Connecting Position, Velocity, and Acceleration - AP Calculus AB

1. Determine if the particle is speeding up or slowing down at $t = 1$ given $v(t)= t^2 - 4t$ and $a(t) = 2t - 4$.

   Speeding up at $t = 1$. $v(1) = -3 < 0$ and $a(1) = -2 < 0$, same signs means speeding up.

2. Identify the integral representing displacement from $t_1$ to $t_2$.

   $\text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt$. Integrating velocity over time gives net displacement.

3. What is the effect of constant positive acceleration?

   Velocity increases linearly. Constant acceleration produces linear increase in velocity.

4. What is the relationship between velocity and acceleration?

   Acceleration is the derivative of velocity. Rate of change of velocity gives instantaneous acceleration.

5. Find the velocity when $a(t) = 3t$ and $v(0) = 2$.

   $v(t) = \frac{3}{2}t^2 + 2$. $v(t) = \int 3t \, dt = \frac{3t^2}{2} + C$, with $v(0) = 2$.

6. What is the graphical representation of velocity?

   Slope of the tangent line on position-time graph. Velocity equals the slope of the position curve at any point.

7. What does the area under a velocity-time graph represent?

   Displacement. Area under velocity curve equals net displacement.

8. Find the total distance traveled from $t=0$ to $t=3$ if $v(t) = t^2$.

   Distance = $9$. Since $v(t) \geq 0$ on $[0,3]$, distance equals $\int_0^3 t^2 \, dt = 9$.

9. Find the velocity at $t = 3$ for $s(t) = 2t^2 - 3t + 4$.

   $v(3) = 9$. $v(t) = s'(t) = 4t - 3$, so $v(3) = 12 - 3 = 9$.

10. What is the interpretation of zero acceleration?

    The velocity is constant. No change in velocity means constant speed and direction.

11. Find the velocity function given $a(t) = 6$ and $v(0) = 4$.

    $v(t) = 6t + 4$. $v(t) = \int 6 \, dt = 6t + C$, with $v(0) = 4$.

12. What does a velocity graph's horizontal tangent indicate?

    Zero acceleration. Horizontal tangent means zero slope, so zero acceleration.

13. Calculate $s(t)$ given $v(t) = 2t + 1$ and $s(0) = 3$.

    $s(t) = t^2 + t + 3$. $s(t) = \int (2t + 1) \, dt = t^2 + t + C$, with $s(0) = 3$.

14. Identify the derivative representing instantaneous velocity.

    $v(t) = s'(t)$. First derivative of position gives instantaneous velocity.

15. Find the change in velocity over $[0, 3]$ for $a(t) = 4t$.

    $\text{Change} = 18$. $\int_0^3 4t \, dt = 2t^2 |_0^3 = 18$.

16. State the formula for average velocity over time interval $[a, b]$.

    $v_{avg} = \frac{s(b) - s(a)}{b - a}$. Change in position divided by change in time.

17. What does a constant velocity imply about acceleration?

    Acceleration is zero. Constant velocity means no change in velocity, so $a = 0$.

18. What is the formula for speed given velocity $v(t)$?

    Speed = $|v(t)|$. Speed is the magnitude (absolute value) of velocity.

19. What is the interpretation of a negative velocity?

    The object is moving in the opposite direction. Direction is opposite to positive coordinate direction.

20. What does it mean if velocity and acceleration have opposite signs?

    The object is slowing down. Opposite signs indicate velocity and acceleration work against each other.

21. What mathematical concept describes the rate of change of velocity?

    Acceleration. Acceleration measures how quickly velocity changes.

22. State the condition for a particle to be at rest.

    Velocity $v(t) = 0$. Zero velocity means the particle is momentarily stationary.

23. Given $s(t) = t^3 - 6t^2 + 9t + 1$, find $v(2)$.

    $v(2) = -3$. $v(t) = s'(t) = 3t^2 - 12t + 9$, so $v(2) = 12 - 24 + 9 = -3$.

24. What condition indicates a change in direction of motion?

    Velocity changes sign. Sign change in velocity indicates direction reversal.

25. What is the derivative of velocity?

    Acceleration. First derivative of velocity equals acceleration.

26. Find the position at $t = 4$ if $v(t) = 3t$ and $s(0) = 2$.

    $s(4) = 26$. $s(t) = \int 3t \, dt = \frac{3t^2}{2} + C$, with $s(0) = 2$ gives $s(4) = 26$.

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

    Velocity $v(t)$. Integrating acceleration gives velocity function.

28. What does a zero velocity at an instant imply about motion?

    Object is momentarily at rest. Zero velocity indicates no motion at that instant.

29. What is the significance of a velocity-time graph crossing the time axis?

    Change in direction of motion. Crossing indicates velocity changes sign, reversing direction.

30. What is the relationship between position and velocity?

    Velocity is the derivative of position. Rate of change of position gives instantaneous velocity.

31. What is the second derivative of position $s(t)$?

    Acceleration $a(t) = s''(t)$. Second derivative of position gives acceleration function.

32. What is the effect of zero velocity over a time interval?

    No displacement occurs. Zero velocity means no change in position occurs.

33. State the sign of acceleration when velocity is increasing.

    Acceleration is positive. Positive acceleration means velocity is increasing over time.

34. Find the acceleration at $t = 2$ for $v(t) = 3t^2 + 2t$.

    $a(2) = 14$. $a(t) = v'(t) = 6t + 2$, so $a(2) = 12 + 2 = 14$.

35. If $s(t) = 4t^3 - 3t^2$, find the velocity function $v(t)$.

    $v(t) = 12t^2 - 6t$. Velocity is the first derivative of position function.

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

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

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

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

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