This problem tests understanding of straight-line motion using trigonometric functions. At t=3π/4, we find v(3π/4)=sin(3π/4)=√2/2>0, indicating rightward motion. To determine if the car is speeding up or slowing down, we need acceleration: a(t)=v'(t)=cos(t), so a(3π/4)=cos(3π/4)=-√2/2<0. Since velocity is positive and acceleration is negative (opposite signs), the car is slowing down. The trap answer "moving right and speeding up" incorrectly assumes positive velocity means speeding up. Key strategy: check if v and a have the same sign (speeding up) or opposite signs (slowing down).

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.

This straight-line motion problem analyzes speed changes over an interval. When velocity is negative on $(1,2)$, the particle moves left throughout this interval. Since acceleration is positive on $(1,2)$, it points right, opposing the leftward motion. When velocity and acceleration have opposite signs, the magnitude of velocity (speed) decreases. Students might think "speed is increasing" because acceleration is positive, but positive acceleration opposes negative velocity. The key motion analysis strategy is that speed decreases when velocity and acceleration have opposite signs, and speed increases when they have the same sign.

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.

This straight-line motion analysis examines speed changes when velocity and acceleration have opposite signs. Since $v(2) > 0$, the particle moves right at $t = 2$. Given $a(2) < 0$, the acceleration points left, opposing the rightward motion. When velocity and acceleration have opposite signs, the speed decreases because acceleration opposes the direction of motion. Students might think positive velocity always means increasing speed, but negative acceleration reduces rightward speed. The key motion analysis strategy is that opposite signs of velocity and acceleration always mean decreasing speed.

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.

This straight-line motion analysis uses explicit notation showing that acceleration equals the derivative of velocity. Since $v(2) > 0$, the object moves right at $t = 2$. Given $a(2) = v'(2) < 0$, the acceleration is negative, pointing left and opposing the rightward motion. When velocity and acceleration have opposite signs, the object slows down. Students might think positive velocity always means speeding up, but negative acceleration reduces the rightward speed. The key motion analysis strategy is to check if velocity and acceleration have the same sign (speeding up) or opposite signs (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).

This straight-line motion analysis examines position function behavior when both derivatives are negative. 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 negative, meaning the position function is concave down. A decreasing and concave down function falls while curving downward, creating a steep downward curve. Students might struggle with visualizing negative concavity combined with decreasing behavior. The key strategy is that $s'(t) < 0$ means decreasing, and $s''(t) < 0$ means concave down.

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.