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

A particle starts at the origin and its velocity is given by v(t)=tsec⁡2(t)v(t) = t \sec^2(t) for 0≤t<π/20 \le t < \pi/2. What is its position at t=π/3t = \pi/3?

3−1\sqrt{3} - 1
π33\frac{\pi\sqrt{3}}{3}
π33+ln⁡(2)\frac{\pi\sqrt{3}}{3} + \ln(2)
π33−ln⁡(2)\frac{\pi\sqrt{3}}{3} - \ln(2)
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Calculus 1 Quiz

Calculus 1 Quiz: Position Velocity Acceleration Via Integrals

Practice Position Velocity Acceleration Via Integrals in Calculus 1 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 1.

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Question 1

A particle starts at the origin and its velocity is given by v(t)=tsec⁡2(t)v(t) = t \sec^2(t) for 0≤t<π/20 \le t < \pi/2. What is its position at t=π/3t = \pi/3?

  1. 3−1\sqrt{3} - 1
  2. π33\frac{\pi\sqrt{3}}{3}
  3. π33+ln⁡(2)\frac{\pi\sqrt{3}}{3} + \ln(2)
  4. π33−ln⁡(2)\frac{\pi\sqrt{3}}{3} - \ln(2) (correct answer)
Explanation: The particle's position s(t)s(t) is the integral of its velocity v(t)v(t). Since it starts at the origin, s(0)=0s(0)=0, so its position at t=π/3t=\pi/3 is the definite integral s(π/3)=∫0π/3tsec⁡2(t)dts(\pi/3) = \int_{0}^{\pi/3} t \sec^2(t) dt. We use integration by parts: ∫udv=uv−∫vdu\int u dv = uv - \int v du. Let u=tu=t and dv=sec⁡2(t)dtdv = \sec^2(t) dt. Then du=dtdu = dt and v=tan⁡(t)v = \tan(t). So, ∫tsec⁡2(t)dt=ttan⁡(t)−∫tan⁡(t)dt\int t \sec^2(t) dt = t \tan(t) - \int \tan(t) dt. The integral of tan⁡(t)\tan(t) is −ln⁡∣cos⁡(t)∣-\ln|\cos(t)| or ln⁡∣sec⁡(t)∣\ln|\sec(t)|. So the antiderivative is ttan⁡(t)−(−ln⁡∣cos⁡(t)∣)=ttan⁡(t)+ln⁡∣cos⁡(t)∣t \tan(t) - (-\ln|\cos(t)|) = t \tan(t) + \ln|\cos(t)|. Now, evaluate the definite integral from 0 to π/3\pi/3: [ttan⁡(t)+ln⁡(cos⁡(t))]0π/3[t \tan(t) + \ln(\cos(t))]_0^{\pi/3} (we can drop the absolute value as cos⁡(t)>0\cos(t)>0 on this interval). =(π3tan⁡(π3)+ln⁡(cos⁡(π3)))−(0⋅tan⁡(0)+ln⁡(cos⁡(0)))= (\frac{\pi}{3} \tan(\frac{\pi}{3}) + \ln(\cos(\frac{\pi}{3}))) - (0 \cdot \tan(0) + \ln(\cos(0))) =(π3⋅3+ln⁡(12))−(0+ln⁡(1))= (\frac{\pi}{3} \cdot \sqrt{3} + \ln(\frac{1}{2})) - (0 + \ln(1)) =π33−ln⁡(2)−0=π33−ln⁡(2)= \frac{\pi\sqrt{3}}{3} - \ln(2) - 0 = \frac{\pi\sqrt{3}}{3} - \ln(2). A: This may arise from misremembering derivatives or integrals related to trig functions. B: This is just the first term of the evaluation, ttan⁡(t)t\tan(t), ignoring the ∫tan⁡(t)dt\int \tan(t) dt part. C: This results from a sign error in the integral of tan⁡(t)\tan(t).

Question 2

The acceleration of a particle is a(t)=(t+1)−3a(t) = (t+1)^{-3} for t≥0t \ge 0. The particle starts from rest. What is the particle's velocity at t=1t=1?

  1. 38\frac{3}{8} (correct answer)
  2. 12\frac{1}{2}
  3. −38-\frac{3}{8}
  4. −18-\frac{1}{8}
Explanation: First, find the velocity v(t)v(t) by integrating the acceleration a(t)a(t). 'Starts from rest' means v(0)=0v(0)=0. v(t)=∫(t+1)−3dt=(t+1)−2−2+C=−12(t+1)2+Cv(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)=0v(0)=0 to find CC. 0=−12(0+1)2+C⇒0=−12+C⇒C=120 = -\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)=12−12(t+1)2v(t) = \frac{1}{2} - \frac{1}{2(t+1)^2}. Finally, evaluate the velocity at t=1t=1. v(1)=12−12(1+1)2=12−12(4)=12−18=48−18=38v(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, CC, not the velocity at t=1t=1. C: This would be the result of a sign error when finding CC, leading to C=−1/2C = -1/2. D: This is the value of the integral part of v(1)v(1) if the constant CC is forgotten.

Question 3

The acceleration of a particle is a(t)=1ta(t) = \frac{1}{t} for t>0t>0. The particle's velocity at t=1t=1 is v(1)=2v(1)=2. What is the change in position of the particle from t=1t=1 to t=e2t=e^2?

  1. e2+1e^2+1
  2. 3e2−13e^2-1 (correct answer)
  3. 3e23e^2
  4. e2e^2
Explanation: First, find the velocity function v(t)v(t) by integrating a(t)a(t): v(t)=∫1tdt=ln⁡(t)+Cv(t) = \int \frac{1}{t} dt = \ln(t) + C (since t>0t>0). Use the condition v(1)=2v(1)=2 to find CC: 2=ln⁡(1)+C⇒2=0+C⇒C=22 = \ln(1) + C \Rightarrow 2 = 0 + C \Rightarrow C=2. So, v(t)=ln⁡(t)+2v(t) = \ln(t) + 2. The change in position (displacement) from t=1t=1 to t=e2t=e^2 is ∫1e2v(t)dt=∫1e2(ln⁡(t)+2)dt\int_{1}^{e^2} v(t) dt = \int_{1}^{e^2} (\ln(t) + 2) dt. To integrate ln⁡(t)\ln(t), we use integration by parts, which yields ∫ln⁡(t)dt=tln⁡(t)−t\int \ln(t) dt = t\ln(t) - t. So, the antiderivative of ln⁡(t)+2\ln(t)+2 is tln⁡(t)−t+2t=tln⁡(t)+tt\ln(t) - t + 2t = t\ln(t) + t. Now evaluate the definite integral: [tln⁡(t)+t]1e2=(e2ln⁡(e2)+e2)−(1ln⁡(1)+1)=(e2(2)+e2)−(0+1)=3e2−1[t\ln(t) + t]_{1}^{e^2} = (e^2\ln(e^2) + e^2) - (1\ln(1) + 1) = (e^2(2) + e^2) - (0 + 1) = 3e^2 - 1. A: This is the displacement if one forgets the constant of integration C=2C=2 and integrates just v(t)=ln⁡(t)v(t)=\ln(t). C: This results from an error when evaluating at the lower bound, 1ln⁡(1)+1=11\ln(1)+1=1, likely by neglecting the +1+1 term. D: This could result from multiple calculation errors, such as incorrectly evaluating ln⁡(e2)\ln(e^2) as 1.

Question 4

A particle moves with velocity v(t)=3tv(t) = 3\sqrt{t} for t≥0t \ge 0. Its position at t=1t=1 is s(1)=−1s(1) = -1. What is the particle's position at t=4t=4?

  1. 1212
  2. 1313 (correct answer)
  3. 1414
  4. 1515
Explanation: We can find the position s(4)s(4) using the Fundamental Theorem of Calculus: s(4)=s(1)+∫14v(t)dts(4) = s(1) + \int_{1}^{4} v(t) dt. First, calculate the definite integral: ∫143tdt=∫143t1/2dt=[3⋅t3/23/2]14=[2t3/2]14\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=142(4)^{3/2} - 2(1)^{3/2} = 2(8) - 2(1) = 16 - 2 = 14. This is the displacement from t=1t=1 to t=4t=4. Now, find the final position: s(4)=s(1)+14=−1+14=13s(4) = s(1) + 14 = -1 + 14 = 13. A: This is the result of calculating the displacement from t=1t=1 to t=4t=4 as 1313 and then subtracting s(1)s(1) instead of adding it. C: This is the displacement from t=1t=1 to t=4t=4, not the final position. D: This is the final position if the displacement were calculated from t=0t=0 to t=4t=4, which is s(4)=s(0)+16s(4) = s(0) + 16. Assuming s(0)=s(1)=−1s(0)=s(1)=-1 yields s(4)=15s(4)=15, a common mistake.

Question 5

A particle moves with velocity v(t)=cos⁡(t)v(t) = \cos(t). For which of the following values of c>0c > 0 is the particle's average velocity on the interval [0,c][0, c] equal to 2π\frac{2}{\pi}?

  1. π4\frac{\pi}{4}
  2. π3\frac{\pi}{3}
  3. π2\frac{\pi}{2} (correct answer)
  4. π\pi
Explanation: The average velocity on [0,c][0, c] is given by the formula for the average value of a function: 1c−0∫0cv(t)dt\frac{1}{c-0} \int_{0}^{c} v(t) dt. In this case, it is 1c∫0ccos⁡(t)dt=1c[sin⁡(t)]0c=1c(sin⁡(c)−sin⁡(0))=sin⁡(c)c\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 cc satisfies sin⁡(c)c=2π\frac{\sin(c)}{c} = \frac{2}{\pi}. We can test the choices: A) For c=π4c=\frac{\pi}{4}: sin⁡(π/4)π/4=2/2π/4=22π≠2π\frac{\sin(\pi/4)}{\pi/4} = \frac{\sqrt{2}/2}{\pi/4} = \frac{2\sqrt{2}}{\pi} \neq \frac{2}{\pi}. B) For c=π3c=\frac{\pi}{3}: sin⁡(π/3)π/3=3/2π/3=332π≠2π\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=π2c=\frac{\pi}{2}: sin⁡(π/2)π/2=1π/2=2π\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}. This is correct. D) For c=πc=\pi: sin⁡(π)π=0π=0≠2π\frac{\sin(\pi)}{\pi} = \frac{0}{\pi} = 0 \neq \frac{2}{\pi}.

Question 6

The velocity of a particle is given by v(t)=et−3v(t) = e^t - 3. What is the total distance traveled by the particle from t=0t=0 to t=2t=2?

  1. e2−7e^2 - 7
  2. e2+6ln⁡(3)−11e^2 + 6\ln(3) - 11 (correct answer)
  3. e2+6ln⁡(3)−12e^2 + 6\ln(3) - 12
  4. e2−5e^2 - 5
Explanation: To find the total distance, we must integrate the speed, ∣v(t)∣|v(t)|. First, find where the velocity is zero: et−3=0⇒et=3⇒t=ln⁡(3)e^t - 3 = 0 \Rightarrow e^t = 3 \Rightarrow t = \ln(3). Since 1<ln⁡(3)<21 < \ln(3) < 2, the velocity changes sign in the interval [0,2][0, 2]. For t<ln⁡(3)t < \ln(3), v(t)<0v(t) < 0, and for t>ln⁡(3)t > \ln(3), v(t)>0v(t) > 0. The total distance is ∫02∣et−3∣dt=∫0ln⁡(3)−(et−3)dt+∫ln⁡(3)2(et−3)dt\int_{0}^{2} |e^t - 3| dt = \int_{0}^{\ln(3)} -(e^t - 3) dt + \int_{\ln(3)}^{2} (e^t - 3) dt. ∫(3−et)dt=3t−et\int (3 - e^t) dt = 3t - e^t. ∫(et−3)dt=et−3t\int (e^t - 3) dt = e^t - 3t. First integral: [3t−et]0ln⁡(3)=(3ln⁡(3)−eln⁡(3))−(0−e0)=3ln⁡(3)−3−(−1)=3ln⁡(3)−2[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: [et−3t]ln⁡(3)2=(e2−6)−(eln⁡(3)−3ln⁡(3))=e2−6−(3−3ln⁡(3))=e2−9+3ln⁡(3)[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 = (3ln⁡(3)−2)+(e2−9+3ln⁡(3))=e2+6ln⁡(3)−11(3\ln(3) - 2) + (e^2 - 9 + 3\ln(3)) = e^2 + 6\ln(3) - 11. A: This is the displacement ∫02(et−3)dt=e2−7\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 e0=0e^0=0 instead of e0=1e^0=1. D: This could result from various calculation errors, such as miscalculating the antiderivative or the bounds of integration.

Question 7

A rocket has a constant acceleration a(t)=−10a(t) = -10 m/s2^2. It is launched upward from the ground (s(0)=0s(0)=0) and is observed to be at a height of 240 meters at t=4t=4 seconds. What was its initial launch velocity, v(0)v(0)?

  1. 60 m/s
  2. 70 m/s
  3. 80 m/s (correct answer)
  4. 90 m/s
Explanation: First, integrate the acceleration to find the velocity function: v(t)=∫−10dt=−10t+C1v(t) = \int -10 dt = -10t + C_1. At t=0t=0, the velocity is the initial velocity v(0)v(0), so v(0)=−10(0)+C1  ⟹  C1=v(0)v(0) = -10(0) + C_1 \implies C_1 = v(0). Thus, v(t)=−10t+v(0)v(t) = -10t + v(0). Next, integrate the velocity to find the position function: s(t)=∫(−10t+v(0))dt=−5t2+v(0)t+C2s(t) = \int (-10t + v(0)) dt = -5t^2 + v(0)t + C_2. The rocket is launched from the ground, so s(0)=0s(0)=0. This gives −5(0)2+v(0)(0)+C2=0  ⟹  C2=0-5(0)^2 + v(0)(0) + C_2 = 0 \implies C_2=0. So, s(t)=−5t2+v(0)ts(t) = -5t^2 + v(0)t. We are given that s(4)=240s(4) = 240. Substitute t=4t=4 into the position function: 240=−5(4)2+v(0)(4)=−5(16)+4v(0)=−80+4v(0)240 = -5(4)^2 + v(0)(4) = -5(16) + 4v(0) = -80 + 4v(0). Now, solve for v(0)v(0): 320=4v(0)  ⟹  v(0)=80320 = 4v(0) \implies v(0) = 80 m/s.

Question 8

The velocity of a particle is v(t)=t2−4t+3v(t) = t^2 - 4t + 3. The particle is at position s(1)=10s(1) = 10. What is the maximum position of the particle on the interval [0,4][0, 4]?

  1. 44
  2. 263\frac{26}{3}
  3. 1010 (correct answer)
  4. 343\frac{34}{3}
Explanation: To find the maximum position, we must check the critical points (where s′(t)=v(t)=0s'(t) = v(t) = 0) and the endpoints of the interval. v(t)=t2−4t+3=(t−1)(t−3)=0v(t) = t^2 - 4t + 3 = (t-1)(t-3) = 0, so critical points are at t=1t=1 and t=3t=3. The candidates for the time of maximum position on [0,4][0, 4] are t=0,1,3,4t=0, 1, 3, 4. Specifically, since v(t)v(t) is positive on [0,1)[0,1), negative on (1,3)(1,3), and positive on (3,4](3,4], the potential maxima are at t=1t=1 (local max) and t=4t=4 (endpoint). First, find the position function: s(t)=∫(t2−4t+3)dt=13t3−2t2+3t+Cs(t) = \int (t^2 - 4t + 3) dt = \frac{1}{3}t^3 - 2t^2 + 3t + C. Use s(1)=10s(1) = 10 to find C: 10=13−2+3+C⇒10=43+C⇒C=26310 = \frac{1}{3} - 2 + 3 + C \Rightarrow 10 = \frac{4}{3} + C \Rightarrow C = \frac{26}{3}. So, s(t)=13t3−2t2+3t+263s(t) = \frac{1}{3}t^3 - 2t^2 + 3t + \frac{26}{3}. Now evaluate the position at the candidates: s(0)=263≈8.67s(0) = \frac{26}{3} \approx 8.67. s(1)=10s(1) = 10. s(4)=13(4)3−2(4)2+3(4)+263=643−32+12+263=903−20=30−20=10s(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][1,4], not the maximum position. B: This is the position at t=0t=0 and t=3t=3 (a local minimum), not the maximum. D: This value, 10+4/310 + 4/3, might result from adding the displacement on [3,4][3,4] to the maximum at t=1t=1 without considering the intermediate minimum.

Question 9

A particle has acceleration a(t)=6a(t) = 6. At time t=0t=0, its velocity is v(0)=−18v(0)=-18 and its position is s(0)=0s(0)=0. What is the total distance traveled by the particle during the first 6 seconds?

  1. 00
  2. 2727
  3. 5454 (correct answer)
  4. 108108
Explanation: First, determine the velocity and position functions. v(t)=∫a(t)dt=∫6dt=6t+C1v(t) = \int a(t) dt = \int 6 dt = 6t + C_1. Given v(0)=−18v(0)=-18, C1=−18C_1=-18. So, v(t)=6t−18v(t) = 6t - 18. s(t)=∫v(t)dt=∫(6t−18)dt=3t2−18t+C2s(t) = \int v(t) dt = \int (6t - 18) dt = 3t^2 - 18t + C_2. Given s(0)=0s(0)=0, C2=0C_2=0. So, s(t)=3t2−18ts(t) = 3t^2 - 18t. To find total distance, we must find when the particle changes direction, which is when v(t)=0v(t) = 0. 6t−18=0⇒t=36t - 18 = 0 \Rightarrow t = 3. This is within our interval [0,6][0, 6]. We calculate the distance traveled on the sub-intervals [0,3][0, 3] and [3,6][3, 6]. We need the positions at t=0,3,6t=0, 3, 6. s(0)=0s(0) = 0. s(3)=3(3)2−18(3)=27−54=−27s(3) = 3(3)^2 - 18(3) = 27 - 54 = -27. s(6)=3(6)2−18(6)=3(36)−108=108−108=0s(6) = 3(6)^2 - 18(6) = 3(36) - 108 = 108 - 108 = 0. Distance on [0,3][0, 3] is ∣s(3)−s(0)∣=∣−27−0∣=27|s(3) - s(0)| = |-27 - 0| = 27. Distance on [3,6][3, 6] is ∣s(6)−s(3)∣=∣0−(−27)∣=27|s(6) - s(3)| = |0 - (-27)| = 27. The total distance traveled is 27+27=5427 + 27 = 54. A: This is the displacement on the interval [0,6][0, 6], not the total distance. B: This is the distance traveled during only the first half of the interval, from t=0t=0 to t=3t=3. D: This could result from an arithmetic error, for instance, adding 54+5454+54 instead of 27+2727+27.

Question 10

A particle starts at position s(0)=5s(0) = 5 and has a continuous velocity function v(t)v(t). Which of the following expressions represents the total distance traveled by the particle on the time interval [0,T][0, T]?

  1. 5+∫0Tv(t)dt5 + \int_{0}^{T} v(t) dt
  2. ∫0T∣v(t)∣dt\int_{0}^{T} |v(t)| dt (correct answer)
  3. ∣∫0Tv(t)dt∣|\int_{0}^{T} v(t) dt|
  4. 5+∫0T∣v(t)∣dt5 + \int_{0}^{T} |v(t)| dt
Explanation: Total distance traveled is the integral of the speed, which is the absolute value of velocity, ∣v(t)∣|v(t)|, over the given time interval. Therefore, the total distance is ∫0T∣v(t)∣dt\int_{0}^{T} |v(t)| dt. The initial position s(0)=5s(0)=5 does not affect the total distance traveled, only the final position. A: This expression represents the particle's final position at time TT, s(T)=s(0)+displacements(T) = s(0) + \text{displacement}. C: This expression represents the magnitude of the displacement, which is the absolute value of the net change in position. This is not the same as total distance unless the velocity never changes sign. D: This expression adds the initial position to the total distance traveled, which is not a standard physical quantity and does not represent the total distance traveled.

Question 11

The acceleration of a particle is given by a(t)=6ta(t) = 6t. The velocity of the particle is 12 at t=2t=2. The particle's initial position is s(0)=5s(0) = 5. What is the total distance traveled by the particle from t=0t=0 to t=2t=2?

  1. 88 (correct answer)
  2. 1212
  3. 1313
  4. 1717
Explanation: First, find the velocity function v(t)v(t) by integrating a(t)a(t). v(t)=∫6tdt=3t2+Cv(t) = \int 6t dt = 3t^2 + C. Use the condition v(2)=12v(2)=12 to find CC: 12=3(22)+C⇒12=12+C⇒C=012 = 3(2^2) + C \Rightarrow 12 = 12 + C \Rightarrow C = 0. So, v(t)=3t2v(t) = 3t^2. To find the total distance traveled on the interval [0,2][0, 2], we need to integrate the speed, ∣v(t)∣|v(t)|. On the interval [0,2][0, 2], v(t)=3t2v(t) = 3t^2 is always non-negative, so ∣v(t)∣=v(t)|v(t)| = v(t). Therefore, the total distance is equal to the displacement. Total Distance =∫023t2dt=[t3]02=23−03=8= \int_{0}^{2} 3t^2 dt = [t^3]_{0}^{2} = 2^3 - 0^3 = 8. The initial position s(0)=5s(0)=5 is extra information not needed for calculating total distance. B: This is the velocity at t=2t=2, not the total distance. C: This is the final position of the particle, s(2)=s(0)+displacement=5+8=13s(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.

Question 12

A particle moves along the x-axis with velocity given by v(t)=3t2−12v(t) = 3t^2 - 12 for t≥0t \ge 0. What is the total distance traveled by the particle during the time interval 0≤t≤30 \le t \le 3?

  1. -9
  2. 9
  3. 16
  4. 23 (correct answer)
Explanation: Total distance is the integral of the absolute value of velocity, ∫ab∣v(t)∣dt\int_a^b |v(t)| dt. First, find where the velocity is zero: 3t2−12=0  ⟹  t2=4  ⟹  t=23t^2 - 12 = 0 \implies t^2 = 4 \implies t = 2 (since t≥0t \ge 0). The velocity is negative on [0,2)[0, 2) and positive on (2,3](2, 3]. The integral must be split at t=2t=2. Total distance = ∫03∣3t2−12∣dt=∫02−(3t2−12)dt+∫23(3t2−12)dt\int_0^3 |3t^2 - 12| dt = \int_0^2 -(3t^2 - 12) dt + \int_2^3 (3t^2 - 12) dt. Integrating gives [12t−t3]02+[t3−12t]23[12t - t^3]_0^2 + [t^3 - 12t]_2^3. Evaluating the first part: (24−8)−0=16(24 - 8) - 0 = 16. Evaluating the second part: (27−36)−(8−24)=−9−(−16)=7(27 - 36) - (8 - 24) = -9 - (-16) = 7. The total distance is 16+7=2316 + 7 = 23.

Question 13

A particle moves along a line with acceleration a(t)=6t−2a(t) = 6t - 2. At time t=0t=0, its velocity is v(0)=3v(0) = 3. What is the displacement of the particle over the time interval [1,4][1, 4]?

  1. 39
  2. 43
  3. 48
  4. 57 (correct answer)
Explanation: First, find the velocity function v(t)v(t) by integrating a(t)a(t). v(t)=∫(6t−2)dt=3t2−2t+Cv(t) = \int (6t - 2) dt = 3t^2 - 2t + C. Use the initial condition v(0)=3v(0) = 3 to find CC: 3(0)2−2(0)+C=3  ⟹  C=33(0)^2 - 2(0) + C = 3 \implies C = 3. So, v(t)=3t2−2t+3v(t) = 3t^2 - 2t + 3. Displacement is the integral of velocity. Displacement = ∫14(3t2−2t+3)dt=[t3−t2+3t]14\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((4)^3 - (4)^2 + 3(4)) - ((1)^3 - (1)^2 + 3(1)) = (64 - 16 + 12) - (1 - 1 + 3) = 60 - 3 = 57.

Question 14

The velocity of a particle moving along a line is v(t)=2t−4v(t) = 2t - 4 meters per second for 0≤t≤50 \le t \le 5. What is the average speed of the particle over this interval?

  1. 1.0 m/s
  2. 2.6 m/s (correct answer)
  3. 5.0 m/s
  4. 13.0 m/s
Explanation: Average speed is the total distance traveled divided by the time interval. Total distance is ∫05∣v(t)∣dt\int_0^5 |v(t)| dt. The velocity function v(t)=2t−4v(t) = 2t-4 is zero at t=2t=2. It's negative for t<2t<2 and positive for t>2t>2. We must split the integral: Total distance = ∫02−(2t−4)dt+∫25(2t−4)dt=[−t2+4t]02+[t2−4t]25\int_0^2 -(2t-4) dt + \int_2^5 (2t-4) dt = [-t^2+4t]_0^2 + [t^2-4t]_2^5. Evaluating the parts: (−4+8)−0=4(-4+8)-0 = 4 and (25−20)−(4−8)=5−(−4)=9(25-20)-(4-8) = 5 - (-4) = 9. Total distance = 4+9=134+9=13 meters. The time interval is 5−0=55-0=5 seconds. Average speed = 135=2.6\frac{13}{5} = 2.6 m/s.

Question 15

A particle's acceleration is given by a(t)=12ta(t) = 12t. At time t=1t=1, its velocity is v(1)=8v(1)=8 and its position is s(1)=−2s(1)=-2. What is the position of the particle at t=2t=2?

  1. 12
  2. 14 (correct answer)
  3. 16
  4. 20
Explanation: First, find v(t)v(t) by integrating a(t)a(t): v(t)=∫12tdt=6t2+C1v(t) = \int 12t dt = 6t^2 + C_1. Use v(1)=8v(1)=8 to find C1C_1: 6(1)2+C1=8  ⟹  C1=26(1)^2 + C_1 = 8 \implies C_1 = 2. So, v(t)=6t2+2v(t) = 6t^2 + 2. Next, find s(t)s(t) by integrating v(t)v(t): s(t)=∫(6t2+2)dt=2t3+2t+C2s(t) = \int (6t^2 + 2) dt = 2t^3 + 2t + C_2. Use s(1)=−2s(1)=-2 to find C2C_2: 2(1)3+2(1)+C2=−2  ⟹  4+C2=−2  ⟹  C2=−62(1)^3 + 2(1) + C_2 = -2 \implies 4 + C_2 = -2 \implies C_2 = -6. So, s(t)=2t3+2t−6s(t) = 2t^3 + 2t - 6. Finally, find the position at t=2t=2: s(2)=2(2)3+2(2)−6=16+4−6=14s(2) = 2(2)^3 + 2(2) - 6 = 16 + 4 - 6 = 14.

Question 16

A particle has acceleration a(t)a(t), initial velocity v(0)v(0), and initial position s(0)s(0). Which of the following expressions represents the displacement of the particle from t=0t=0 to t=3t=3?

  1. ∫03a(t)dt\int_0^3 a(t) dt
  2. v(0)+∫03a(t)dtv(0) + \int_0^3 a(t) dt
  3. ∫03(v(0)+∫0ta(τ)dτ)dt\int_0^3 \left( v(0) + \int_0^t a(\tau) d\tau \right) dt (correct answer)
  4. s(0)+∫03(v(0)+∫0ta(τ)dτ)dts(0) + \int_0^3 \left( v(0) + \int_0^t a(\tau) d\tau \right) dt
Explanation: Displacement is the change in position, s(3)−s(0)s(3) - s(0), which is given by the definite integral of the velocity function, ∫03v(t)dt\int_0^3 v(t) dt. The velocity function v(t)v(t) can be expressed in terms of the initial velocity and acceleration: v(t)=v(0)+∫0ta(τ)dτv(t) = v(0) + \int_0^t a(\tau) d\tau. Substituting this expression for v(t)v(t) into the displacement integral gives ∫03(v(0)+∫0ta(τ)dτ)dt\int_0^3 \left( v(0) + \int_0^t a(\tau) d\tau \right) dt.

Question 17

A particle moves along a line with acceleration a(t)=2eta(t) = 2e^t. Its velocity at t=0t=0 is v(0)=−2v(0) = -2. What is the total distance traveled by the particle from t=0t=0 to t=ln⁡(3)t=\ln(3)?

  1. 8ln⁡(2)−4ln⁡(3)8\ln(2) - 4\ln(3) (correct answer)
  2. 4ln⁡(2)−4ln⁡(3)+24\ln(2) - 4\ln(3) + 2
  3. 2−4ln⁡(3)2 - 4\ln(3)
  4. 2
Explanation: First, find the velocity function: v(t)=∫2etdt=2et+Cv(t) = \int 2e^t dt = 2e^t + C. Using v(0)=−2v(0)=-2, we get 2e0+C=−2  ⟹  2+C=−2  ⟹  C=−42e^0 + C = -2 \implies 2+C=-2 \implies C=-4. So, v(t)=2et−4v(t) = 2e^t - 4. Total distance requires integrating ∣v(t)∣|v(t)|. Find when v(t)=0v(t)=0: 2et−4=0  ⟹  et=2  ⟹  t=ln⁡(2)2e^t - 4 = 0 \implies e^t = 2 \implies t = \ln(2). This is within the interval [0,ln⁡(3)][0, \ln(3)]. For t∈[0,ln⁡(2))t \in [0, \ln(2)), v(t)<0v(t) < 0. For t∈(ln⁡(2),ln⁡(3)]t \in (\ln(2), \ln(3)], v(t)>0v(t) > 0. We split the integral: Total distance = ∫0ln⁡(2)−(2et−4)dt+∫ln⁡(2)ln⁡(3)(2et−4)dt\int_0^{\ln(2)} -(2e^t - 4) dt + \int_{\ln(2)}^{\ln(3)} (2e^t - 4) dt. First integral: [4t−2et]0ln⁡(2)=(4ln⁡(2)−2eln⁡(2))−(0−2e0)=4ln⁡(2)−4−(−2)=4ln⁡(2)−2[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: [2et−4t]ln⁡(2)ln⁡(3)=(2eln⁡(3)−4ln⁡(3))−(2eln⁡(2)−4ln⁡(2))=(6−4ln⁡(3))−(4−4ln⁡(2))=2−4ln⁡(3)+4ln⁡(2)[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 = (4ln⁡(2)−2)+(2−4ln⁡(3)+4ln⁡(2))=8ln⁡(2)−4ln⁡(3)(4\ln(2) - 2) + (2 - 4\ln(3) + 4\ln(2)) = 8\ln(2) - 4\ln(3).

Question 18

A particle moves on a line with velocity v(t)=sin⁡(t)−cos⁡(t)v(t) = \sin(t) - \cos(t) for 0≤t≤π0 \le t \le \pi. If the particle's initial position is s(0)=2s(0) = 2, what is its maximum position on the interval?

  1. 3−23 - \sqrt{2}
  2. 22
  3. 3+23 + \sqrt{2}
  4. 44 (correct answer)
Explanation: First, find the position function s(t)s(t) by integrating v(t)v(t): s(t)=∫(sin⁡(t)−cos⁡(t))dt=−cos⁡(t)−sin⁡(t)+Cs(t) = \int (\sin(t) - \cos(t)) dt = -\cos(t) - \sin(t) + C. Using s(0)=2s(0)=2, we get −cos⁡(0)−sin⁡(0)+C=2  ⟹  −1−0+C=2  ⟹  C=3-\cos(0) - \sin(0) + C = 2 \implies -1 - 0 + C = 2 \implies C=3. So, s(t)=3−cos⁡(t)−sin⁡(t)s(t) = 3 - \cos(t) - \sin(t). To find the maximum position, we must check the critical points and the endpoints. Critical points occur when v(t)=0v(t)=0: sin⁡(t)−cos⁡(t)=0  ⟹  tan⁡(t)=1\sin(t) - \cos(t) = 0 \implies \tan(t) = 1. In the interval [0,π][0, \pi], this occurs at t=π/4t = \pi/4. Now evaluate s(t)s(t) at the endpoints and the critical point: s(0)=2s(0) = 2, s(π)=3−cos⁡(π)−sin⁡(π)=3−(−1)−0=4s(\pi) = 3 - \cos(\pi) - \sin(\pi) = 3 - (-1) - 0 = 4, and s(π/4)=3−cos⁡(π/4)−sin⁡(π/4)=3−22−22=3−2≈1.586s(\pi/4) = 3 - \cos(\pi/4) - \sin(\pi/4) = 3 - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 3 - \sqrt{2} \approx 1.586. Comparing these values, the maximum position is 4.

Question 19

A particle starts at the origin at t=0t=0 and moves along the x-axis with velocity v(t)=t2−8t+12v(t) = t^2 - 8t + 12. What is the first time t>0t > 0 that the particle returns to the origin?

  1. t=2t=2
  2. t=4t=4
  3. t=6t=6 (correct answer)
  4. t=12t=12
Explanation: To find the position function s(t)s(t), we integrate the velocity function: s(t)=∫(t2−8t+12)dt=13t3−4t2+12t+Cs(t) = \int (t^2 - 8t + 12) dt = \frac{1}{3}t^3 - 4t^2 + 12t + C. Since the particle starts at the origin at t=0t=0, we have s(0)=0s(0)=0, which implies C=0C=0. So, s(t)=13t3−4t2+12ts(t) = \frac{1}{3}t^3 - 4t^2 + 12t. To find when the particle returns to the origin, we set s(t)=0s(t) = 0 and solve for t>0t>0: 13t3−4t2+12t=0  ⟹  t(13t2−4t+12)=0\frac{1}{3}t^3 - 4t^2 + 12t = 0 \implies t(\frac{1}{3}t^2 - 4t + 12) = 0. One solution is t=0t=0. For the other solutions, we solve 13t2−4t+12=0\frac{1}{3}t^2 - 4t + 12 = 0. Multiplying by 3 gives t2−12t+36=0t^2 - 12t + 36 = 0, which factors as (t−6)2=0(t-6)^2 = 0. The only solution is t=6t=6. Thus, the first time t>0t>0 the particle returns to the origin is at t=6t=6.

Question 20

A particle's acceleration is a(t)=6t−2a(t) = 6t - 2. Its velocity at t=1t=1 is v(1)=0v(1)=0 and its position at t=1t=1 is s(1)=4s(1)=4. What is the particle's position at t=0t=0?

  1. 00
  2. 33
  3. 44
  4. 55 (correct answer)
Explanation: First, integrate a(t)a(t) to find v(t)v(t): v(t)=∫(6t−2)dt=3t2−2t+C1v(t) = \int (6t-2) dt = 3t^2 - 2t + C_1. Use v(1)=0v(1)=0 to find C1C_1: 0=3(1)2−2(1)+C1⇒0=1+C1⇒C1=−10 = 3(1)^2 - 2(1) + C_1 \Rightarrow 0 = 1 + C_1 \Rightarrow C_1 = -1. So, v(t)=3t2−2t−1v(t) = 3t^2 - 2t - 1. Next, integrate v(t)v(t) to find s(t)s(t): s(t)=∫(3t2−2t−1)dt=t3−t2−t+C2s(t) = \int (3t^2 - 2t - 1) dt = t^3 - t^2 - t + C_2. Use s(1)=4s(1)=4 to find C2C_2: 4=13−12−1+C2⇒4=−1+C2⇒C2=54 = 1^3 - 1^2 - 1 + C_2 \Rightarrow 4 = -1 + C_2 \Rightarrow C_2 = 5. So, the position function is s(t)=t3−t2−t+5s(t) = t^3 - t^2 - t + 5. Finally, find the position at t=0t=0: s(0)=03−02−0+5=5s(0) = 0^3 - 0^2 - 0 + 5 = 5. A: This happens if one correctly finds C1=−1C_1=-1 but forgets the second constant of integration C2C_2, assuming C2=0C_2=0. B: This happens if a sign error is made when finding C1C_1 (i.e., C1=1C_1=1), leading to C2=3C_2=3 and s(0)=3s(0)=3. C: This happens if one forgets the first constant C1C_1 (assuming C1=0C_1=0), which leads to C2=4C_2=4 and s(0)=4s(0)=4.