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 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.

This straight-line motion analysis focuses on what happens when velocity changes sign at a specific time. Since $v(3) = 0$, the particle is momentarily at rest at $t = 3$. The fact that velocity changes from negative to positive means the particle changes from moving left to moving right. This represents a change in direction of motion. Students might think about speed maxima or zero acceleration, but the key insight is the direction change. The correct motion analysis strategy is to recognize that when velocity changes sign (passes through zero), the particle changes direction.

This straight-line motion analysis examines what must be true when velocity is consistently positive over an interval. Since $v(t) > 0$ for all $t$ in $(1,4)$, the velocity is always positive on this interval. Positive velocity means the position function $s(t)$ is always increasing on $(1,4)$ because $s'(t) = v(t) > 0$. We cannot determine acceleration, speed changes, or specific values without more information. Students might think about acceleration or speed, but only position behavior is guaranteed. The key strategy is that positive velocity always means increasing position.

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.

This problem requires analyzing straight-line motion using velocity and acceleration signs. Given $v(2) < 0$, the particle is moving left (negative direction) at $t = 2$. Since $a(2) > 0$, the acceleration is positive, pointing right. When velocity and acceleration have opposite signs, the particle is slowing down because acceleration opposes the direction of motion. Many students incorrectly choose "moving left and speeding up" by thinking positive acceleration always means speeding up. The key strategy is to check if velocity and acceleration have the same sign (speeding up) or opposite signs (slowing down).

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 $x = c$.

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.

This straight-line motion problem examines speed changes when both velocity and acceleration are negative. Since $v(2) < 0$, the particle moves left at $t = 2$. Given $a(2) < 0$, the acceleration also points left, in the same direction as the velocity. When velocity and acceleration have the same sign (both negative), the speed increases because acceleration enhances the leftward motion. Students might think negative acceleration means decreasing speed, but here both point left together. The correct motion analysis strategy is that same signs of velocity and acceleration mean increasing speed.

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.