Position, Velocity, Acceleration via Integrals - Calculus 2

1. A particle moves along a line such that its velocity at time $t$ is $v(t) = \frac{1}{1+t}$. What is the total distance traveled by the particle from $t=0$ to $t=e-1$?
   Answer: 1 Total distance traveled is the integral of the speed, $|v(t)|$. For $t \ge 0$, the velocity $v(t) = \frac{1}{1+t}$ is always positive, so the speed is equal to the velocity. The total distance is $\int_0^{e-1} \frac{1}{1+t} \,dt$. The integral evaluates to $[\ln|1+t|]_0^{e-1}$. Plugging in the limits of integration gives $\ln|1 + (e-1)| - \ln|1+0| = \ln(e) - \ln(1) = 1 - 0 = 1$.
2. The velocity of a particle is $v(t)$, and its initial position is $s(0)=10$. If $\int_0^5 v(t) \,dt = -7$ and $\int_5^8 v(t) \,dt = 4$, what is the particle's position at $t=8$?
   Answer: 7 The net displacement from $t=0$ to $t=8$ is the sum of the displacements over the subintervals: $s(8) - s(0) = \int_0^8 v(t) \,dt = \int_0^5 v(t) \,dt + \int_5^8 v(t) \,dt$. Using the given values, the total displacement is $-7 + 4 = -3$. So, $s(8) - s(0) = -3$. Given $s(0)=10$, we have $s(8) - 10 = -3$, which gives $s(8) = 7$.
3. A particle moves with velocity $v(t) = \sin(\pi t)$. What is the total distance traveled from $t=0$ to $t=2$?
   Answer: $4/\pi$ Total distance is the integral of speed, $|v(t)|$. The function $v(t) = \sin(\pi t)$ is positive on $(0, 1)$ and negative on $(1, 2)$. Therefore, we must split the integral: Distance = $\int_0^1 \sin(\pi t) \,dt + |\int_1^2 \sin(\pi t) \,dt|$. The antiderivative of $\sin(\pi t)$ is $-\frac{1}{\pi}\cos(\pi t)$. The first integral is $[-\frac{1}{\pi}\cos(\pi t)]_0^1 = -\frac{1}{\pi}(\cos(\pi) - \cos(0)) = -\frac{1}{\pi}(-1 - 1) = \frac{2}{\pi}$. The second integral is $[-\frac{1}{\pi}\cos(\pi t)]_1^2 = -\frac{1}{\pi}(\cos(2\pi) - \cos(\pi)) = -\frac{1}{\pi}(1 - (-1)) = -\frac{2}{\pi}$. The total distance is $\frac{2}{\pi} + |-\frac{2}{\pi}| = \frac{4}{\pi}$.
4. A particle moves on a line with velocity $v(t) = t^2 - 4t + 3$. If the particle's position at $t=0$ is $s(0)=5$, what is its position when it first changes direction for $t>0$?
   Answer: 19/3 The particle changes direction when its velocity changes sign, which occurs when $v(t)=0$. $v(t) = t^2 - 4t + 3 = (t-1)(t-3) = 0$. The first time this happens for $t>0$ is at $t=1$. Now, we need to find the particle's position at $t=1$. First, find the position function $s(t) = \int (t^2-4t+3) \,dt = \frac{t^3}{3} - 2t^2 + 3t + C$. We are given $s(0)=5$, so $C=5$. The position function is $s(t) = \frac{t^3}{3} - 2t^2 + 3t + 5$. Now, evaluate at $t=1$: $s(1) = \frac{1^3}{3} - 2(1)^2 + 3(1) + 5 = \frac{1}{3} - 2 + 3 + 5 = \frac{1}{3} + 6 = \frac{19}{3}$.
5. A particle starting from rest has an acceleration $a(t) = \frac{1}{t+1}$. What is its velocity at $t=e^2-1$?
   Answer: 2 Starting from rest means $v(0)=0$. The velocity function is $v(t) = \int a(t) \,dt = \int \frac{1}{t+1} \,dt = \ln(t+1) + C$ (for $t \ge 0$). Using the initial condition, $v(0) = \ln(0+1) + C = 0 \Rightarrow \ln(1) + C = 0 \Rightarrow C=0$. So the velocity function is $v(t) = \ln(t+1)$. We need to find the velocity at $t = e^2-1$: $v(e^2-1) = \ln((e^2-1)+1) = \ln(e^2) = 2$.
6. Two particles start at the origin and move along the x-axis. For $t \ge 0$, their velocities are given by $v_1(t) = t+1$ and $v_2(t) = t^2+1$. At what time $t>0$ do the particles have the same position again?
   Answer: $t = 3/2$ First, find the position functions for both particles. Since they start at the origin, $s_1(0)=0$ and $s_2(0)=0$. $s_1(t) = \int (t+1) \,dt = \frac{t^2}{2} + t + C_1$. Since $s_1(0)=0$, $C_1=0$. So $s_1(t) = \frac{t^2}{2} + t$. $s_2(t) = \int (t^2+1) \,dt = \frac{t^3}{3} + t + C_2$. Since $s_2(0)=0$, $C_2=0$. So $s_2(t) = \frac{t^3}{3} + t$. Now, set the positions equal to each other to find when they meet: $\frac{t^2}{2} + t = \frac{t^3}{3} + t$. Subtract $t$ from both sides: $\frac{t^2}{2} = \frac{t^3}{3}$. Since we are looking for $t>0$, we can divide by $t^2$: $\frac{1}{2} = \frac{t}{3}$. Solving for $t$ gives $t = \frac{3}{2}$.
7. A particle has velocity $v(t) = 3\sqrt{t}$ for $t \ge 0$. If $s(1)=4$, what is the value of $s(4)$?
   Answer: 18 There are two ways to solve this. Method 1: Find the indefinite integral. $s(t) = \int 3t^{1/2} \,dt = 2t^{3/2} + C$. Use $s(1)=4$ to find $C$: $2(1)^{3/2} + C = 4 \Rightarrow 2+C=4 \Rightarrow C=2$. So $s(t) = 2t^{3/2} + 2$. Now evaluate at $t=4$: $s(4) = 2(4)^{3/2} + 2 = 2(8) + 2 = 18$. Method 2: Use the Fundamental Theorem of Calculus. $s(4) = s(1) + \int_1^4 v(t) \,dt$. The integral is $\int_1^4 3t^{1/2} \,dt = [2t^{3/2}]_1^4 = 2(4^{3/2}) - 2(1^{3/2}) = 2(8) - 2(1) = 14$. Therefore, $s(4) = s(1) + 14 = 4 + 14 = 18$.
8. The velocity of a particle moving on the x-axis is $v(t) = t^2 - 8t + 12$. The particle is at $x=5$ when $t=0$. At which of the following times is the particle farthest to the left?
   Answer: $t = 6$ The particle changes direction when $v(t)=0$, which is $t^2 - 8t + 12 = (t-2)(t-6) = 0$, so at $t=2$ and $t=6$. The velocity is positive for $t<2$ (moving right), negative for $2<t<6$ (moving left), and positive for $t>6$ (moving right). The particle moves left between $t=2$ and $t=6$, so the minimum position (farthest left) will occur at $t=6$. To confirm, we can find the position function $s(t) = \int(t^2 - 8t + 12)dt = \frac{t^3}{3} - 4t^2 + 12t + C$. Since $s(0)=5$, $C=5$. The positions at the critical times are: $s(0)=5$, $s(2)=\frac{8}{3}-16+24+5 = \frac{59}{3} \approx 19.67$, $s(6)=\frac{216}{3}-4(36)+12(6)+5 = 72-144+72+5=5$. Wait, the position is the same at t=0 and t=6. Ah, so it returns to the position x=5 at t=6. It moved left from t=2 to t=6. Thus, the minimum position must be at t=6. I must have made a mistake in calculation. $s(6) = 72 - 144 + 72 + 5 = 5$. This is correct. Let's re-evaluate $s(2)$. $s(2) = 8/3 - 16 + 24 + 5 = 8/3 + 13 = (8+39)/3 = 47/3 \approx 15.67$. My arithmetic was off. So it moves right from 0 to 2, then left from 2 to 6. The position at t=6 is 5. It started at 5. This means it's a local minimum. What about farther out? At t=8, it is moving right again. $s(8) = 512/3 - 4(64) + 12(8) + 5 = 512/3 - 256 + 96 + 5 = 512/3 - 155 = (512 - 465)/3 = 47/3$. So the positions are s(0)=5, s(2)=47/3, s(6)=5, s(8)=47/3. The minimum position on this list is 5, which occurs at t=0 and t=6. The question asks for farthest left, which is minimum position. So t=6 is a valid answer. Let me re-read. Ah, I see. It moves right from 0 to 2, reaching a local max at t=2. Then it moves left from t=2 to t=6, reaching a local min at t=6. Then it moves right again. So t=6 is the only local minimum for t>0. It must be the correct answer. The distractors are t=2 (local max), t=4 (in the middle of moving left), t=8 (after it has turned around and moved right again). Yes, t=6 is the correct choice.
9. A particle moves along a line with velocity given by $v(t) = 3t^2 - 6t$ for $t \ge 0$. What is the total distance traveled by the particle on the time interval $t \in [0, 4]$?
   Answer: 24 To find the total distance, we must integrate the speed, $|v(t)|$, over the interval. First, find where the velocity is zero: $3t^2 - 6t = 3t(t-2) = 0$, which gives $t=0$ and $t=2$. The particle changes direction at $t=2$. The velocity is negative on $(0, 2)$ and positive on $(2, 4)$. The total distance is the sum of the absolute values of the integrals over these subintervals: $\text{Distance} = \int_0^4 |3t^2 - 6t| \,dt = |\int_0^2 (3t^2 - 6t)\,dt| + \int_2^4 (3t^2 - 6t)\,dt$. The antiderivative is $s(t) = t^3 - 3t^2$. The first integral is $[t^3 - 3t^2]_0^2 = (8 - 12) - 0 = -4$. The second integral is $[t^3 - 3t^2]_2^4 = (64 - 48) - (8 - 12) = 16 - (-4) = 20$. The total distance is $|-4| + 20 = 24$.
10. A particle moves along the x-axis with velocity $v(t) = 3t^2 - 12$. What is the average velocity of the particle on the interval $[0, 3]$?
    Answer: -3 Average velocity is the net displacement divided by the time interval. The displacement is $\int_0^3 v(t) \,dt$. $\int_0^3 (3t^2 - 12) \,dt = [t^3 - 12t]_0^3 = (3^3 - 12(3)) - (0) = 27 - 36 = -9$. The time interval is $3 - 0 = 3$. The average velocity is $\frac{-9}{3} = -3$.
11. A particle's acceleration is described by $a(t) = 6t - 2$. If the particle has a velocity of 4 at $t=1$ and a position of 0 at $t=1$, what is the position of the particle at $t=2$?
    Answer: 7 First, integrate acceleration to find velocity: $v(t) = \int (6t - 2) \,dt = 3t^2 - 2t + C_1$. Use $v(1)=4$ to find $C_1$: $3(1)^2 - 2(1) + C_1 = 4 \Rightarrow 1 + C_1 = 4 \Rightarrow C_1 = 3$. So, $v(t) = 3t^2 - 2t + 3$. Next, integrate velocity to find position: $s(t) = \int (3t^2 - 2t + 3) \,dt = t^3 - t^2 + 3t + C_2$. Use $s(1)=0$ to find $C_2$: $1^3 - 1^2 + 3(1) + C_2 = 0 \Rightarrow 3 + C_2 = 0 \Rightarrow C_2 = -3$. The position function is $s(t) = t^3 - t^2 + 3t - 3$. Finally, evaluate at $t=2$: $s(2) = 2^3 - 2^2 + 3(2) - 3 = 8 - 4 + 6 - 3 = 7$.
12. The acceleration of a particle moving along the x-axis is $a(t) = -4$. The velocity of the particle at $t=0$ is 12. What is the total distance traveled by the particle from $t=0$ to $t=4$?
    Answer: 20 First, find the velocity function: $v(t) = \int -4 \,dt = -4t + C$. Given $v(0)=12$, we find $C=12$. So, $v(t) = -4t + 12$. To find the total distance, we must find where the particle changes direction by setting $v(t)=0$. This gives $-4t+12=0$, so $t=3$. The particle moves in the positive direction for $t \in [0, 3)$ and in the negative direction for $t \in (3, 4]$. The total distance is the sum of the absolute values of the displacements on these intervals: Distance = $\int_0^3 (-4t+12) \,dt + |\int_3^4 (-4t+12) \,dt|$. The antiderivative is $-2t^2+12t$. The first integral is $[-2t^2+12t]_0^3 = (-2(9)+12(3)) - 0 = -18+36 = 18$. The second integral is $[-2t^2+12t]_3^4 = (-2(16)+12(4)) - (-2(9)+12(3)) = (-32+48) - (18) = 16 - 18 = -2$. The total distance is $18 + |-2| = 20$.
13. A particle has acceleration $a(t) = \cos(t)$. Its velocity at $t=0$ is 0, and its position at $t=0$ is 2. What is the position of the particle at $t=\pi$?
    Answer: 4 First, find the velocity function: $v(t) = \int \cos(t) \,dt = \sin(t) + C_1$. Use $v(0) = 0$: $\sin(0) + C_1 = 0 \Rightarrow C_1 = 0$. So, $v(t) = \sin(t)$. Now, find the position function: $s(t) = \int \sin(t) \,dt = -\cos(t) + C_2$. Use $s(0) = 2$: $-\cos(0) + C_2 = 2 \Rightarrow -1 + C_2 = 2 \Rightarrow C_2 = 3$. Thus, the position function is $s(t) = 3 - \cos(t)$. Finally, find the position at $t=\pi$: $s(\pi) = 3 - \cos(\pi) = 3 - (-1) = 4$.
14. The acceleration of a particle is $a(t) = k\cos(t)$ for some constant $k$. The particle starts from rest at $t=0$ and has a position of $s(0)=5$. If the particle's position at $t=\pi$ is $s(\pi)=9$, what is the value of $k$?
    Answer: 2 First, find $v(t)$ by integrating $a(t)$: $v(t) = \int k\cos(t) \,dt = k\sin(t) + C_1$. The particle starts from rest, so $v(0)=0$. $k\sin(0) + C_1 = 0 \Rightarrow C_1 = 0$. So, $v(t) = k\sin(t)$. Next, find $s(t)$ by integrating $v(t)$: $s(t) = \int k\sin(t) \,dt = -k\cos(t) + C_2$. We are given $s(0)=5$. $-k\cos(0) + C_2 = 5 \Rightarrow -k + C_2 = 5 \Rightarrow C_2 = 5+k$. So, $s(t) = -k\cos(t) + 5 + k$. Now use the condition $s(\pi)=9$: $s(\pi) = -k\cos(\pi) + 5 + k = -k(-1) + 5 + k = k + 5 + k = 2k+5$. We set this equal to 9: $2k+5=9 \Rightarrow 2k=4 \Rightarrow k=2$.
15. The velocity $v(t)$ of a particle moving along a line is a continuous function. If $\int_2^5 v(t) \,dt = 0$, which of the following statements must be true?
    Answer: The particle's position at $t=5$ is the same as its position at $t=2$. The definite integral of the velocity function, $\int_a^b v(t) \,dt$, represents the net displacement of the particle from time $t=a$ to $t=b$, which is $s(b) - s(a)$. Given $\int_2^5 v(t) \,dt = 0$, it means $s(5) - s(2) = 0$, or $s(5) = s(2)$. This means the particle's final position is the same as its initial position. Distractor A is false because the velocity could be positive and negative within the interval, with the areas canceling out. Distractor C is false because total distance is $\int_2^5 |v(t)| \,dt$, which would be positive if $v(t)$ is not identically zero. Distractor D is not necessarily true; for example, if $v(t)=(t-3.5)$, the integral from 2 to 5 is 0, but $a(t)=1$ is never zero.
16. A particle's acceleration is $a(t) = t e^{-t}$. If the particle is at rest at $t=0$ and its initial position is $s(0) = -1$, what is its position at $t=1$?
    Answer: $3e^{-1} - 2$ First, find $v(t)$ by integrating $a(t)$ using integration by parts ($u=t, dv=e^{-t}dt$): $v(t) = \int t e^{-t} \,dt = -t e^{-t} - \int -e^{-t} \,dt = -t e^{-t} - e^{-t} + C_1$. Given $v(0)=0$, $-0 - e^0 + C_1 = 0 \Rightarrow -1 + C_1 = 0 \Rightarrow C_1 = 1$. So $v(t) = 1 - t e^{-t} - e^{-t}$. Next, find $s(t)$ by integrating $v(t)$. The integral of $-t e^{-t} - e^{-t}$ is $t e^{-t} + 2e^{-t}$. So $s(t) = t + (t e^{-t} + e^{-t}) + C_2$. Let me re-integrate. $\int(1 - te^{-t} - e^{-t})dt = t - \int(te^{-t}+e^{-t})dt$. The derivative of $te^{-t}$ is $e^{-t}-te^{-t}$. So $te^{-t}$ is not the antiderivative. The antiderivative of $-te^{-t}-e^{-t}$ is $te^{-t}$. No, that's not right either. Let's re-do the $s(t)$ integral. $s(t) = \int (1 - (t+1)e^{-t}) \,dt = t - \int (t+1)e^{-t} \,dt$. Use parts for the integral: $u=t+1, dv=e^{-t}dt$. Then $du=dt, v=-e^{-t}$. So $\int (t+1)e^{-t} \,dt = -(t+1)e^{-t} - \int -e^{-t} \,dt = -(t+1)e^{-t} - e^{-t} = -te^{-t}-2e^{-t}$. So $s(t) = t - (-te^{-t}-2e^{-t}) + D = t + te^{-t} + 2e^{-t} + D$. Check derivative: $s'(t) = 1 + (e^{-t}-te^{-t}) - 2e^{-t} = 1 - te^{-t} - e^{-t} = v(t)$. Correct. Now use $s(0)=-1$: $0 + 0 + 2e^0 + D = -1 \Rightarrow 2+D=-1 \Rightarrow D=-3$. So $s(t) = t + te^{-t} + 2e^{-t} - 3$. Finally, $s(1) = 1 + 1e^{-1} + 2e^{-1} - 3 = 3e^{-1} - 2$.
17. The acceleration of a particle is given by $a(t) = 2t - 6$. At $t=0$, its velocity is $v(0) = 5$. What is the total distance traveled by the particle from $t=0$ to $t=4$?
    Answer: $\frac{34}{3}$ First, find $v(t)$ by integrating $a(t)$: $v(t) = \int (2t - 6) \,dt = t^2 - 6t + C$. Using $v(0)=5$, we find $C=5$, so $v(t) = t^2 - 6t + 5$. To find total distance, we must find where $v(t)=0$: $(t-1)(t-5)=0$, so $t=1$ and $t=5$. In the interval $[0, 4]$, the velocity changes sign at $t=1$. Total distance is $\int_0^1 (t^2 - 6t + 5) \,dt + |\int_1^4 (t^2 - 6t + 5) \,dt|$. The antiderivative is $s(t) = \frac{t^3}{3} - 3t^2 + 5t$. The first integral is $[s(t)]_0^1 = (\frac{1}{3} - 3 + 5) - 0 = \frac{7}{3}$. The second integral is $[s(t)]_1^4 = (\frac{64}{3} - 48 + 20) - (\frac{7}{3}) = \frac{64}{3} - 28 - \frac{7}{3} = \frac{57}{3} - 28 = 19 - 28 = -9$. Total distance = $\frac{7}{3} + |-9| = \frac{7}{3} + 9 = \frac{7+27}{3} = \frac{34}{3}$.
18. The acceleration of an object is given by $a(t) = 12t^2 - 2$. The change in the object's velocity from $t=1$ to $t=3$ is 90. What is its position at $t=1$ if $s(0)=5$ and its velocity at $t=0$ is 3?
    Answer: 7 This question contains extra information designed to be a distractor. The change in velocity from $t=1$ to $t=3$ can be calculated as $\int_1^3 (12t^2-2)dt = [4t^3-2t]_1^3 = (4(27)-6)-(4(1)-2) = (108-6)-(2) = 102-2 = 100$. The problem states this value is 90, which is a contradiction, but we don't need this information to solve the problem. We only need $a(t)$, $v(0)$, and $s(0)$. First, find $v(t)$: $v(t) = \int (12t^2 - 2) \,dt = 4t^3 - 2t + C$. Given $v(0)=3$, we have $C=3$. So, $v(t) = 4t^3 - 2t + 3$. Next, find $s(t)$: $s(t) = \int (4t^3 - 2t + 3) \,dt = t^4 - t^2 + 3t + D$. Given $s(0)=5$, we have $D=5$. So, $s(t) = t^4 - t^2 + 3t + 5$. The question asks for $s(1)$: $s(1) = 1^4 - 1^2 + 3(1) + 5 = 1 - 1 + 3 + 5 = 7$.
19. The rate of change of velocity of a particle is proportional to the square of the time $t$, i.e., $a(t) = kt^2$. The particle starts at position $s(0)=2$ with velocity $v(0)=2$. At $t=1$, the velocity is $v(1)=3$. What is the position $s(2)$?
    Answer: 10 First, we use the given information to find $k$. $v(t) = \int kt^2 \,dt = \frac{k}{3}t^3 + C$. We know $v(0)=2$, so $C=2$. Thus $v(t) = \frac{k}{3}t^3 + 2$. We are given $v(1)=3$, so $\frac{k}{3}(1)^3 + 2 = 3 \Rightarrow \frac{k}{3} = 1 \Rightarrow k=3$. The velocity function is $v(t) = t^3 + 2$. Now we find the position function $s(t) = \int (t^3+2) \,dt = \frac{t^4}{4} + 2t + D$. We are given $s(0)=2$, so $D=2$. The position function is $s(t) = \frac{t^4}{4} + 2t + 2$. Finally, we find $s(2)$: $s(2) = \frac{2^4}{4} + 2(2) + 2 = \frac{16}{4} + 4 + 2 = 4 + 4 + 2 = 10$.
20. A particle's position is given by $s(t) = \int_0^t (x^2 - 2x) \,dx$. What is the total distance traveled by the particle from $t=0$ to $t=3$?
    Answer: 8/3 By the Fundamental Theorem of Calculus, the particle's velocity is $v(t) = s'(t) = t^2 - 2t$. To find the total distance, we must integrate the speed $|v(t)|$ from 0 to 3. First, find where $v(t) = 0$: $t(t-2) = 0$, so $t=0$ and $t=2$. The velocity is negative on $(0, 2)$ and positive on $(2, 3)$. The total distance is $\int_0^3 |t^2 - 2t| \,dt = |\int_0^2 (t^2 - 2t)\,dt| + \int_2^3 (t^2 - 2t)\,dt$. The antiderivative is $\frac{t^3}{3} - t^2$. The first integral is $[\frac{t^3}{3} - t^2]_0^2 = (\frac{8}{3} - 4) - 0 = -\frac{4}{3}$. The second integral is $[\frac{t^3}{3} - t^2]_2^3 = (\frac{27}{3} - 9) - (\frac{8}{3} - 4) = (9 - 9) - (-\frac{4}{3}) = \frac{4}{3}$. The total distance is $|-\frac{4}{3}| + \frac{4}{3} = \frac{8}{3}$.

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