AP Calculus AB Flashcards: Position Velocity And Acceleration Using Integrals

What this deck covers

This deck focuses on Position Velocity And Acceleration Using Integrals, 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: What is the derivative of position with respect to time?

Answer: Velocity. Rate of change of position is velocity by definition.

Flashcard 2: If $a(t) = 0$, what can be said about $v(t)$?

Answer: $v(t)$ is a constant function. Zero acceleration implies constant velocity function.

Flashcard 3: What does the area under a velocity-time graph represent?

Answer: Displacement. Area under curve equals change in position.

Flashcard 4: Given $s(t) = 4t^2 - 2t + 1$, find $v(t)$ and $a(t)$.

Answer: $v(t) = 8t - 2$, $a(t) = 8$. Differentiate position once for $v(t)$, twice for $a(t)$.

Flashcard 5: What is the initial condition used for when integrating velocity to find position?

Answer: To find the constant of integration. Determines the arbitrary constant $C$ from integration.

Flashcard 6: How do you find average velocity over an interval $[a, b]$?

Answer: Average velocity = $\frac{s(b) - s(a)}{b - a}$. Displacement divided by time elapsed.

Flashcard 7: If velocity is constant, what is acceleration?

Answer: Zero. Derivative of constant velocity is zero.

Flashcard 8: Given $v(t) = 6t - 3t^2$, find $s(t)$ if $s(0) = 5$.

Answer: $s(t) = 3t^2 - t^3 + 5$. Integrate velocity and use initial condition.

Flashcard 9: If $s(t) = t^2 - 3t + 2$, what is $v(t)$?

Answer: $v(t) = 2t - 3$. Differentiate position function using power rule.

Flashcard 10: Given $v(t) = 5t - 4$, find the integral to determine position.

Answer: Integrate $v(t)$ to get $s(t) = \frac{5t^2}{2} - 4t + C$. Apply antiderivative rules for polynomial functions.

Flashcard 11: If $v(t) = t^3 - 6t^2 + 9t$, find $a(t)$.

Answer: $a(t) = 3t^2 - 12t + 9$. Take the derivative of velocity using power rule.

Flashcard 12: Given $v(t) = 4t - 7$, find the acceleration $a(t)$.

Answer: $a(t) = 4$. Take the derivative of velocity function.

Flashcard 13: If $a(t) = 3$, find $v(t)$ given $v(0) = 5$.

Answer: $v(t) = 3t + 5$. Integrate constant acceleration and apply initial condition.

Flashcard 14: Given $s(t) = 2t^3 - 5t^2 + 3t + 1$, find $v(t)$.

Answer: $v(t) = 6t^2 - 10t + 3$. Take the derivative of position function.

Flashcard 15: How do you find total distance traveled from a velocity function $v(t)$?

Answer: Integrate $|v(t)|$ over the interval. Absolute value accounts for direction changes.

Flashcard 16: What is the integral of zero acceleration over time?

Answer: Constant velocity. Integrating zero gives constant velocity.

Flashcard 17: If $s(t) = 4t - \text{ln}(t)$, what is $v(t)$?

Answer: $v(t) = 4 - \frac{1}{t}$. Differentiate using power and logarithmic rules.

Flashcard 18: If $v(t) = 0$ for all $t$, what can be said about $s(t)$?

Answer: $s(t)$ is constant. Zero velocity means no change in position.

Flashcard 19: What is the implication of $a(t) = 0$ over an interval?

Answer: Velocity is constant. Zero acceleration means no change in velocity.

Flashcard 20: If $v(t) = 3t$, find $s(t)$ given $s(0) = 0$.

Answer: $s(t) = \frac{3t^2}{2}$. Integrate velocity and apply zero initial position.

Flashcard 21: What condition must be checked when integrating to determine total distance traveled?

Answer: Check for sign changes in $v(t)$. Must split integral where velocity changes sign.

Flashcard 22: If $v(t) = 2t + 1$, what is the integral to find total distance?

Answer: Integrate $|2t + 1|$ over the interval. Use absolute value when velocity changes sign.

Flashcard 23: How do you determine if an object is speeding up or slowing down?

Answer: Check if $v(t)$ and $a(t)$ have the same sign. Same sign means speeding up, opposite means slowing down.

Flashcard 24: What is the physical interpretation of $v(t) = 0$?

Answer: Object is momentarily at rest. Zero velocity means no instantaneous motion.

Flashcard 25: What is the formula for displacement from time $t_1$ to $t_2$?

Answer: $s(t_2) - s(t_1) = \text{integral of } v(t) \text{ from } t_1 \text{ to } t_2$. Net change in position equals integral of velocity.

Flashcard 26: If $a(t) = 8 - 2t$, find $v(t)$ given $v(0) = 0$.

Answer: $v(t) = 8t - t^2$. Integrate acceleration and apply zero initial velocity.

Flashcard 27: For $v(t) = 2t^3 + 3$, find the acceleration $a(t)$.

Answer: $a(t) = 6t^2$. Differentiate velocity using power rule.

Flashcard 28: What is the algebraic expression for displacement?

Answer: $\text{integral of } v(t) \text{ from } t_1 \text{ to } t_2$. Definite integral of velocity over time interval.

Flashcard 29: What is the significance of the constant of integration when finding $s(t)$?

Answer: It represents initial position. The constant $C$ represents position at $t = 0$.

Flashcard 30: If $s(t) = \text{ln}(t)$, what is $v(t)$?

Answer: $v(t) = \frac{1}{t}$. Differentiate logarithmic function.