Calculus 2 Quiz: Position Velocity Acceleration Via Integrals
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Position Velocity Acceleration Via IntegralsQuestion 1 of 20

The velocity of a particle is v(t)v(t), and its initial position is s(0)=10s(0)=10. If ∫05v(t) dt=−7\int_0^5 v(t) \,dt = -7 and ∫58v(t) dt=4\int_5^8 v(t) \,dt = 4, what is the particle's position at t=8t=8?

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Calculus 2 Quiz

Calculus 2 Quiz: Position Velocity Acceleration Via Integrals

Practice Position Velocity Acceleration Via Integrals in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Position Velocity Acceleration Via Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

The velocity of a particle is v(t)v(t), and its initial position is s(0)=10s(0)=10. If ∫05v(t) dt=−7\int_0^5 v(t) \,dt = -7 and ∫58v(t) dt=4\int_5^8 v(t) \,dt = 4, what is the particle's position at t=8t=8?

  1. 1
  2. 3
  3. 7 (correct answer)
  4. 13
Explanation: The net displacement from t=0t=0 to t=8t=8 is the sum of the displacements over the subintervals: s(8)−s(0)=∫08v(t) dt=∫05v(t) dt+∫58v(t) dts(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=−3-7 + 4 = -3. So, s(8)−s(0)=−3s(8) - s(0) = -3. Given s(0)=10s(0)=10, we have s(8)−10=−3s(8) - 10 = -3, which gives s(8)=7s(8) = 7.

Question 2

A particle moves along the x-axis with velocity v(t)=3t2−12v(t) = 3t^2 - 12. What is the average velocity of the particle on the interval [0,3][0, 3]?

  1. -3 (correct answer)
  2. 3
  3. 5
  4. -9
Explanation: Average velocity is the net displacement divided by the time interval. The displacement is ∫03v(t) dt\int_0^3 v(t) \,dt. ∫03(3t2−12) dt=[t3−12t]03=(33−12(3))−(0)=27−36=−9\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=33 - 0 = 3. The average velocity is −93=−3\frac{-9}{3} = -3.

Question 3

A particle starting from rest has an acceleration a(t)=1t+1a(t) = \frac{1}{t+1}. What is its velocity at t=e2−1t=e^2-1?

  1. 1
  2. 2 (correct answer)
  3. e
  4. e-1
Explanation: Starting from rest means v(0)=0v(0)=0. The velocity function is v(t)=∫a(t) dt=∫1t+1 dt=ln⁡(t+1)+Cv(t) = \int a(t) \,dt = \int \frac{1}{t+1} \,dt = \ln(t+1) + C (for t≥0t \ge 0). Using the initial condition, v(0)=ln⁡(0+1)+C=0⇒ln⁡(1)+C=0⇒C=0v(0) = \ln(0+1) + C = 0 \Rightarrow \ln(1) + C = 0 \Rightarrow C=0. So the velocity function is v(t)=ln⁡(t+1)v(t) = \ln(t+1). We need to find the velocity at t=e2−1t = e^2-1: v(e2−1)=ln⁡((e2−1)+1)=ln⁡(e2)=2v(e^2-1) = \ln((e^2-1)+1) = \ln(e^2) = 2.

Question 4

A particle moves on a line with velocity v(t)=t2−4t+3v(t) = t^2 - 4t + 3. If the particle's position at t=0t=0 is s(0)=5s(0)=5, what is its position when it first changes direction for t>0t>0?

  1. 5/3
  2. 19/3 (correct answer)
  3. 5
  4. 11/3
Explanation: The particle changes direction when its velocity changes sign, which occurs when v(t)=0v(t)=0. v(t)=t2−4t+3=(t−1)(t−3)=0v(t) = t^2 - 4t + 3 = (t-1)(t-3) = 0. The first time this happens for t>0t>0 is at t=1t=1. Now, we need to find the particle's position at t=1t=1. First, find the position function s(t)=∫(t2−4t+3) dt=t33−2t2+3t+Cs(t) = \int (t^2-4t+3) \,dt = \frac{t^3}{3} - 2t^2 + 3t + C. We are given s(0)=5s(0)=5, so C=5C=5. The position function is s(t)=t33−2t2+3t+5s(t) = \frac{t^3}{3} - 2t^2 + 3t + 5. Now, evaluate at t=1t=1: s(1)=133−2(1)2+3(1)+5=13−2+3+5=13+6=193s(1) = \frac{1^3}{3} - 2(1)^2 + 3(1) + 5 = \frac{1}{3} - 2 + 3 + 5 = \frac{1}{3} + 6 = \frac{19}{3}.

Question 5

The velocity v(t)v(t) of a particle moving along a line is a continuous function. If ∫25v(t) dt=0\int_2^5 v(t) \,dt = 0, which of the following statements must be true?

  1. The particle was at rest for the entire interval [2,5][2, 5].
  2. The particle's position at t=5t=5 is the same as its position at t=2t=2. (correct answer)
  3. The total distance traveled by the particle from t=2t=2 to t=5t=5 is zero.
  4. The particle's acceleration is zero at some time t∈(2,5)t \in (2, 5).
Explanation: The definite integral of the velocity function, ∫abv(t) dt\int_a^b v(t) \,dt, represents the net displacement of the particle from time t=at=a to t=bt=b, which is s(b)−s(a)s(b) - s(a). Given ∫25v(t) dt=0\int_2^5 v(t) \,dt = 0, it means s(5)−s(2)=0s(5) - s(2) = 0, or s(5)=s(2)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 ∫25∣v(t)∣ dt\int_2^5 |v(t)| \,dt, which would be positive if v(t)v(t) is not identically zero. Distractor D is not necessarily true; for example, if v(t)=(t−3.5)v(t)=(t-3.5), the integral from 2 to 5 is 0, but a(t)=1a(t)=1 is never zero.

Question 6

A particle moves along a line with velocity v(t)=e−tv(t) = e^{-t} for t≥0t \ge 0. If the position of the particle at t=0t=0 is s(0)=2s(0)=2, what is the limiting position of the particle as t→∞t \to \infty?

  1. 0
  2. 1
  3. 2
  4. 3 (correct answer)
Explanation: To find the position function s(t)s(t), we integrate the velocity function v(t)v(t): s(t)=∫e−t dt=−e−t+Cs(t) = \int e^{-t} \,dt = -e^{-t} + C. We are given the initial condition s(0)=2s(0) = 2. We use this to solve for the constant of integration CC: s(0)=−e−0+C=−1+C=2s(0) = -e^{-0} + C = -1 + C = 2, which implies C=3C = 3. Therefore, the position function is s(t)=3−e−ts(t) = 3 - e^{-t}. To find the limiting position, we take the limit as t→∞t \to \infty: lim⁡t→∞s(t)=lim⁡t→∞(3−e−t)=3−0=3\lim_{t \to \infty} s(t) = \lim_{t \to \infty} (3 - e^{-t}) = 3 - 0 = 3.

Question 7

The acceleration of a particle is a(t)=kcos⁡(t)a(t) = k\cos(t) for some constant kk. The particle starts from rest at t=0t=0 and has a position of s(0)=5s(0)=5. If the particle's position at t=πt=\pi is s(π)=9s(\pi)=9, what is the value of kk?

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 4/π\pi
Explanation: First, find v(t)v(t) by integrating a(t)a(t): v(t)=∫kcos⁡(t) dt=ksin⁡(t)+C1v(t) = \int k\cos(t) \,dt = k\sin(t) + C_1. The particle starts from rest, so v(0)=0v(0)=0. ksin⁡(0)+C1=0⇒C1=0k\sin(0) + C_1 = 0 \Rightarrow C_1 = 0. So, v(t)=ksin⁡(t)v(t) = k\sin(t). Next, find s(t)s(t) by integrating v(t)v(t): s(t)=∫ksin⁡(t) dt=−kcos⁡(t)+C2s(t) = \int k\sin(t) \,dt = -k\cos(t) + C_2. We are given s(0)=5s(0)=5. −kcos⁡(0)+C2=5⇒−k+C2=5⇒C2=5+k-k\cos(0) + C_2 = 5 \Rightarrow -k + C_2 = 5 \Rightarrow C_2 = 5+k. So, s(t)=−kcos⁡(t)+5+ks(t) = -k\cos(t) + 5 + k. Now use the condition s(π)=9s(\pi)=9: s(π)=−kcos⁡(π)+5+k=−k(−1)+5+k=k+5+k=2k+5s(\pi) = -k\cos(\pi) + 5 + k = -k(-1) + 5 + k = k + 5 + k = 2k+5. We set this equal to 9: 2k+5=9⇒2k=4⇒k=22k+5=9 \Rightarrow 2k=4 \Rightarrow k=2.

Question 8

A particle moves along a line with velocity v(t)=t2−4v(t) = t^2 - 4. For t>0t > 0, when is the particle's speed increasing?

  1. (0,2)(0, 2)
  2. (2,∞)(2, \infty) (correct answer)
  3. (0,∞)(0, \infty)
  4. (0,2)∪(2,∞)(0, 2) \cup (2, \infty)
Explanation: Speed is increasing when velocity v(t)v(t) and acceleration a(t)a(t) have the same sign. First, find the acceleration: a(t)=v′(t)=2ta(t) = v'(t) = 2t. For t>0t>0, a(t)a(t) is always positive. Therefore, speed is increasing when v(t)v(t) is also positive. We need to solve v(t)>0v(t) > 0: t2−4>0⇒(t−2)(t+2)>0t^2 - 4 > 0 \Rightarrow (t-2)(t+2) > 0. Since we are only considering t>0t>0, this inequality holds for t>2t > 2. Thus, the speed is increasing on the interval (2,∞)(2, \infty).

Question 9

Two particles start at the origin and move along the x-axis. For t≥0t \ge 0, their velocities are given by v1(t)=t+1v_1(t) = t+1 and v2(t)=t2+1v_2(t) = t^2+1. At what time t>0t>0 do the particles have the same position again?

  1. t=3/2t = 3/2 (correct answer)
  2. t=2t = 2
  3. t=3t = 3
  4. They never have the same position for t>0t>0.
Explanation: First, find the position functions for both particles. Since they start at the origin, s1(0)=0s_1(0)=0 and s2(0)=0s_2(0)=0. s1(t)=∫(t+1) dt=t22+t+C1s_1(t) = \int (t+1) \,dt = \frac{t^2}{2} + t + C_1. Since s1(0)=0s_1(0)=0, C1=0C_1=0. So s1(t)=t22+ts_1(t) = \frac{t^2}{2} + t. s2(t)=∫(t2+1) dt=t33+t+C2s_2(t) = \int (t^2+1) \,dt = \frac{t^3}{3} + t + C_2. Since s2(0)=0s_2(0)=0, C2=0C_2=0. So s2(t)=t33+ts_2(t) = \frac{t^3}{3} + t. Now, set the positions equal to each other to find when they meet: t22+t=t33+t\frac{t^2}{2} + t = \frac{t^3}{3} + t. Subtract tt from both sides: t22=t33\frac{t^2}{2} = \frac{t^3}{3}. Since we are looking for t>0t>0, we can divide by t2t^2: 12=t3\frac{1}{2} = \frac{t}{3}. Solving for tt gives t=32t = \frac{3}{2}.

Question 10

The rate of change of velocity of a particle is proportional to the square of the time tt, i.e., a(t)=kt2a(t) = kt^2. The particle starts at position s(0)=2s(0)=2 with velocity v(0)=2v(0)=2. At t=1t=1, the velocity is v(1)=3v(1)=3. What is the position s(2)s(2)?

  1. 10 (correct answer)
  2. 11
  3. 12
  4. 13
Explanation: First, we use the given information to find kk. v(t)=∫kt2 dt=k3t3+Cv(t) = \int kt^2 \,dt = \frac{k}{3}t^3 + C. We know v(0)=2v(0)=2, so C=2C=2. Thus v(t)=k3t3+2v(t) = \frac{k}{3}t^3 + 2. We are given v(1)=3v(1)=3, so k3(1)3+2=3⇒k3=1⇒k=3\frac{k}{3}(1)^3 + 2 = 3 \Rightarrow \frac{k}{3} = 1 \Rightarrow k=3. The velocity function is v(t)=t3+2v(t) = t^3 + 2. Now we find the position function s(t)=∫(t3+2) dt=t44+2t+Ds(t) = \int (t^3+2) \,dt = \frac{t^4}{4} + 2t + D. We are given s(0)=2s(0)=2, so D=2D=2. The position function is s(t)=t44+2t+2s(t) = \frac{t^4}{4} + 2t + 2. Finally, we find s(2)s(2): s(2)=244+2(2)+2=164+4+2=4+4+2=10s(2) = \frac{2^4}{4} + 2(2) + 2 = \frac{16}{4} + 4 + 2 = 4 + 4 + 2 = 10.

Question 11

The rate of change of velocity of a particle is proportional to the square of the time tt, i.e., a(t)=kt2a(t) = kt^2. The particle starts at position s(0)=2s(0)=2 with velocity v(0)=2v(0)=2. At t=1t=1, the velocity is v(1)=3v(1)=3. What is the position s(2)s(2)?

  1. 9
  2. 10 (correct answer)
  3. 11
  4. 12
Explanation: First, use the given information to find the constant kk. The velocity function is v(t)=∫kt2 dt=k3t3+Cv(t) = \int kt^2 \,dt = \frac{k}{3}t^3 + C. We know v(0)=2v(0)=2, so C=2C=2. This gives v(t)=k3t3+2v(t) = \frac{k}{3}t^3 + 2. We are also given v(1)=3v(1)=3, so k3(1)3+2=3⇒k3=1⇒k=3\frac{k}{3}(1)^3 + 2 = 3 \Rightarrow \frac{k}{3} = 1 \Rightarrow k=3. Therefore, the velocity function is v(t)=t3+2v(t) = t^3 + 2. Now, find the position function: s(t)=∫(t3+2) dt=t44+2t+Ds(t) = \int (t^3+2) \,dt = \frac{t^4}{4} + 2t + D. We are given the initial position s(0)=2s(0)=2, so D=2D=2. The position function is s(t)=t44+2t+2s(t) = \frac{t^4}{4} + 2t + 2. Finally, evaluate the position at t=2t=2: s(2)=244+2(2)+2=164+4+2=4+4+2=10s(2) = \frac{2^4}{4} + 2(2) + 2 = \frac{16}{4} + 4 + 2 = 4 + 4 + 2 = 10.

Question 12

The acceleration of an object is given by a(t)=12t2−2a(t) = 12t^2 - 2. The change in the object's velocity from t=1t=1 to t=3t=3 is 90. What is its position at t=1t=1 if s(0)=5s(0)=5 and its velocity at t=0t=0 is 3?

  1. 7 (correct answer)
  2. 8
  3. 12
  4. 90
Explanation: This question contains extra information designed to be a distractor. The change in velocity from t=1t=1 to t=3t=3 can be calculated as ∫13(12t2−2)dt=[4t3−2t]13=(4(27)−6)−(4(1)−2)=(108−6)−(2)=102−2=100\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)a(t), v(0)v(0), and s(0)s(0). First, find v(t)v(t): v(t)=∫(12t2−2) dt=4t3−2t+Cv(t) = \int (12t^2 - 2) \,dt = 4t^3 - 2t + C. Given v(0)=3v(0)=3, we have C=3C=3. So, v(t)=4t3−2t+3v(t) = 4t^3 - 2t + 3. Next, find s(t)s(t): s(t)=∫(4t3−2t+3) dt=t4−t2+3t+Ds(t) = \int (4t^3 - 2t + 3) \,dt = t^4 - t^2 + 3t + D. Given s(0)=5s(0)=5, we have D=5D=5. So, s(t)=t4−t2+3t+5s(t) = t^4 - t^2 + 3t + 5. The question asks for s(1)s(1): s(1)=14−12+3(1)+5=1−1+3+5=7s(1) = 1^4 - 1^2 + 3(1) + 5 = 1 - 1 + 3 + 5 = 7.

Question 13

A particle moves with velocity v(t)=sin⁡(πt)v(t) = \sin(\pi t). What is the total distance traveled from t=0t=0 to t=2t=2?

  1. 0
  2. 2/π2/\pi
  3. 4/π4/\pi (correct answer)
  4. 1/π1/\pi
Explanation: Total distance is the integral of speed, ∣v(t)∣|v(t)|. The function v(t)=sin⁡(πt)v(t) = \sin(\pi t) is positive on (0,1)(0, 1) and negative on (1,2)(1, 2). Therefore, we must split the integral: Distance = ∫01sin⁡(πt) dt+∣∫12sin⁡(πt) dt∣\int_0^1 \sin(\pi t) \,dt + |\int_1^2 \sin(\pi t) \,dt|. The antiderivative of sin⁡(πt)\sin(\pi t) is −1πcos⁡(πt)-\frac{1}{\pi}\cos(\pi t). The first integral is [−1πcos⁡(πt)]01=−1π(cos⁡(π)−cos⁡(0))=−1π(−1−1)=2π[-\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 [−1πcos⁡(πt)]12=−1π(cos⁡(2π)−cos⁡(π))=−1π(1−(−1))=−2π[-\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 2π+∣−2π∣=4π\frac{2}{\pi} + |-\frac{2}{\pi}| = \frac{4}{\pi}.

Question 14

The acceleration of a particle moving along the x-axis is a(t)=−4a(t) = -4. The velocity of the particle at t=0t=0 is 12. What is the total distance traveled by the particle from t=0t=0 to t=4t=4?

  1. 16
  2. 20 (correct answer)
  3. 24
  4. 32
Explanation: First, find the velocity function: v(t)=∫−4 dt=−4t+Cv(t) = \int -4 \,dt = -4t + C. Given v(0)=12v(0)=12, we find C=12C=12. So, v(t)=−4t+12v(t) = -4t + 12. To find the total distance, we must find where the particle changes direction by setting v(t)=0v(t)=0. This gives −4t+12=0-4t+12=0, so t=3t=3. The particle moves in the positive direction for t∈[0,3)t \in [0, 3) and in the negative direction for t∈(3,4]t \in (3, 4]. The total distance is the sum of the absolute values of the displacements on these intervals: Distance = ∫03(−4t+12) dt+∣∫34(−4t+12) dt∣\int_0^3 (-4t+12) \,dt + |\int_3^4 (-4t+12) \,dt|. The antiderivative is −2t2+12t-2t^2+12t. The first integral is [−2t2+12t]03=(−2(9)+12(3))−0=−18+36=18[-2t^2+12t]_0^3 = (-2(9)+12(3)) - 0 = -18+36 = 18. The second integral is [−2t2+12t]34=(−2(16)+12(4))−(−2(9)+12(3))=(−32+48)−(18)=16−18=−2[-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∣=2018 + |-2| = 20.

Question 15

A particle moves along a line with velocity given by v(t)=3t2−6tv(t) = 3t^2 - 6t for t≥0t \ge 0. What is the total distance traveled by the particle on the time interval t∈[0,4]t \in [0, 4]?

  1. 16
  2. 20
  3. 4
  4. 24 (correct answer)
Explanation: To find the total distance, we must integrate the speed, ∣v(t)∣|v(t)|, over the interval. First, find where the velocity is zero: 3t2−6t=3t(t−2)=03t^2 - 6t = 3t(t-2) = 0, which gives t=0t=0 and t=2t=2. The particle changes direction at t=2t=2. The velocity is negative on (0,2)(0, 2) and positive on (2,4)(2, 4). The total distance is the sum of the absolute values of the integrals over these subintervals: Distance=∫04∣3t2−6t∣ dt=∣∫02(3t2−6t) dt∣+∫24(3t2−6t) dt\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)=t3−3t2s(t) = t^3 - 3t^2. The first integral is [t3−3t2]02=(8−12)−0=−4[t^3 - 3t^2]_0^2 = (8 - 12) - 0 = -4. The second integral is [t3−3t2]24=(64−48)−(8−12)=16−(−4)=20[t^3 - 3t^2]_2^4 = (64 - 48) - (8 - 12) = 16 - (-4) = 20. The total distance is ∣−4∣+20=24|-4| + 20 = 24.

Question 16

A particle's acceleration is described by a(t)=6t−2a(t) = 6t - 2. If the particle has a velocity of 4 at t=1t=1 and a position of 0 at t=1t=1, what is the position of the particle at t=2t=2?

  1. 4
  2. 10
  3. 7 (correct answer)
  4. 13
Explanation: First, integrate acceleration to find velocity: v(t)=∫(6t−2) dt=3t2−2t+C1v(t) = \int (6t - 2) \,dt = 3t^2 - 2t + C_1. Use v(1)=4v(1)=4 to find C1C_1: 3(1)2−2(1)+C1=4⇒1+C1=4⇒C1=33(1)^2 - 2(1) + C_1 = 4 \Rightarrow 1 + C_1 = 4 \Rightarrow C_1 = 3. So, v(t)=3t2−2t+3v(t) = 3t^2 - 2t + 3. Next, integrate velocity to find position: s(t)=∫(3t2−2t+3) dt=t3−t2+3t+C2s(t) = \int (3t^2 - 2t + 3) \,dt = t^3 - t^2 + 3t + C_2. Use s(1)=0s(1)=0 to find C2C_2: 13−12+3(1)+C2=0⇒3+C2=0⇒C2=−31^3 - 1^2 + 3(1) + C_2 = 0 \Rightarrow 3 + C_2 = 0 \Rightarrow C_2 = -3. The position function is s(t)=t3−t2+3t−3s(t) = t^3 - t^2 + 3t - 3. Finally, evaluate at t=2t=2: s(2)=23−22+3(2)−3=8−4+6−3=7s(2) = 2^3 - 2^2 + 3(2) - 3 = 8 - 4 + 6 - 3 = 7.

Question 17

A particle has acceleration a(t)=cos⁡(t)a(t) = \cos(t). Its velocity at t=0t=0 is 0, and its position at t=0t=0 is 2. What is the position of the particle at t=πt=\pi?

  1. π+2\pi + 2
  2. 2
  3. 3
  4. 4 (correct answer)
Explanation: First, find the velocity function: v(t)=∫cos⁡(t) dt=sin⁡(t)+C1v(t) = \int \cos(t) \,dt = \sin(t) + C_1. Use v(0)=0v(0) = 0: sin⁡(0)+C1=0⇒C1=0\sin(0) + C_1 = 0 \Rightarrow C_1 = 0. So, v(t)=sin⁡(t)v(t) = \sin(t). Now, find the position function: s(t)=∫sin⁡(t) dt=−cos⁡(t)+C2s(t) = \int \sin(t) \,dt = -\cos(t) + C_2. Use s(0)=2s(0) = 2: −cos⁡(0)+C2=2⇒−1+C2=2⇒C2=3-\cos(0) + C_2 = 2 \Rightarrow -1 + C_2 = 2 \Rightarrow C_2 = 3. Thus, the position function is s(t)=3−cos⁡(t)s(t) = 3 - \cos(t). Finally, find the position at t=πt=\pi: s(π)=3−cos⁡(π)=3−(−1)=4s(\pi) = 3 - \cos(\pi) = 3 - (-1) = 4.

Question 18

A particle moves along a line such that its velocity at time tt is v(t)=11+tv(t) = \frac{1}{1+t}. What is the total distance traveled by the particle from t=0t=0 to t=e−1t=e-1?

  1. 1 (correct answer)
  2. e
  3. ln⁡(e−1)\ln(e-1)
  4. −1e2\frac{-1}{e^2}
Explanation: Total distance traveled is the integral of the speed, ∣v(t)∣|v(t)|. For t≥0t \ge 0, the velocity v(t)=11+tv(t) = \frac{1}{1+t} is always positive, so the speed is equal to the velocity. The total distance is ∫0e−111+t dt\int_0^{e-1} \frac{1}{1+t} \,dt. The integral evaluates to [ln⁡∣1+t∣]0e−1[\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\ln|1 + (e-1)| - \ln|1+0| = \ln(e) - \ln(1) = 1 - 0 = 1.

Question 19

The acceleration of a particle is given by a(t)=2t−6a(t) = 2t - 6. At t=0t=0, its velocity is v(0)=5v(0) = 5. What is the total distance traveled by the particle from t=0t=0 to t=4t=4?

  1. −203-\frac{20}{3}
  2. 203\frac{20}{3}
  3. 803\frac{80}{3}
  4. 343\frac{34}{3} (correct answer)
Explanation: First, find v(t)v(t) by integrating a(t)a(t): v(t)=∫(2t−6) dt=t2−6t+Cv(t) = \int (2t - 6) \,dt = t^2 - 6t + C. Using v(0)=5v(0)=5, we find C=5C=5, so v(t)=t2−6t+5v(t) = t^2 - 6t + 5. To find total distance, we must find where v(t)=0v(t)=0: (t−1)(t−5)=0(t-1)(t-5)=0, so t=1t=1 and t=5t=5. In the interval [0,4][0, 4], the velocity changes sign at t=1t=1. Total distance is ∫01(t2−6t+5) dt+∣∫14(t2−6t+5) dt∣\int_0^1 (t^2 - 6t + 5) \,dt + |\int_1^4 (t^2 - 6t + 5) \,dt|. The antiderivative is s(t)=t33−3t2+5ts(t) = \frac{t^3}{3} - 3t^2 + 5t. The first integral is [s(t)]01=(13−3+5)−0=73[s(t)]_0^1 = (\frac{1}{3} - 3 + 5) - 0 = \frac{7}{3}. The second integral is [s(t)]14=(643−48+20)−(73)=643−28−73=573−28=19−28=−9[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 = 73+∣−9∣=73+9=7+273=343\frac{7}{3} + |-9| = \frac{7}{3} + 9 = \frac{7+27}{3} = \frac{34}{3}.

Question 20

A particle starts at position s(1)=2s(1) = 2 and moves with velocity v(t)=2t−6t2v(t) = 2t - \frac{6}{t^2} for t>0t > 0. At what time t>1t > 1 does the particle first return to its starting position?

  1. t=3t = \sqrt{3}
  2. t=2t = 2 (correct answer)
  3. t=3t = 3
  4. The particle never returns to its starting position.
Explanation: First, find the position function s(t)s(t) by integrating v(t)v(t): s(t)=∫(2t−6t−2) dt=t2+6t−1+C=t2+6t+Cs(t) = \int (2t - 6t^{-2}) \,dt = t^2 + 6t^{-1} + C = t^2 + \frac{6}{t} + C. Use the initial condition s(1)=2s(1) = 2 to find CC: 12+61+C=2⇒7+C=2⇒C=−51^2 + \frac{6}{1} + C = 2 \Rightarrow 7 + C = 2 \Rightarrow C = -5. So, s(t)=t2+6t−5s(t) = t^2 + \frac{6}{t} - 5. We want to find when the particle returns to the position s=2s=2 for t>1t>1. Set s(t)=2s(t) = 2: t2+6t−5=2⇒t2+6t−7=0t^2 + \frac{6}{t} - 5 = 2 \Rightarrow t^2 + \frac{6}{t} - 7 = 0. Multiply by tt to get t3−7t+6=0t^3 - 7t + 6 = 0. We know t=1t=1 is a root. By the Rational Root Theorem, we can test other integer factors of 6. For t=2t=2, we have 23−7(2)+6=8−14+6=02^3 - 7(2) + 6 = 8 - 14 + 6 = 0. Thus, the particle returns to its starting position at t=2t=2.