AP Calculus BC 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 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: Find the value of $\int_2^5 1 \, dt$.

Answer: $3$. Integral of constant 1 over interval of length $5-2=3$.

Flashcard 2: What is the acceleration at $t = 5$ for $v(t) = t^2 - 4t + 3$?

Answer: $a(5) = 6$. Take derivative: $\frac{d}{dt}(t^2-4t+3) = 2t-4$, so $a(5) = 6$.

Flashcard 3: What is the velocity at $t = 3$ if $v(t) = 8t - 5$?

Answer: $v(3) = 19$. Substitute $t=3$ into the velocity function.

Flashcard 4: What is the integral of acceleration function $a(t)$ with respect to $t$?

Answer: Velocity function $v(t) = \text{integral of } a(t) \text{ dt}$. Integration reverses differentiation to get velocity from acceleration.

Flashcard 5: How do you find displacement from time $t_1$ to $t_2$?

Answer: Displacement = $\int_{t_1}^{t_2} v(t) \, dt$. Definite integral of velocity over time interval gives displacement.

Flashcard 6: State the fundamental theorem of calculus for definite integrals.

Answer: If $F$ is antiderivative of $f$, $\int_a^b f(x) \, dx = F(b) - F(a)$. Evaluates definite integrals using antiderivatives.

Flashcard 7: What is the formula for finding total distance traveled using velocity?

Answer: Total distance = $\text{integral of } |v(t)| \text{ dt}$. Absolute value ensures all movement counts as positive distance.

Flashcard 8: What is the integral of velocity function $v(t)$ with respect to $t$?

Answer: Position function $s(t) = \text{integral of } v(t) \text{ dt}$. Integration reverses differentiation to find position from velocity.

Flashcard 9: Find the position function for $v(t) = 2t^2 - 3t$ with $s(0) = 1$.

Answer: $s(t) = \frac{2}{3}t^3 - \frac{3}{2}t^2 + 1$. Integrate velocity and apply initial condition.

Flashcard 10: Find the velocity function given acceleration $a(t) = 5$ and initial velocity $v(0) = 3$.

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

Flashcard 11: Determine the position function $s(t)$ for $v(t) = 4t - 1$ and $s(0) = 2$.

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

Flashcard 12: What is the initial velocity if $v(t) = 4t + 6$ at $t = 0$?

Answer: $v(0) = 6$. Evaluate velocity function at $t=0$.

Flashcard 13: Determine the acceleration at $t = 2$ for $v(t) = 3t^2 - 5t$.

Answer: $a(2) = 7$. Take derivative: $\frac{d}{dt}(3t^2-5t) = 6t-5$, so $a(2) = 7$.

Flashcard 14: If $v(t) = 5t$, what is the total distance from $t = 0$ to $t = 3$?

Answer: $22.5$ units. $\int_0^3 5t dt = [\frac{5t^2}{2}]_0^3 = \frac{45}{2} = 22.5$.

Flashcard 15: Find the velocity function for $a(t) = 2t + 3$ with $v(0) = -1$.

Answer: $v(t) = t^2 + 3t - 1$. Integrate acceleration and apply initial velocity condition.

Flashcard 16: State the relationship between velocity and acceleration in terms of derivatives.

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

Flashcard 17: How is velocity $v(t)$ related to position $s(t)$ through differentiation?

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

Flashcard 18: Calculate the net change in position from $t = 0$ to $t = 3$ for $v(t) = t^2$.

Answer: $9$ units. $\int_0^3 t^2 dt = [\frac{t^3}{3}]_0^3 = 9$.

Flashcard 19: What is the velocity at $t = 0$ for $v(t) = 3t^2 - 2t + 1$?

Answer: $v(0) = 1$. Substitute $t=0$ into velocity function.

Flashcard 20: Evaluate the change in position from $t = 1$ to $t = 4$ for $v(t) = 2t$.

Answer: $15$ units. $\int_1^4 2t dt = [t^2]_1^4 = 16-1 = 15$.

Flashcard 21: Find the acceleration function if $v(t) = 6t^2 - 4t + 1$.

Answer: $a(t) = 12t - 4$. Take the derivative of velocity to get acceleration.

Flashcard 22: Given $v(t) = 4t - 1$, find the position function if $s(0) = 0$.

Answer: $s(t) = 2t^2 - t$. Integrate velocity: $\int(4t-1)dt = 2t^2-t+C$, with $C=0$.

Flashcard 23: Calculate the velocity at $t = 1$ for $v(t) = 7 - t^2$.

Answer: $v(1) = 6$. Substitute $t=1$ into velocity function: $7-1^2 = 6$.

Flashcard 24: Evaluate $\text{integral from } 1 \text{ to } 4 \text{ of } 3 \text{ dt}$.

Answer: $9$. Integral of constant over interval $(4-1) = 3 \times 3 = 9$.

Flashcard 25: Calculate the total distance for $v(t) = t^2 - 4t$ from $t = 0$ to $t = 4$.

Answer: $\frac{32}{3}$ units. Velocity changes sign at $t=4$, so split integral using $|v(t)|$.

Flashcard 26: What is the relationship between total distance and displacement?

Answer: Total distance considers absolute value of velocity; displacement does not. Distance uses $|v(t)|$ while displacement uses $v(t)$ directly.

Flashcard 27: How do you express change in position from time $t_1$ to $t_2$?

Answer: $s(t_2) - s(t_1) = \int_{t_1}^{t_2} v(t) \, dt$. Change in position equals the integral of velocity over time.

Flashcard 28: What is the formula for finding total distance traveled using velocity?

Answer: Total distance $= \text{integral of } |v(t)| \text{ dt}$. Absolute value ensures all movement counts as positive distance.

Flashcard 29: Find the position function for $v(t) = 2t^2 - 3t$ with $s(0) = 1$.

Answer: $s(t) = \frac{2}{3}t^3 - \frac{3}{2}t^2 + 1$. Integrate velocity and apply initial condition.

Flashcard 30: Evaluate $\text{integral from } 1 \text{ to } 4 \text{ of } 3 \text{ dt}$.

Answer: $9$. Integral of constant over interval $(4-1)$ = $3 \times 3$ = 9