AP Calculus AB Flashcards: Connecting Position Velocity And Acceleration

What this deck covers

This deck focuses on Connecting Position Velocity And Acceleration, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find the position at $t = 4$ if $v(t) = 3t$ and $s(0) = 2$.

Answer: $s(4) = 26$. $s(t) = \int 3t \, dt = \frac{3t^2}{2} + C$, with $s(0) = 2$ gives $s(4) = 26$.

Flashcard 2: What is the formula for speed given velocity $v(t)$?

Answer: Speed = $|v(t)|$. Speed is the magnitude (absolute value) of velocity.

Flashcard 3: Identify the derivative representing instantaneous velocity.

Answer: $v(t) = s'(t)$. First derivative of position gives instantaneous velocity.

Flashcard 4: What is the formula for velocity given position function $s(t)$?

Answer: $v(t) = s'(t)$. Velocity is the rate of change of position with respect to time.

Flashcard 5: What is the interpretation of zero acceleration?

Answer: The velocity is constant. No change in velocity means constant speed and direction.

Flashcard 6: What mathematical concept describes the rate of change of velocity?

Answer: Acceleration. Acceleration measures how quickly velocity changes.

Flashcard 7: What condition indicates a change in direction of motion?

Answer: Velocity changes sign. Sign change in velocity indicates direction reversal.

Flashcard 8: What is the graphical representation of velocity?

Answer: Slope of the tangent line on position-time graph. Velocity equals the slope of the position curve at any point.

Flashcard 9: What is the formula for acceleration given velocity function $v(t)$?

Answer: $a(t) = v'(t)$. Acceleration is the rate of change of velocity with respect to time.

Flashcard 10: State the sign of acceleration when velocity is increasing.

Answer: Acceleration is positive. Positive acceleration means velocity is increasing over time.

Flashcard 11: What is the significance of a velocity-time graph crossing the time axis?

Answer: Change in direction of motion. Crossing indicates velocity changes sign, reversing direction.

Flashcard 12: Find the velocity at $t = 3$ for $s(t) = 2t^2 - 3t + 4$.

Answer: $v(3) = 9$. $v(t) = s'(t) = 4t - 3$, so $v(3) = 12 - 3 = 9$.

Flashcard 13: What does a constant velocity imply about acceleration?

Answer: Acceleration is zero. Constant velocity means no change in velocity, so $a = 0$.

Flashcard 14: What is the integral of acceleration $a(t)$ with respect to $t$?

Answer: Velocity $v(t)$. Integrating acceleration gives velocity function.

Flashcard 15: State the formula for average velocity over time interval $[a, b]$.

Answer: $v_{avg} = \frac{s(b) - s(a)}{b - a}$. Change in position divided by change in time.

Flashcard 16: Find the acceleration at $t = 2$ for $v(t) = 3t^2 + 2t$.

Answer: $a(2) = 14$. $a(t) = v'(t) = 6t + 2$, so $a(2) = 12 + 2 = 14$.

Flashcard 17: Find the velocity function given $a(t) = 6$ and $v(0) = 4$.

Answer: $v(t) = 6t + 4$. $v(t) = \int 6 \, dt = 6t + C$, with $v(0) = 4$.

Flashcard 18: Find the change in velocity over $[0, 3]$ for $a(t) = 4t$.

Answer: $\text{Change} = 18$. $\int_0^3 4t \, dt = 2t^2 |_0^3 = 18$.

Flashcard 19: What is the relationship between velocity and acceleration?

Answer: Acceleration is the derivative of velocity. Rate of change of velocity gives instantaneous acceleration.

Flashcard 20: Find the velocity when $a(t) = 3t$ and $v(0) = 2$.

Answer: $v(t) = \frac{3}{2}t^2 + 2$. $v(t) = \int 3t \, dt = \frac{3t^2}{2} + C$, with $v(0) = 2$.

Flashcard 21: What is the relationship between position and velocity?

Answer: Velocity is the derivative of position. Rate of change of position gives instantaneous velocity.

Flashcard 22: What is the effect of zero velocity over a time interval?

Answer: No displacement occurs. Zero velocity means no change in position occurs.

Flashcard 23: If $s(t) = 4t^3 - 3t^2$, find the velocity function $v(t)$.

Answer: $v(t) = 12t^2 - 6t$. Velocity is the first derivative of position function.

Flashcard 24: What does a velocity graph's horizontal tangent indicate?

Answer: Zero acceleration. Horizontal tangent means zero slope, so zero acceleration.

Flashcard 25: What does it mean if velocity and acceleration have opposite signs?

Answer: The object is slowing down. Opposite signs indicate velocity and acceleration work against each other.

Flashcard 26: What is the derivative of velocity?

Answer: Acceleration. First derivative of velocity equals acceleration.

Flashcard 27: What is the second derivative of position $s(t)$?

Answer: Acceleration $a(t) = s''(t)$. Second derivative of position gives acceleration function.

Flashcard 28: Calculate $s(t)$ given $v(t) = 2t + 1$ and $s(0) = 3$.

Answer: $s(t) = t^2 + t + 3$. $s(t) = \int (2t + 1) \, dt = t^2 + t + C$, with $s(0) = 3$.

Flashcard 29: What is the effect of constant positive acceleration?

Answer: Velocity increases linearly. Constant acceleration produces linear increase in velocity.

Flashcard 30: What does a zero velocity at an instant imply about motion?

Answer: Object is momentarily at rest. Zero velocity indicates no motion at that instant.