Connecting Position, Velocity, and Acceleration - AP Calculus AB

1. A particle's acceleration is positive at $t=3$ and its velocity is zero at $t=3$; which is correct at $t=3$?
   Answer: At rest and speeding up This straight-line motion problem involves a particle at rest with positive acceleration. Since the velocity is zero at $t = 3$, the particle is momentarily at rest. Given that acceleration is positive at $t = 3$, the particle will begin moving in the positive direction (right). When at rest with non-zero acceleration, the particle is effectively "speeding up" from zero velocity. Students might think about current motion, but the particle is transitioning from rest to motion. The key motion analysis strategy is that being at rest with non-zero acceleration means accelerating away from rest in the direction of the acceleration.
2. A particle has $v(4)>0$ and $a(4)>0$; which statement about its position $s(t)$ is true at $t=4$?
   Answer: $s(t)$ is increasing at $t=4$ This straight-line motion analysis connects velocity and acceleration to position function behavior. Since $v(4) > 0$, the velocity is positive, meaning the position function $s(t)$ is increasing at $t = 4$. The fact that $a(4) > 0$ tells us about the acceleration but doesn't change that positive velocity means increasing position. Students might think about local maxima or minima, but those occur when velocity equals zero. The key motion analysis strategy is to remember that positive velocity means the position function is increasing, and negative velocity means it's decreasing.
3. For $0<t<3$, an object has $v(t)>0$ and $a(t)<0$; which best describes its motion on that interval?
   Answer: Moving right and slowing down This straight-line motion analysis covers motion over an interval with specific velocity and acceleration signs. Since $v(t) > 0$ on $(0,3)$, the object moves right throughout this interval. Given $a(t) < 0$ on $(0,3)$, the acceleration is negative, pointing left and opposing the rightward motion. When velocity and acceleration have opposite signs, the object slows down. Students might incorrectly choose "moving right and speeding up" by focusing only on the positive velocity. The correct motion analysis strategy is to check if velocity and acceleration have the same sign (speeding up) or opposite signs (slowing down).
4. An object has $s'(t)>0$ and $s''(t)<0$ on $(0,1)$; what is true about $s(t)$ there?
   Answer: Increasing and concave down This straight-line motion problem analyzes position function characteristics using derivative information. Since $s'(t) > 0$ on $(0,1)$, the position function is increasing throughout this interval. Given $s''(t) < 0$ on $(0,1)$, the second derivative is negative, meaning the position function is concave down. An increasing and concave down function rises while curving downward (like a regular smile). Students might confuse the signs of derivatives with function behavior. The key strategy is that $s'(t) > 0$ means increasing, and $s''(t) < 0$ means concave down.
5. At $t=2$, $v(2)<0$ and $a(2)=v'(2)>0$; which description is correct at $t=2$?
   Answer: Moving left and slowing down This straight-line motion problem explicitly shows acceleration as the derivative of velocity. Since $v(2) < 0$, the object moves left at $t = 2$. Given $a(2) = v'(2) > 0$, the acceleration is positive, pointing right and opposing the leftward motion. When velocity and acceleration have opposite signs, the object slows down. Students might think positive acceleration always means speeding up, but here it opposes the negative velocity. The correct motion analysis strategy is to determine if velocity and acceleration have the same sign (speeding up) or opposite signs (slowing down).
6. Velocity is positive and decreasing at $t=5$; which statement about acceleration at $t=5$ is correct?
   Answer: Acceleration is negative This straight-line motion problem connects velocity behavior to acceleration using derivative relationships. When velocity is positive and decreasing at $t = 5$, we have $v(5) > 0$ and $v'(5) < 0$. Since acceleration is the derivative of velocity, $a(5) = v'(5) < 0$, so acceleration is negative. Students might think positive velocity means positive acceleration, but decreasing velocity always means negative acceleration. The key motion analysis strategy is that acceleration equals the derivative of velocity, so decreasing velocity always indicates negative acceleration.
7. A particle has $v(5)=0$ and $a(5)>0$; which statement about motion at $t=5$ is correct?
   Answer: At rest and speeding up This straight-line motion problem involves a particle with zero velocity but non-zero acceleration. Since $v(5) = 0$, the particle is momentarily at rest at $t = 5$. However, $a(5) > 0$ means the acceleration is positive, indicating the particle will begin moving in the positive direction. When at rest with positive acceleration, the particle is "speeding up" from zero. Students might think "at rest and slowing down" makes sense, but slowing down requires the particle to already be moving. The correct motion analysis strategy is to recognize that at rest with non-zero acceleration means the particle is about to start moving in the direction of acceleration.
8. A particle has velocity $v(t)=(t+1)(t-3)$. At what time is the particle at rest?
   Answer: $t=-1$ and $t=3$ This problem requires finding when a particle is at rest by solving for when velocity equals zero. Given v(t)=(t+1)(t-3), we set this equal to zero: (t+1)(t-3)=0. Using the zero product property, either t+1=0 or t-3=0, giving us t=-1 or t=3. Both values are valid times when the particle is at rest. Students might miss one solution by not fully factoring or might include t=0 by confusing it with a root. Key strategy: to find when a particle is at rest, set v(t)=0 and solve completely, checking all factors.
9. An object has $v(t)<0$ and $a(t)>0$ for all $t$ in $(2,5)$; which is true on $(2,5)$?
   Answer: Moving left and slowing down This straight-line motion problem analyzes motion over an interval with consistent velocity and acceleration signs. Since $v(t) < 0$ on $(2,5)$, the object moves left throughout this interval. Given $a(t) > 0$ on $(2,5)$, the acceleration points right, opposing the leftward motion. When velocity and acceleration have opposite signs, the object slows down. Students might think "moving left and speeding up" because they misinterpret positive acceleration, but positive acceleration opposes negative velocity. The correct motion analysis strategy is that opposite signs of velocity and acceleration always mean slowing down.
10. A particle's position $s(t)$ satisfies $s'(1)=0$ and $s''(1)>0$; what best describes $s(t)$ at $t=1$?
    Answer: $s(t)$ has a local minimum at $t=1$ This straight-line motion analysis uses the second derivative test to analyze position function behavior. Since $s'(1) = 0$, the velocity is zero, creating a critical point for the position function at $t = 1$. Given $s''(1) > 0$, the second derivative is positive, indicating the position function is concave up at this critical point. By the second derivative test, this means $s(t)$ has a local minimum at $t = 1$. Students might confuse this with a maximum by misremembering the second derivative test. The key strategy is that $s'(c) = 0$ and $s''(c) > 0$ means a local minimum at $x = c$.
11. An object has $s'(t)<0$ and $s''(t)>0$ on $(0,1)$; what is true about $s(t)$ there?
    Answer: Decreasing and concave up This straight-line motion analysis examines position function behavior with specific derivative signs. Since $s'(t) < 0$ on $(0,1)$, the position function is decreasing throughout this interval. Given $s''(t) > 0$ on $(0,1)$, the second derivative is positive, meaning the position function is concave up. A decreasing and concave up function falls while curving upward (like an upside-down smile). Students might confuse how concavity combines with increasing/decreasing behavior. The key strategy is that $s'(t) < 0$ means decreasing, and $s''(t) > 0$ means concave up.
12. At $t=8$, $v(8)<0$ and $a(8)>0$; which statement about speed is correct at $t=8$?
    Answer: Speed is decreasing This straight-line motion problem focuses on speed changes when velocity and acceleration have specific signs. Since $v(8) < 0$, the particle moves left at $t = 8$. Given $a(8) > 0$, the acceleration points right, opposing the leftward motion. When velocity and acceleration have opposite signs, the speed (magnitude of velocity) decreases. Students might think "speed is increasing" because acceleration is positive, but positive acceleration actually opposes negative velocity here. The correct motion analysis strategy is that speed decreases when velocity and acceleration have opposite signs.
13. Velocity is negative and increasing at $t=5$; which statement about acceleration at $t=5$ is correct?
    Answer: Acceleration is positive This straight-line motion analysis connects velocity changes to acceleration through derivative relationships. When velocity is negative and increasing at $t = 5$, we have $v(5) < 0$ and $v'(5) > 0$. Since acceleration equals the derivative of velocity, $a(5) = v'(5) > 0$, so acceleration is positive. Students might think negative velocity implies negative acceleration, but increasing velocity (even if negative) always means positive acceleration. The key strategy is that acceleration is the derivative of velocity, so increasing velocity always indicates positive acceleration regardless of velocity's sign.
14. A particle has $v(2)>0$ and $a(2)>0$; which statement about speed at $t=2$ is correct?
    Answer: Speed is increasing This straight-line motion analysis focuses on speed changes when both velocity and acceleration are positive. Since $v(2) > 0$, the particle moves right at $t = 2$. Given $a(2) > 0$, the acceleration also points right, in the same direction as the velocity. When velocity and acceleration have the same sign (both positive), the speed increases because acceleration enhances the motion. Students rarely make errors with this straightforward case. The key motion analysis strategy is that when velocity and acceleration have the same sign, speed always increases.
15. A particle has $v(3)=0$ and changes from $v>0$ to $v<0$ at $t=3$; what occurs at $t=3$?
    Answer: A change in direction This straight-line motion problem examines what occurs when velocity changes from positive to negative. Since $v(3) = 0$, the particle is momentarily at rest at $t = 3$. The change from $v > 0$ to $v < 0$ means the particle changes from moving right to moving left, representing a change in direction. Students might think this creates a minimum in position, but direction change is the primary characteristic. The correct motion analysis strategy is to recognize that when velocity changes sign at a point, the particle changes direction at that point.
16. A particle moves with $v(t)<0$ for all $t$ in $(1,4)$; which statement must be true on $(1,4)$?
    Answer: Position is decreasing This straight-line motion problem examines the consequences of consistently negative velocity over an interval. Since $v(t) < 0$ for all $t$ in $(1,4)$, the velocity is always negative on this interval. Negative velocity means the position function $s(t)$ is always decreasing on $(1,4)$ because $s'(t) = v(t) < 0$. We cannot determine acceleration, speed changes, or other properties without additional information. Students might think about acceleration or speed, but only position behavior is determined. The key strategy is that negative velocity always means decreasing position.
17. A particle's acceleration is negative and its velocity is positive at $t=1$; which is true at that instant?
    Answer: Moving right and slowing down This straight-line motion analysis examines the relationship between velocity and acceleration signs. Since the particle has positive velocity, it's moving right at $t = 1$. Given that acceleration is negative, it points left, opposing the rightward motion. When velocity and acceleration have opposite signs, the particle slows down. Students might incorrectly choose "moving right and speeding up" by thinking positive velocity always means speeding up. The key motion analysis strategy is to check if velocity and acceleration have the same sign (speeding up) or opposite signs (slowing down).
18. A particle moves with $v(0)>0$ and $a(0)=0$; which conclusion is necessarily true at $t=0$?
    Answer: Moving right This straight-line motion problem focuses on what can be definitively concluded from given information. Since $v(0) > 0$, the particle has positive velocity, meaning it's moving right at $t = 0$. The fact that $a(0) = 0$ means acceleration is zero, so the particle is neither speeding up nor slowing down at that instant. With zero acceleration, we cannot determine if the particle is speeding up or slowing down. Students might think "speeding up" or "slowing down" are necessarily true, but zero acceleration means neither applies. The key strategy is to recognize that only the sign of velocity (positive means moving right) can be definitively determined when acceleration is zero.
19. An object moves with $v(t)>0$ and $a(t)>0$ for $t$ in $(0,2)$; which is true on $(0,2)$?
    Answer: Moving right and speeding up This straight-line motion analysis examines consistent positive velocity and acceleration over an interval. Since $v(t) > 0$ on $(0,2)$, the object moves right throughout this interval. Given $a(t) > 0$ on $(0,2)$, the acceleration also points right, in the same direction as the motion. When velocity and acceleration have the same sign (both positive), the object speeds up. Students rarely make errors with this case since both quantities point in the same direction. The key motion analysis strategy is that same signs of velocity and acceleration always mean speeding up.
20. An object moves along a line with $v(t)<0$ and $a(t)<0$ for $0<t<2$. On this interval, the object is
    Answer: moving left and speeding up This problem tests understanding straight-line motion when velocity and acceleration have the same sign. With v(t)<0, the object moves left (negative direction). Since a(t)<0 as well, velocity becomes more negative over time, meaning the speed (absolute value of velocity) increases. When velocity and acceleration have the same sign (both negative here), the object speeds up in the direction of motion. The trap is thinking negative acceleration always means slowing down, but this is only true when velocity is positive. Remember: same signs (both positive or both negative) mean speeding up; opposite signs mean slowing down.

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