Displacement, Velocity, and Acceleration - AP Physics C Mechanics

1. A particle moves in the xy-plane with a position vector given by $\vec{r}(t) = (2t^3 - 3t)\hat{i} + (t^2 + 1)\hat{j}$. What is the magnitude of the particle's velocity at $t=2 \text{ s}$?
   Answer: $\sqrt{457}$ The velocity vector is the time derivative of the position vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t^2 - 3)\hat{i} + (2t)\hat{j}$. At $t=2 \text{ s}$, the velocity vector is $\vec{v}(2) = (6(2)^2 - 3)\hat{i} + (2(2))\hat{j} = (24 - 3)\hat{i} + 4\hat{j} = 21\hat{i} + 4\hat{j}$. The magnitude of the velocity is $|\vec{v}(2)| = \sqrt{21^2 + 4^2} = \sqrt{441 + 16} = \sqrt{457}$.
2. The velocity of a particle as a function of time is given by $v(t) = 5t - t^2$. What is the average acceleration of the particle over the time interval from $t=1 \text{ s}$ to $t=4 \text{ s}$?
   Answer: $0 \text{ m/s}^2$ Average acceleration is the change in velocity divided by the time interval: $a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v(4) - v(1)}{4-1}$. First, calculate the velocities: $v(4) = 5(4) - (4)^2 = 20 - 16 = 4 \text{ m/s}$. $v(1) = 5(1) - (1)^2 = 5 - 1 = 4 \text{ m/s}$. Then, $a_{avg} = \frac{4 - 4}{3} = 0 \text{ m/s}^2$.
3. Two objects A and B move along parallel tracks. Object A has constant velocity $v_A = 10$ m/s. Object B starts from rest with constant acceleration $a_B = 2.0$ m/s². At $t = 0$, object A is 50 m behind object B. At what time do they have the same velocity, and what is their relative position at that moment?
   Answer: $t = 5.0$ s; A is 25 m behind B Object B's velocity is $v_B(t) = 2.0t$. They have equal velocities when $10 = 2.0t$, so $t = 5.0$ s. At $t = 5.0$ s: Object A's position is $x_A = x_{A0} + 10(5) = (-50) + 50 = 0$ m (taking B's initial position as origin). Object B's position is $x_B = 0 + \frac{1}{2}(2.0)(5)^2 = 25$ m. Therefore A is at 0 m and B is at 25 m, meaning A is 25 m behind B. Choice B incorrectly claims A is ahead. Choice C uses wrong distance calculation. Choice D uses incorrect time calculation.
4. A particle moves along a straight line such that its position is given by $x(t) = 2t^3 - 9t^2 + 12t + 5$ meters, where $t$ is in seconds. At what time(s) does the particle momentarily come to rest?
   Answer: $t = 1$ s and $t = 2$ s The particle comes to rest when its velocity equals zero. Taking the derivative: $v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12$. Setting $v(t) = 0$: $6t^2 - 18t + 12 = 0$, which simplifies to $t^2 - 3t + 2 = 0$. Factoring: $(t-1)(t-2) = 0$, giving $t = 1$ s and $t = 2$ s. Choice B incorrectly uses the acceleration equation. Choice C uses $t = 2$ correctly but incorrectly includes $t = 3$. Choice D correctly identifies $t = 1$ but incorrectly includes $t = 3$.
5. A car starts from rest and its speed is given by $v(t) = kt^2$, where $k$ is a positive constant. How does its average speed $v_{avg}$ over the interval from $t=0$ to $t=T$ compare to its instantaneous speed $v(T/2)$ at the midpoint of the interval?
   Answer: $v_{avg} > v(T/2)$ First, find the instantaneous speed at $t=T/2$: $v(T/2) = k(T/2)^2 = \frac{1}{4}kT^2$. Next, find the average speed. Average speed is total distance divided by time. Distance is the integral of speed: $x(T) = \int_0^T kt^2 dt = [\frac{1}{3}kt^3]_0^T = \frac{1}{3}kT^3$. The average speed is $v_{avg} = \frac{x(T)}{T} = \frac{\frac{1}{3}kT^3}{T} = \frac{1}{3}kT^2$. Comparing the two, $\frac{1}{3}kT^2 > \frac{1}{4}kT^2$, so $v_{avg} > v(T/2)$.
6. An object starts from rest and experiences an acceleration given by $a(t) = A\cos(\frac{\pi}{2}t)$, where $A$ is a constant. What is the object's change in velocity during the time interval from $t=0$ to $t=1$ s?
   Answer: $\frac{2A}{\pi}$ The change in velocity is the definite integral of acceleration over the time interval. $\Delta v = \int_0^1 a(t) dt = \int_0^1 A\cos(\frac{\pi}{2}t) dt$. The integral of $\cos(kt)$ is $\frac{1}{k}\sin(kt)$. So, $\Delta v = A [\frac{2}{\pi}\sin(\frac{\pi}{2}t)]_0^1 = \frac{2A}{\pi}[\sin(\frac{\pi}{2}) - \sin(0)] = \frac{2A}{\pi}[1 - 0] = \frac{2A}{\pi}$.
7. A particle moves along the x-axis. At a particular instant, its velocity is negative and its acceleration is positive. Which of the following statements correctly describes the particle's motion at that instant?
   Answer: The particle is moving in the negative direction and slowing down. The sign of velocity indicates the direction of motion. Since velocity is negative, the particle is moving in the negative direction. When velocity and acceleration have opposite signs, the object's speed is decreasing. Therefore, the particle is moving in the negative direction and slowing down.
8. An object's position is given by the function $x(t) = 2t^3 - 15t^2 + 36t - 8$. At which of the following times is the object momentarily at rest?
   Answer: $t=2 \text{ s}$ and $t=3 \text{ s}$ The object is momentarily at rest when its velocity is zero. We find the velocity function by differentiating the position function: $v(t) = \frac{dx}{dt} = 6t^2 - 30t + 36$. Set $v(t) = 0$ and solve for $t$: $6(t^2 - 5t + 6) = 0$, which factors to $6(t-2)(t-3) = 0$. The solutions are $t=2 \text{ s}$ and $t=3 \text{ s}$.
9. A particle moves along the x-axis with acceleration $a(x) = -kx$, where $k > 0$ is a constant and $x$ is the position. If the particle starts from rest at position $x_0$, what is its speed when it reaches position $x = 0$?
   Answer: $x_0\sqrt{k}$ Since acceleration depends on position, use $v\frac{dv}{dx} = a(x) = -kx$. Integrating: $\int v dv = \int -kx dx$, giving $\frac{v^2}{2} = -\frac{kx^2}{2} + C$. Using initial condition $v = 0$ at $x = x_0$: $0 = -\frac{kx_0^2}{2} + C$, so $C = \frac{kx_0^2}{2}$. Therefore $\frac{v^2}{2} = \frac{k}{2}(x_0^2 - x^2)$. At $x = 0$: $\frac{v^2}{2} = \frac{kx_0^2}{2}$, so $v^2 = kx_0^2$ and $v = x_0\sqrt{k}$. Choice A includes an unnecessary factor of 1/2. Choice B incorrectly includes the factor 1/2 under the square root. Choice C has incorrect algebraic form.
10. The velocity of a particle is given by $v(t) = \frac{1}{3}t^3 - 2t^2 + 4t$. At what time $t>0$ is the particle's acceleration equal to zero?
    Answer: $t=2 \text{ s}$ Acceleration is the derivative of velocity with respect to time. $a(t) = \frac{dv}{dt} = t^2 - 4t + 4 = (t-2)^2$. To find when acceleration is zero, we set $a(t)=0$. This gives $(t-2)^2 = 0$, so $t=2 \text{ s}$.
11. A ball is thrown vertically upward with initial velocity $v_0$. At the instant when the ball returns to its starting height, which statement correctly compares the ball's displacement, velocity, and acceleration to their values at the moment of release?
    Answer: Displacement is zero, velocity has the same magnitude but opposite direction, acceleration is unchanged When the ball returns to its starting height, the displacement is zero by definition. Due to energy conservation and symmetry, the velocity magnitude equals the initial velocity but points downward (opposite direction). Acceleration due to gravity remains constant at $-g$ throughout the motion. Choice B incorrectly states velocity is zero (that occurs only at maximum height). Choice C also incorrectly states velocity is zero. Choice D incorrectly states displacement is maximum and velocity direction is the same.
12. A particle moves along the x-axis such that its position is given by the function $x(t) = At^3 - Bt^2 + C$, where $A=2.0 \text{ m/s}^3$, $B=3.0 \text{ m/s}^2$, and $C=1.0 \text{ m}$. What is the instantaneous velocity of the particle at $t = 2.0 \text{ s}$?
    Answer: $12 \text{ m/s}$ Instantaneous velocity is the time derivative of the position function, $v(t) = \frac{dx}{dt}$. Differentiating $x(t)$ with respect to time gives $v(t) = 3At^2 - 2Bt$. Substituting the given values and $t=2.0 \text{ s}$ yields $v(2.0) = 3(2.0)(2.0)^2 - 2(3.0)(2.0) = 24 - 12 = 12 \text{ m/s}$.
13. The velocity of a particle moving in one dimension is described by the equation $v(t) = (12t - 3t^2) \text{ m/s}$. What is the particle's acceleration at the instant it momentarily stops?
    Answer: $-12 \text{ m/s}^2$ First, find the time when the particle momentarily stops by setting $v(t) = 0$. This gives $12t - 3t^2 = 3t(4 - t) = 0$, so $t=0 \text{ s}$ and $t=4 \text{ s}$. Acceleration is the derivative of velocity: $a(t) = \frac{dv}{dt} = 12 - 6t$. At $t=4 \text{ s}$, the acceleration is $a(4) = 12 - 6(4) = 12 - 24 = -12 \text{ m/s}^2$.
14. Which of the following describes a situation where an object has an instantaneous velocity of zero but a non-zero instantaneous acceleration?
    Answer: A ball thrown vertically upward at the highest point of its trajectory. At the peak of its trajectory, a vertically thrown ball momentarily stops moving upwards before it begins to fall back down, so its instantaneous velocity is zero. However, the acceleration due to gravity is still acting on it, so its acceleration is approximately $-9.8 \text{ m/s}^2$. Therefore, it has zero velocity and non-zero acceleration.
15. An object's acceleration along the x-axis is given by $a(t) = 6t - 4$, in $m/s^2$. If the object's initial velocity at $t=0$ is $v(0) = 5 \text{ m/s}$, what is its velocity at $t=3 \text{ s}$?
    Answer: $20 \text{ m/s}$ Velocity is the integral of acceleration with respect to time. $v(t) = \int a(t) dt = \int (6t - 4) dt = 3t^2 - 4t + C$. We use the initial condition $v(0) = 5 \text{ m/s}$ to find the constant of integration $C$. $v(0) = 3(0)^2 - 4(0) + C = 5$, so $C=5$. The velocity function is $v(t) = 3t^2 - 4t + 5$. At $t=3 \text{ s}$, $v(3) = 3(3)^2 - 4(3) + 5 = 27 - 12 + 5 = 20 \text{ m/s}$.
16. A particle moves along the x-axis with a velocity given by $v_x(t) = 4t^3 - 2t$. If the particle is at $x=1$ at time $t=1$, what is its position at time $t=2$?
    Answer: $13$ Position is the integral of velocity. $x(t) = \int v_x(t) dt = \int (4t^3 - 2t) dt = t^4 - t^2 + C$. Use the condition $x(1)=1$ to find $C$. $1 = (1)^4 - (1)^2 + C$, which gives $1 = 0 + C$, so $C=1$. The position function is $x(t) = t^4 - t^2 + 1$. At $t=2$, the position is $x(2) = (2)^4 - (2)^2 + 1 = 16 - 4 + 1 = 13$.
17. The position of an object is given by $x(t) = 2t^3 - 3t^2 + 4$. What is the average velocity of the object over the time interval from $t=1 \text{ s}$ to $t=3 \text{ s}$?
    Answer: $14 \text{ m/s}$ Average velocity is defined as displacement divided by the time interval: $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x(3) - x(1)}{3-1}$. First, calculate the positions: $x(3) = 2(3)^3 - 3(3)^2 + 4 = 54 - 27 + 4 = 31 \text{ m}$. $x(1) = 2(1)^3 - 3(1)^2 + 4 = 2 - 3 + 4 = 3 \text{ m}$. Then, $v_{avg} = \frac{31 - 3}{2} = \frac{28}{2} = 14 \text{ m/s}$.
18. A particle moves along the x-axis with a velocity given by $v(t) = 3t^2 - 6t$. What is the particle's displacement between $t=1 \text{ s}$ and $t=3 \text{ s}$?
    Answer: $2 \text{ m}$ Displacement is the definite integral of velocity over the time interval. $\Delta x = \int_{1}^{3} v(t) dt = \int_{1}^{3} (3t^2 - 6t) dt$. Evaluating the integral gives $[t^3 - 3t^2]_{1}^{3} = ((3)^3 - 3(3)^2) - ((1)^3 - 3(1)^2) = (27 - 27) - (1 - 3) = 0 - (-2) = 2 \text{ m}$.
19. A particle starts from rest at the origin and moves along the x-axis with an acceleration $a_x(t) = C t^{1/2}$, where $C$ is a positive constant. Which of the following represents the particle's position $x(t)$ as a function of time?
    Answer: $\frac{4C}{15}t^{5/2}$ To find position from acceleration, we must integrate twice. First, find velocity: $v_x(t) = \int a_x(t) dt = \int C t^{1/2} dt = C \frac{t^{3/2}}{3/2} + v_0$. Since the particle starts from rest, $v_0=0$, so $v_x(t) = \frac{2C}{3}t^{3/2}$. Next, find position: $x(t) = \int v_x(t) dt = \int \frac{2C}{3}t^{3/2} dt = \frac{2C}{3} \frac{t^{5/2}}{5/2} + x_0$. Since it starts at the origin, $x_0=0$. Thus, $x(t) = \frac{4C}{15}t^{5/2}$.
20. The velocity of a particle in the xy-plane is given by $\vec{v}(t) = (4t)\hat{i} + (3t^2 - 1)\hat{j}$. What is the magnitude of the particle's acceleration at $t=1 \text{ s}$?
    Answer: $\sqrt{52}$ The acceleration vector is the time derivative of the velocity vector: $\vec{a}(t) = \frac{d\vec{v}}{dt} = 4\hat{i} + (6t)\hat{j}$. At $t=1 \text{ s}$, the acceleration vector is $\vec{a}(1) = 4\hat{i} + 6(1)\hat{j} = 4\hat{i} + 6\hat{j}$. The magnitude of the acceleration is $|\vec{a}(1)| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$.

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