Motion: Position, Velocity, Acceleration - Calculus 1

1. A particle moves along a straight line. For a certain time interval, the particle's velocity is decreasing, but its speed is increasing. Which of the following statements must be true during that interval?
   Answer: The particle's velocity is negative and its acceleration is negative. 'Velocity is decreasing' means that the derivative of velocity, acceleration, is negative ($a(t) < 0$). 'Speed is increasing' means that velocity and acceleration have the same sign. Since we know acceleration is negative, velocity must also be negative ($v(t) < 0$). Therefore, the particle's velocity is negative and its acceleration is negative. This corresponds to moving in the negative direction and speeding up.
2. A particle moves along the x-axis with velocity $v(t) = 3(t-1)(t-3)$ for $t \ge 0$. For which of the following time intervals is the particle's displacement equal to the total distance it has traveled?
   Answer: $[0, 1]$ A particle's displacement is equal to the total distance traveled over an interval if and only if the particle does not change direction on that interval. This means its velocity, $v(t)$, must be either always non-negative or always non-positive throughout the interval. The velocity function is $v(t) = 3(t-1)(t-3)$. The sign of $v(t)$ changes at $t=1$ and $t=3$. $v(t) \ge 0$ for $t \in [0, 1] \cup [3, \infty)$. $v(t) \le 0$ for $t \in [1, 3]$. Let's check the given intervals: - On $[0, 1]$, $v(t) \ge 0$, so the particle only moves in the positive direction. Displacement equals total distance. - On $[1, 3]$, $v(t) \le 0$, so the particle only moves in the negative direction. Displacement is negative while total distance is positive, so they are not equal. - On $[0, 2]$, the particle moves in the positive direction for $[0, 1]$ and in the negative direction for $[1, 2]$. Since it changes direction, displacement is not equal to total distance. - On $[2, 4]$, the particle moves in the negative direction for $[2, 3]$ and in the positive direction for $[3, 4]$. Since it changes direction, displacement is not equal to total distance.
3. A particle moves along the x-axis with velocity $v(t) = e^{-t} (2t - t^2)$ for $t > 0$. The particle's acceleration is zero at the exact moment its velocity reaches a local maximum. At what time $t$ does this occur?
   Answer: $t = 2 + \sqrt{2}$ The particle's acceleration is zero when its velocity reaches a local maximum. We need to find the time $t$ when $a(t) = v'(t) = 0$. Using the product rule for $v(t) = e^{-t} (2t - t^2)$: $a(t) = (e^{-t})' (2t - t^2) + e^{-t} (2t - t^2)' = -e^{-t}(2t - t^2) + e^{-t}(2 - 2t)$. Set $a(t) = 0$: $e^{-t} [-(2t - t^2) + (2 - 2t)] = 0$. Since $e^{-t} > 0$, we solve $-2t + t^2 + 2 - 2t = 0 \implies t^2 - 4t + 2 = 0$. Using the quadratic formula: $t = \frac{4 \pm \sqrt{16 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}$. To determine which is a maximum, we can use the second derivative test for velocity, $a'(t)$, or analyze the sign of $a(t)$. The parabola $y=t^2-4t+2$ opens up, so it changes from + to - to + at its roots. The acceleration $a(t)$ goes from positive to negative at the larger root, $t = 2 + \sqrt{2}$, which indicates a local maximum for velocity.
4. The position of a particle on the x-axis is $s(t) = At^2 + Bt + C$, where A, B, and C are non-zero constants. Which of the following statements about the particle's motion is true for all $t > 0$?
   Answer: The particle's acceleration is always constant. Let's analyze the motion based on the position function $s(t) = At^2 + Bt + C$. The velocity is $v(t) = s'(t) = 2At + B$. The acceleration is $a(t) = v'(t) = 2A$. Since A is a non-zero constant, the acceleration $2A$ is also a non-zero constant. Thus, statement B is true. Statement A is false because $v(t)$ is a linear function of $t$, so speed $|v(t)|$ is not constant unless $A=0$, which is ruled out. Statement C is false because $v(t)=0$ at $t = -B/(2A)$. This change of direction may or may not occur for $t>0$, depending on the signs of A and B. Statement D is false because the particle is speeding up only when $v(t)$ and $a(t)$ have the same sign. It will slow down if their signs are different.
5. A particle's velocity is given by $v(t) = t^2 - 8t + 15$. Which of the following statements correctly describes the motion of the particle at the exact moment its acceleration is zero?
   Answer: The particle has reached its minimum velocity. First, find the acceleration function $a(t) = v'(t)$. $a(t) = 2t - 8$. Set the acceleration to zero to find the time: $2t - 8 = 0 \implies t = 4$. Now we analyze the particle's motion at $t=4$. - Velocity at $t=4$: $v(4) = (4)^2 - 8(4) + 15 = 16 - 32 + 15 = -1$. Since $v(4) \neq 0$, the particle is not at rest. - Change of direction: The particle changes direction when $v(t)$ changes sign, which occurs when $v(t)=0$. $t^2 - 8t + 15 = (t-3)(t-5)=0$ at $t=3$ and $t=5$. The particle is not changing direction at $t=4$. - The velocity function $v(t)$ is a parabola that opens upward. Its minimum value occurs at its vertex. The time at which the vertex occurs is found by setting the derivative of $v(t)$ to zero, which is $v'(t) = a(t) = 0$. We already found this occurs at $t=4$. Therefore, at $t=4$, the particle has reached its minimum velocity. - Speed is $|v(t)|$. The speed at $t=4$ is $|-1|=1$. For large $t$, $v(t)$ grows infinitely large, so the speed is not at a maximum.
6. A particle moves along a line such that its position is given by $s(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$. The particle is momentarily at rest at two different times. What is the magnitude of the difference in its acceleration at these two times?
   Answer: $12$ First, find the velocity function $v(t) = s'(t)$. $v(t) = 3t^2 - 12t + 9$. The particle is momentarily at rest when $v(t) = 0$. $3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t-1)(t-3) = 0$ The particle is at rest at $t=1$ and $t=3$. Next, find the acceleration function $a(t) = v'(t)$. $a(t) = 6t - 12$. Now, calculate the acceleration at the two times when the particle is at rest: $a(1) = 6(1) - 12 = -6$. $a(3) = 6(3) - 12 = 18 - 12 = 6$. The difference in the accelerations is $a(3) - a(1) = 6 - (-6) = 12$. The magnitude of this difference is $|12| = 12$.
7. The position of a particle is $s(t) = 2\sin(t) - t$ for $0 \le t \le 2\pi$. For which of the following values of $t$ is the particle moving to the right and slowing down?
   Answer: $t = \pi/6$ Moving right means $v(t) > 0$. Slowing down means $v(t)$ and $a(t)$ have opposite signs. So we need $v(t) > 0$ and $a(t) < 0$. First, find the functions: $v(t) = s'(t) = 2\cos(t) - 1$. $a(t) = v'(t) = -2\sin(t)$. Now we test the given points: A) At $t=\pi/6$: $v(\pi/6) = 2\cos(\pi/6) - 1 = 2(\sqrt{3}/2) - 1 = \sqrt{3}-1 > 0$. $a(\pi/6) = -2\sin(\pi/6) = -2(1/2) = -1 < 0$. Here, $v>0$ and $a<0$, so the particle is moving right and slowing down. B) At $t=\pi/3$, $v=0$. C) At $t=2\pi/3$, $v<0$. D) At $t=\pi$, $v<0$.
8. A particle moves on a line with velocity $v(t) = t^2 - kt + 4$, where $k$ is a constant. The particle is momentarily at rest at $t=2$. What is the acceleration of the particle at this instant?
   Answer: 0 The particle is momentarily at rest when its velocity is zero. We are given this occurs at $t=2$, so $v(2)=0$. Substitute $t=2$ into the velocity function: $v(2) = (2)^2 - k(2) + 4 = 0$. This gives $4 - 2k + 4 = 0$, so $8 - 2k = 0$, which means $k=4$. Now we know the full velocity function is $v(t) = t^2 - 4t + 4$. To find the acceleration, we differentiate the velocity: $a(t) = v'(t) = 2t - 4$. Finally, we evaluate the acceleration at $t=2$: $a(2) = 2(2) - 4 = 0$.
9. The velocity of a particle is $v(t) = (t-1)(t-3)(t-5)$ for $t \ge 0$. Which statement best describes the particle's motion at $t=2$?
   Answer: Moving right and slowing down. We need the signs of velocity $v(t)$ and acceleration $a(t)$ at $t=2$. At $t=2$, $v(2) = (2-1)(2-3)(2-5) = (1)(-1)(-3) = 3$. Since $v(2) > 0$, the particle is moving right. For acceleration, we first find the expression for $v(t) = t^3 - 9t^2 + 23t - 15$. Differentiating gives $a(t) = v'(t) = 3t^2 - 18t + 23$. Now evaluate at $t=2$: $a(2) = 3(2^2) - 18(2) + 23 = 12 - 36 + 23 = -1$. Since $v(2) > 0$ and $a(2) < 0$, the signs are opposite, so the particle is slowing down. Thus, it is moving right and slowing down.
10. The acceleration of a particle moving on a line is $a(t) = 12t^2 - 18t$. At $t=0$, the particle's velocity is $v(0)=5$ and its position is $s(0)=2$. At $t=1$, what is the particle's velocity?
    Answer: 0 This question requires antidifferentiation. To find velocity $v(t)$, we integrate acceleration $a(t)$: $v(t) = \int (12t^2 - 18t) dt = 4t^3 - 9t^2 + C$. We use the initial condition $v(0)=5$ to find C: $v(0) = 4(0)^3 - 9(0)^2 + C = 5 \implies C=5$. So the velocity function is $v(t) = 4t^3 - 9t^2 + 5$. Finally, we find the velocity at $t=1$: $v(1) = 4(1)^3 - 9(1)^2 + 5 = 4 - 9 + 5 = 0$. The information about position $s(0)$ is extra information not needed to solve for velocity.
11. For a particle in rectilinear motion, its velocity $v(t)$ is a differentiable function of time $t$. For a specific time interval $(a, b)$, it is known that $v(t) < 0$ and the acceleration $a(t) > 0$. Which of the following statements must be true for the particle on the interval $(a, b)$?
    Answer: The particle's speed is decreasing. The speed of the particle is given by $|v(t)|$. A particle is speeding up if its velocity and acceleration have the same sign, and it is slowing down if they have opposite signs. - We are given that $v(t) < 0$, which means the particle is moving in the negative direction. Consequently, its position is decreasing, not increasing. - We are given that $a(t) > 0$. - Since $v(t)$ and $a(t)$ have opposite signs, the particle is slowing down. Its speed, $|v(t)|$, is decreasing. - For the particle to change direction, its velocity $v(t)$ must change sign. We are given that $v(t) < 0$ for the entire interval $(a, b)$, so it does not change direction.
12. The position of a particle moving along a line is $s(t) = t^4 - 8t^3 + 18t^2$. For what value of $t>0$ is the particle's velocity at a minimum?
    Answer: $t=3$ To find when velocity is at a minimum, we need to find the critical points of the velocity function by finding where its derivative (acceleration) is zero. First, find velocity: $v(t) = s'(t) = 4t^3 - 24t^2 + 36t$. Now, find acceleration: $a(t) = v'(t) = 12t^2 - 48t + 36$. Set $a(t)=0$: $12(t^2 - 4t + 3) = 0 \implies 12(t-1)(t-3)=0$. The critical points for velocity are at $t=1$ and $t=3$. To determine which is a minimum, we use the second derivative test on velocity, which means finding the sign of $a'(t)$. $a'(t) = 24t - 48$. At $t=1$, $a'(1) = 24-48 = -24 < 0$, indicating a local maximum for velocity. At $t=3$, $a'(3) = 24(3)-48 = 72-48 = 24 > 0$, indicating a local minimum for velocity. Thus, velocity is at a minimum at $t=3$.
13. A particle moves on the x-axis with position given by $s(t) = t^3 - 6t^2 + 5$ for $t \ge 0$. At which time $t>0$ is the particle momentarily at rest after having moved in the negative direction?
    Answer: $t=4$ First, find when the particle is at rest by setting the velocity $v(t)=0$. $v(t) = s'(t) = 3t^2 - 12t = 3t(t-4)$. The particle is at rest at $t=0$ and $t=4$. We are interested in $t>0$, so we consider $t=4$. Next, we must determine if the particle was moving in the negative direction just before $t=4$. Moving in the negative direction means $v(t) < 0$. The sign of $v(t) = 3t(t-4)$ is negative for $t$ between 0 and 4. So, on the interval $(0, 4)$, the particle is moving in the negative direction. Therefore, at $t=4$, it comes to rest after having moved in the negative direction.
14. The position of a particle is given by $s(t) = \frac{8t}{t^2+4}$ for $t \ge 0$. Which statement accurately describes the particle's motion at the instant it is farthest from the origin in the positive direction?
    Answer: Its velocity is zero and its acceleration is negative. The particle is farthest from the origin in the positive direction when its position $s(t)$ is at a local maximum. This occurs when $s'(t) = v(t) = 0$ and $s''(t) = a(t) < 0$ (by the Second Derivative Test). First, find the velocity: $v(t) = s'(t) = \frac{8(t^2+4) - 8t(2t)}{(t^2+4)^2} = \frac{32-8t^2}{(t^2+4)^2}$. Setting $v(t)=0$ gives $32-8t^2=0$, so $t^2=4$ and $t=2$ (since $t \ge 0$). Now, find the acceleration at $t=2$. $a(t)=v'(t) = \frac{-16t(t^2+4)^2 - (32-8t^2) \cdot 2(t^2+4)(2t)}{(t^2+4)^4}$. At $t=2$, the numerator of $v(t)$ is 0, so the second term of the numerator of $a(t)$ is 0. Thus, $a(2) = \frac{-16(2)(2^2+4)^2}{(2^2+4)^4} = \frac{-32}{(8)^2} = -1/2$. Since $v(2)=0$ and $a(2)<0$, the particle is at a local maximum position.
15. A particle's position is given by $s(t) = 2t^3 - 21t^2 + 60t$ for $t \ge 0$. What is the particle's velocity when its acceleration is zero?
    Answer: -13.5 This is a two-step problem. First, find the time when acceleration is zero. Then, find the velocity at that time. $s(t) = 2t^3 - 21t^2 + 60t$. $v(t) = s'(t) = 6t^2 - 42t + 60$. $a(t) = v'(t) = 12t - 42$. Set $a(t) = 0$: $12t - 42 = 0 \implies t = 42/12 = 7/2 = 3.5$. Now, substitute this time back into the velocity function: $v(3.5) = 6(3.5)^2 - 42(3.5) + 60 = 6(12.25) - 147 + 60 = 73.5 - 147 + 60 = -13.5$.
16. The position of a particle is given by $s(t) = t^3 - 12t^2 + 36t$ for $t \ge 0$. At what time $t$ in the open interval $(0, 3)$ is the particle's instantaneous velocity equal to its average velocity on the closed interval $[0, 3]$?
    Answer: $4 - \sqrt{7}$ This question is an application of the Mean Value Theorem. First, calculate the average velocity on the interval $[0, 3]$. $s(0) = 0^3 - 12(0)^2 + 36(0) = 0$. $s(3) = 3^3 - 12(3)^2 + 36(3) = 27 - 108 + 108 = 27$. The average velocity is $v_{avg} = \frac{s(3) - s(0)}{3 - 0} = \frac{27}{3} = 9$. Next, find the instantaneous velocity function, $v(t) = s'(t)$. $v(t) = 3t^2 - 24t + 36$. Set the instantaneous velocity equal to the average velocity: $3t^2 - 24t + 36 = 9$ $3t^2 - 24t + 27 = 0$ $t^2 - 8t + 9 = 0$ Use the quadratic formula to solve for $t$: $t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(9)}}{2(1)} = \frac{8 \pm \sqrt{64 - 36}}{2} = \frac{8 \pm \sqrt{28}}{2} = \frac{8 \pm 2\sqrt{7}}{2} = 4 \pm \sqrt{7}$. The two possible times are $t = 4 - \sqrt{7}$ and $t = 4 + \sqrt{7}$. We must determine which of these lies in the interval $(0, 3)$. Since $2 < \sqrt{7} < 3$, we have: $t_1 = 4 - \sqrt{7} \approx 4 - 2.65 = 1.35$. This is in $(0, 3)$. $t_2 = 4 + \sqrt{7} \approx 4 + 2.65 = 6.65$. This is not in $(0, 3)$. Therefore, the correct time is $t = 4 - \sqrt{7}$.
17. A particle moves along a line with acceleration $a(t) = 4t - 10$. At time $t=0$, its velocity is $v(0) = 8$. For $t \ge 0$, what is the minimum speed of the particle?
    Answer: $0$ First, we need to find the velocity function $v(t)$ by integrating the acceleration function $a(t)$. $v(t) = \int a(t) \,dt = \int (4t - 10) \,dt = 2t^2 - 10t + C$. We use the initial condition $v(0)=8$ to find the constant $C$. $v(0) = 2(0)^2 - 10(0) + C = 8 \implies C = 8$. So, the velocity function is $v(t) = 2t^2 - 10t + 8$. Speed is the absolute value of velocity, $|v(t)|$. To find the minimum speed, we must find the minimum value of $|2t^2 - 10t + 8|$. The minimum value of an absolute value function can be 0 if the function itself can be 0. Let's check if the particle ever stops, i.e., if $v(t)=0$. $2t^2 - 10t + 8 = 0$ $2(t^2 - 5t + 4) = 0$ $2(t-1)(t-4) = 0$ The velocity is zero at $t=1$ and $t=4$. When the velocity is zero, the speed is also zero. Since speed cannot be negative, the minimum possible speed is 0. This minimum is achieved at $t=1$ and $t=4$.
18. The position of a particle on the x-axis is given by $s(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$. What is the total distance traveled by the particle from $t=0$ to $t=2$?
    Answer: 6 Total distance requires accounting for any changes in direction. First, find the velocity: $v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$. The particle changes direction when $v(t)$ changes sign, which occurs at $t=1$ and $t=3$. Within the interval $[0, 2]$, there is a change of direction at $t=1$. We must calculate the distance traveled on $[0, 1]$ and $[1, 2]$ separately. Positions are: $s(0) = 1$, $s(1) = 1-6+9+1=5$, and $s(2) = 8-24+18+1=3$. The distance traveled from $t=0$ to $t=1$ is $|s(1)-s(0)| = |5-1| = 4$. The distance traveled from $t=1$ to $t=2$ is $|s(2)-s(1)| = |3-5| = 2$. The total distance is $4+2=6$.
19. A particle moves along a line with acceleration $a(t) = 2t$. Its velocity at $t=3$ is $v(3)=5$. What is its average velocity over the interval $[0, 3]$?
    Answer: -1 First, find the velocity function by integrating acceleration: $v(t) = \int 2t dt = t^2 + C$. Use the condition $v(3)=5$ to find C: $3^2+C=5 \implies 9+C=5 \implies C=-4$. So, $v(t) = t^2-4$. Average velocity over $[0, 3]$ is given by $\frac{s(3)-s(0)}{3-0}$, which requires finding the position function $s(t)$. $s(t) = \int (t^2-4) dt = \frac{1}{3}t^3 - 4t + D$. The displacement is $s(3)-s(0) = (\frac{1}{3}(3)^3 - 4(3) + D) - (\frac{1}{3}(0)^3 - 4(0) + D) = (9-12) - 0 = -3$. The average velocity is $\frac{-3}{3-0} = -1$.
20. The position of a particle moving on the x-axis is given by $s(t) = e^t(t^2 - 3)$ for $t \ge 0$. For what open interval of time $t$ is the particle moving toward the origin?
    Answer: $(1, \sqrt{3})$ A particle is moving toward the origin when its position and velocity have opposite signs. This means we need to find where ($s(t) > 0$ and $v(t) < 0$) or ($s(t) < 0$ and $v(t) > 0$). 1. Find the sign of position $s(t) = e^t(t^2 - 3)$. Since $e^t$ is always positive, the sign of $s(t)$ is determined by the sign of $t^2 - 3$. For $t \ge 0$, $s(t) < 0$ when $t^2 < 3$, i.e., on $[0, \sqrt{3})$. $s(t) > 0$ when $t^2 > 3$, i.e., on $(\sqrt{3}, \infty)$. 2. Find the velocity $v(t) = s'(t)$ using the product rule: $v(t) = e^t(t^2 - 3) + e^t(2t) = e^t(t^2 + 2t - 3) = e^t(t+3)(t-1)$. Since $e^t$ and $(t+3)$ are positive for $t \ge 0$, the sign of $v(t)$ is determined by the sign of $(t-1)$. Thus, $v(t) < 0$ on $[0, 1)$ and $v(t) > 0$ on $(1, \infty)$. 3. Combine the results. We need $s(t)$ and $v(t)$ to have opposite signs. - On $[0, 1)$: $s(t) < 0$ and $v(t) < 0$. Same signs. Moving away from the origin. - On $(1, \sqrt{3})$: $s(t) < 0$ and $v(t) > 0$. Opposite signs. Moving toward the origin. - On $(\sqrt{3}, \infty)$: $s(t) > 0$ and $v(t) > 0$. Same signs. Moving away from the origin. The particle is moving toward the origin on the interval $(1, \sqrt{3})$.

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