Position, Velocity, and Acceleration Using Integrals - AP Calculus BC

1. What is the formula for finding total distance traveled using velocity?

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

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

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

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

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

4. What is the relationship between total distance and displacement?

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

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

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

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

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

7. State the relationship between velocity and acceleration in terms of derivatives.

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

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

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

9. What is the integral of acceleration function $a(t)$ with respect to $t$?

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

10. What is the initial velocity if $v(t) = 4t + 6$ at $t = 0$?

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

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

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

12. How do you find displacement from time $t_1$ to $t_2$?

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

13. What is the integral of velocity function $v(t)$ with respect to $t$?

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

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

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

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

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

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

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

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

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

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

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

19. Evaluate $\text{integral from } 1 \text{ to } 4 \text{ of } 3 \text{ dt}$.

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

20. Find the value of $\int_2^5 1 \, dt$.

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

21. Find the value of $\text{integral from } 2 \text{ to } 5 \text{ of } 1 \text{ dt}$.

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

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

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

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

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

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

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

25. How do you express change in position from time $t_1$ to $t_2$?

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

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

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

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

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

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

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

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