Connecting Position, Velocity, and Acceleration - AP Calculus BC

1. What is the velocity function $v(t)$ if $a(t) = 6t$?

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

2. What is the physical interpretation of acceleration?

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

3. What is the physical interpretation of velocity?

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

4. Find $v(2)$ if $s(t) = 5t^2 - t$.

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

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

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

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

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

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

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

8. Identify the unit for velocity if position is in meters.

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

9. Identify the unit for acceleration if velocity is in meters per second.

   Meters per second squared (m/s$^2$). Velocity per time unit gives acceleration units.

10. What is the formula for acceleration given velocity $v(t)$?

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

11. If $v(t) = t^2 + 5$, find $a(t)$.

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

12. Identify the derivative that gives the acceleration function.

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

13. What is the initial velocity if $v(t) = 2t + 3$?

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

14. If $s(t) = 7t - t^2$, what is $v(t)$?

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

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

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

16. What is the formula for velocity given position $s(t)$?

    Velocity $v(t) = \frac{ds}{dt}$. Velocity is the derivative of position with respect to time.

17. Find the acceleration at $t = 2$ if $v(t) = 3t^3 - 5t + 4$.

    $a(2) = 31$. Take the derivative: $a(t) = 9t^2 - 5$, then substitute $t = 2$.

18. Determine $s(t)$ given $v(t) = 4t$ and $s(0) = 0$.

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

19. What is $a(t)$ if $v(t)$ is constant?

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

20. What is the expression for instantaneous velocity?

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

21. State the relationship between position, velocity, and acceleration.

    $v(t) = \frac{ds}{dt}$ and $a(t) = \frac{dv}{dt}$. Position, velocity, and acceleration are related through differentiation.

22. Calculate the average velocity if $s(4) = 10$ and $s(1) = 4$.

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

23. State the formula to find average velocity over $[a, b]$.

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

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

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

25. Evaluate $a(0)$ if $v(t) = 4t^2 - 2t$.

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

26. Evaluate $v(1)$ if $s(t) = t^3 + 2t$.

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

27. If $v(t) = 5t - 3$, find $a(t)$.

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

28. Identify the type of motion when $a(t) = 0$.

    Uniform motion. Zero acceleration means constant velocity motion.

29. If $s(t) = 3t^3$, what is $v(t)$?

    $v(t) = 9t^2$. Differentiate the position function.

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

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

31. State the formula to find average velocity over $ [a, b] $.

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

32. What does a negative velocity indicate?

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

33. What is the velocity function $v(t)$ if $a(t) = 0$?

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

34. Describe the velocity if acceleration is zero.

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

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

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

36. If $s(t) = \frac{1}{3}t^3$, what is $v(t)$?

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

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