AP Calculus BC 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 BC.

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: If $v(t) = 3t - 4$, find $a(t)$.

Answer: $a(t) = 3$. Differentiate the velocity function.

Flashcard 2: If $s(t) = \frac{1}{3}t^3$, what is $v(t)$?

Answer: $v(t) = t^2$. Differentiate the position function.

Flashcard 3: State the formula to find average velocity over $[a, b]$.

Answer: $\frac{s(b) - s(a)}{b - a}$. Average rate of change of position over the interval.

Flashcard 4: Find $a(3)$ if $v(t) = t^2 - 2t + 1$.

Answer: $a(3) = 4$. Differentiate velocity: $a(t) = 2t - 2$, then evaluate at $t = 3$.

Flashcard 5: What is the velocity function $v(t)$ if $a(t) = 0$?

Answer: $v(t) = C$, a constant. Zero acceleration implies velocity remains constant.

Flashcard 6: Identify the type of motion when $a(t) = 0$.

Answer: Uniform motion. Zero acceleration means constant velocity motion.

Flashcard 7: Evaluate $v(1)$ if $s(t) = t^3 + 2t$.

Answer: $v(1) = 5$. Differentiate position: $v(t) = 3t^2 + 2$, then evaluate at $t = 1$.

Flashcard 8: If $v(t) = 5t - 3$, find $a(t)$.

Answer: $a(t) = 5$. Differentiate the velocity function.

Flashcard 9: What does a negative velocity indicate?

Answer: Motion in the opposite direction. Negative velocity means moving in the negative direction.

Flashcard 10: If $v(t) = 3t - 4$, find $a(t)$.

Answer: $a(t) = 3$. Differentiate the velocity function.

Flashcard 11: What is the expression for instantaneous velocity?

Answer: $v(t) = \frac{ds}{dt}$. The derivative of position gives instantaneous velocity.

Flashcard 12: What is $a(t)$ if $v(t)$ is constant?

Answer: $a(t) = 0$. Derivative of a constant velocity is zero.

Flashcard 13: Identify the derivative that gives the acceleration function.

Answer: Derivative of the velocity function. Acceleration is found by differentiating velocity.

Flashcard 14: Describe the velocity if acceleration is zero.

Answer: Velocity is constant. No acceleration means velocity doesn't change.

Flashcard 15: Identify the unit for velocity if position is in meters.

Answer: Meters per second (m/s). Distance per time unit matches the position unit.

Flashcard 16: If $s(t) = t^3 - 3t^2 + t$, find $v(t)$.

Answer: $v(t) = 3t^2 - 6t + 1$. Differentiate the position function to get velocity.

Flashcard 17: What is the physical interpretation of acceleration?

Answer: Rate of change of velocity with respect to time. Describes how velocity changes with time.

Flashcard 18: What is the physical interpretation of velocity?

Answer: Rate of change of position with respect to time. Describes how position changes with time.

Flashcard 19: Calculate the acceleration at $t = 1$ if $v(t) = t^2 + 2t$.

Answer: $a(1) = 4$. Differentiate velocity: $a(t) = 2t + 2$, then evaluate at $t = 1$.

Flashcard 20: What is the initial velocity if $v(t) = 2t + 3$?

Answer: $v(0) = 3$. Substitute $t = 0$ into the velocity function.

Flashcard 21: If $s(t) = 7t - t^2$, what is $v(t)$?

Answer: $v(t) = 7 - 2t$. Differentiate the position function.

Flashcard 22: Find the velocity at $t = 3$ if $s(t) = 4t^2 + 2t + 1$.

Answer: $v(3) = 26$. Take the derivative: $v(t) = 8t + 2$, then substitute $t = 3$.

Flashcard 23: Evaluate $a(0)$ if $v(t) = 4t^2 - 2t$.

Answer: $a(0) = -2$. Differentiate velocity: $a(t) = 8t - 2$, then evaluate at $t = 0$.

Flashcard 24: Find the velocity at $t = 0$ if $v(t) = t^2 - 4t + 3$.

Answer: $v(0) = 3$. Evaluate the velocity function at $t = 0$.

Flashcard 25: If $v(t) = 2t^2 - 4t + 1$, find $a(t)$.

Answer: $a(t) = 4t - 4$. Differentiate the velocity function to get acceleration.

Flashcard 26: What is the formula for acceleration given velocity $v(t)$?

Answer: Acceleration $a(t) = \frac{dv}{dt}$. Acceleration is the derivative of velocity with respect to time.

Flashcard 27: Calculate the average velocity if $s(4) = 10$ and $s(1) = 4$.

Answer: $\frac{10 - 4}{4 - 1} = 2$. Apply the average velocity formula with given values.

Flashcard 28: What is the velocity function $v(t)$ if $a(t) = 6t$?

Answer: $v(t) = 3t^2 + C$. Integrate acceleration to get velocity, adding constant of integration.

Flashcard 29: If $v(t) = t^2 + 5$, find $a(t)$.

Answer: $a(t) = 2t$. Differentiate the velocity function.

Flashcard 30: Find $v(2)$ if $s(t) = 5t^2 - t$.

Answer: $v(2) = 19$. Differentiate position: $v(t) = 10t - 1$, then evaluate at $t = 2$.