we integrate the acceleration function $a(t)$. $v(t) = \int -\frac{1}{(t+1)^2} dt = \int -(t+1)^{-2} dt$. The antiderivative is $- \frac{(t+1)^{-1}}{-1} + C = (t+1)^{-1} + C = \frac{1}{t+1} + C$. We are given the initial condition $v(0)=2$. We use this to solve for $C$. $2 = \frac{1}{0+1} + C \Rightarrow 2 = 1 + C \Rightarrow C = 1$. So, the velocity function is $v(t) = \frac{1}{t+1} + 1$. To find the velocity as $t \to \infty$, we take the limit of $v(t)$. $\lim_{t \to \infty} v(t) = \lim_{t \to \infty} (\frac{1}{t+1} + 1)$. As $t \to \infty$, the term $\frac{1}{t+1}$ approaches 0. So, $\lim_{t \to \infty} v(t) = 0 + 1 = 1$. The velocity approaches a terminal value of 1. A: This would be the limit if the constant of integration $C$ were 0. C: This is the initial velocity, not the terminal velocity. D: This is incorrect because the limit exists and is finite.
13. A particle moves along the x-axis with velocity given by $v(t) = 3t^2 - 12$ for $t \ge 0$. What is the total distance traveled by the particle during the time interval $0 \le t \le 3$?
    Answer: 23 Total distance is the integral of the absolute value of velocity, $\int_a^b |v(t)| dt$. First, find where the velocity is zero: $3t^2 - 12 = 0 \implies t^2 = 4 \implies t = 2$ (since $t \ge 0$). The velocity is negative on $[0, 2)$ and positive on $(2, 3]$. The integral must be split at $t=2$. Total distance = $\int_0^3 |3t^2 - 12| dt = \int_0^2 -(3t^2 - 12) dt + \int_2^3 (3t^2 - 12) dt$. Integrating gives $[12t - t^3]_0^2 + [t^3 - 12t]_2^3$. Evaluating the first part: $(24 - 8) - 0 = 16$. Evaluating the second part: $(27 - 36) - (8 - 24) = -9 - (-16) = 7$. The total distance is $16 + 7 = 23$.
14. A particle moves along a line with acceleration $a(t) = 6t - 2$. At time $t=0$, its velocity is $v(0) = 3$. What is the displacement of the particle over the time interval $[1, 4]$?
    Answer: 57 First, find the velocity function $v(t)$ by integrating $a(t)$. $v(t) = \int (6t - 2) dt = 3t^2 - 2t + C$. Use the initial condition $v(0) = 3$ to find $C$: $3(0)^2 - 2(0) + C = 3 \implies C = 3$. So, $v(t) = 3t^2 - 2t + 3$. Displacement is the integral of velocity. Displacement = $\int_1^4 (3t^2 - 2t + 3) dt = [t^3 - t^2 + 3t]_1^4$. Evaluate at the limits: $((4)^3 - (4)^2 + 3(4)) - ((1)^3 - (1)^2 + 3(1)) = (64 - 16 + 12) - (1 - 1 + 3) = 60 - 3 = 57$.
15. The velocity of a particle is given by $v(t) = e^t - 3$. What is the total distance traveled by the particle from $t=0$ to $t=2$?
    Answer: $e^2 + 6\ln(3) - 11$ To find the total distance, we must integrate the speed, $|v(t)|$. First, find where the velocity is zero: $e^t - 3 = 0 \Rightarrow e^t = 3 \Rightarrow t = \ln(3)$. Since $1 < \ln(3) < 2$, the velocity changes sign in the interval $[0, 2]$. For $t < \ln(3)$, $v(t) < 0$, and for $t > \ln(3)$, $v(t) > 0$. The total distance is $\int_{0}^{2} |e^t - 3| dt = \int_{0}^{\ln(3)} -(e^t - 3) dt + \int_{\ln(3)}^{2} (e^t - 3) dt$. $\int (3 - e^t) dt = 3t - e^t$. $\int (e^t - 3) dt = e^t - 3t$. First integral: $[3t - e^t]_{0}^{\ln(3)} = (3\ln(3) - e^{\ln(3)}) - (0 - e^0) = 3\ln(3) - 3 - (-1) = 3\ln(3) - 2$. Second integral: $[e^t - 3t]_{\ln(3)}^{2} = (e^2 - 6) - (e^{\ln(3)} - 3\ln(3)) = e^2 - 6 - (3 - 3\ln(3)) = e^2 - 9 + 3\ln(3)$. Total distance = $(3\ln(3) - 2) + (e^2 - 9 + 3\ln(3)) = e^2 + 6\ln(3) - 11$. A: This is the displacement $\int_{0}^{2} (e^t - 3) dt = e^2 - 7$, which incorrectly ignores the change in direction. C: This would be the result if one made the common error of evaluating $e^0=0$ instead of $e^0=1$. D: This could result from various calculation errors, such as miscalculating the antiderivative or the bounds of integration.
16. The velocity of a particle is $v(t) = t^2 - 4t + 3$. The particle is at position $s(1) = 10$. What is the maximum position of the particle on the interval $[0, 4]$?
    Answer: $10$ To find the maximum position, we must check the critical points (where $s'(t) = v(t) = 0$) and the endpoints of the interval. $v(t) = t^2 - 4t + 3 = (t-1)(t-3) = 0$, so critical points are at $t=1$ and $t=3$. The candidates for the time of maximum position on $[0, 4]$ are $t=0, 1, 3, 4$. Specifically, since $v(t)$ is positive on $[0,1)$, negative on $(1,3)$, and positive on $(3,4]$, the potential maxima are at $t=1$ (local max) and $t=4$ (endpoint). First, find the position function: $s(t) = \int (t^2 - 4t + 3) dt = \frac{1}{3}t^3 - 2t^2 + 3t + C$. Use $s(1) = 10$ to find C: $10 = \frac{1}{3} - 2 + 3 + C \Rightarrow 10 = \frac{4}{3} + C \Rightarrow C = \frac{26}{3}$. So, $s(t) = \frac{1}{3}t^3 - 2t^2 + 3t + \frac{26}{3}$. Now evaluate the position at the candidates: $s(0) = \frac{26}{3} \approx 8.67$. $s(1) = 10$. $s(4) = \frac{1}{3}(4)^3 - 2(4)^2 + 3(4) + \frac{26}{3} = \frac{64}{3} - 32 + 12 + \frac{26}{3} = \frac{90}{3} - 20 = 30 - 20 = 10$. The maximum position is 10. A: This is the total distance traveled on $[1,4]$, not the maximum position. B: This is the position at $t=0$ and $t=3$ (a local minimum), not the maximum. D: This value, $10 + 4/3$, might result from adding the displacement on $[3,4]$ to the maximum at $t=1$ without considering the intermediate minimum.
17. A particle moves with acceleration $a(t) = \sin(t) - \cos(t)$. If $v(0) = 0$ and $s(0) = 1$, what is the particle's maximum velocity on the interval $0 \le t \le 2\pi$?
    Answer: $1+\sqrt{2}$ First, find the velocity function $v(t)$ by integrating $a(t)$. $v(t) = \int (\sin(t) - \cos(t)) dt = -\cos(t) - \sin(t) + C$. Using $v(0) = 0$, we have $0 = -\cos(0) - \sin(0) + C \Rightarrow 0 = -1 - 0 + C \Rightarrow C = 1$. So, $v(t) = 1 - \cos(t) - \sin(t)$. To find the maximum velocity, we find critical points by setting $v'(t) = a(t) = 0$. $\sin(t) - \cos(t) = 0 \Rightarrow \tan(t) = 1$. On $[0, 2\pi]$, this occurs at $t = \pi/4$ and $t = 5\pi/4$. Evaluating $v(t)$ at critical points and endpoints: $v(0) = 1 - 1 - 0 = 0$ $v(\pi/4) = 1 - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 1 - \sqrt{2}$ (minimum) $v(5\pi/4) = 1 - (-\frac{\sqrt{2}}{2}) - (-\frac{\sqrt{2}}{2}) = 1 + \sqrt{2}$ (maximum) $v(2\pi) = 1 - 1 - 0 = 0$ The maximum velocity is $1 + \sqrt{2}$. B: This would be the maximum if the constant of integration were 2 instead of 1. C: This is twice the initial velocity. D: This is the minimum velocity on the given interval.
18. A particle's acceleration is $a(t) = 6t - 2$. Its velocity at $t=1$ is $v(1)=0$ and its position at $t=1$ is $s(1)=4$. What is the particle's position at $t=0$?
    Answer: $5$ First, integrate $a(t)$ to find $v(t)$: $v(t) = \int (6t-2) dt = 3t^2 - 2t + C_1$. Use $v(1)=0$ to find $C_1$: $0 = 3(1)^2 - 2(1) + C_1 \Rightarrow 0 = 1 + C_1 \Rightarrow C_1 = -1$. So, $v(t) = 3t^2 - 2t - 1$. Next, integrate $v(t)$ to find $s(t)$: $s(t) = \int (3t^2 - 2t - 1) dt = t^3 - t^2 - t + C_2$. Use $s(1)=4$ to find $C_2$: $4 = 1^3 - 1^2 - 1 + C_2 \Rightarrow 4 = -1 + C_2 \Rightarrow C_2 = 5$. So, the position function is $s(t) = t^3 - t^2 - t + 5$. Finally, find the position at $t=0$: $s(0) = 0^3 - 0^2 - 0 + 5 = 5$. A: This happens if one correctly finds $C_1=-1$ but forgets the second constant of integration $C_2$, assuming $C_2=0$. B: This happens if a sign error is made when finding $C_1$ (i.e., $C_1=1$), leading to $C_2=3$ and $s(0)=3$. C: This happens if one forgets the first constant $C_1$ (assuming $C_1=0$), which leads to $C_2=4$ and $s(0)=4$.
19. A rocket has a constant acceleration $a(t) = -10$ m/s$^2$. It is launched upward from the ground ($s(0)=0$) and is observed to be at a height of 240 meters at $t=4$ seconds. What was its initial launch velocity, $v(0)$?
    Answer: 80 m/s First, integrate the acceleration to find the velocity function: $v(t) = \int -10 dt = -10t + C_1$. At $t=0$, the velocity is the initial velocity $v(0)$, so $v(0) = -10(0) + C_1 \implies C_1 = v(0)$. Thus, $v(t) = -10t + v(0)$. Next, integrate the velocity to find the position function: $s(t) = \int (-10t + v(0)) dt = -5t^2 + v(0)t + C_2$. The rocket is launched from the ground, so $s(0)=0$. This gives $-5(0)^2 + v(0)(0) + C_2 = 0 \implies C_2=0$. So, $s(t) = -5t^2 + v(0)t$. We are given that $s(4) = 240$. Substitute $t=4$ into the position function: $240 = -5(4)^2 + v(0)(4) = -5(16) + 4v(0) = -80 + 4v(0)$. Now, solve for $v(0)$: $320 = 4v(0) \implies v(0) = 80$ m/s.
20. The velocity of a particle is $v(t)$. The expression $\frac{1}{4} \int_{1}^{5} v(t) dt$ represents which of the following?
    Answer: The particle's average velocity from $t=1$ to $t=5$. The expression $\frac{1}{b-a} \int_{a}^{b} f(x) dx$ is the definition of the average value of the function $f(x)$ on the interval $[a, b]$. In this case, $f(x)$ is the velocity function $v(t)$, $a=1$, and $b=5$. Therefore, $\frac{1}{5-1} \int_{1}^{5} v(t) dt = \frac{1}{4} \int_{1}^{5} v(t) dt$ represents the average velocity of the particle on the time interval $[1, 5]$. B: Average acceleration is the change in velocity divided by the change in time, $\frac{v(5)-v(1)}{5-1}$, which by the Mean Value Theorem for Integrals is equal to the average value of acceleration $\frac{1}{4}\int_1^5 a(t)dt$, not velocity. C: Total distance traveled is given by $\int_{1}^{5} |v(t)| dt$. D: The net change in position is $\int_{1}^{5} v(t) dt$. The expression divides this by 4 (the duration), not the initial time.

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