Chapter 3: Kinematics in Two Dimensions

Displacement, Velocity, and Acceleration in Two Dimensions

• Position Vector: The position of an object in a two-dimensional plane is defined by its position vector $\vec{\mathbf{r}}$. For an initial position vector $\vec{\mathbf{r}}_0$ at time $t_0$ and a final position vector $\vec{\mathbf{r}}$ at time $t$, the displacement vector $\Delta \vec{\mathbf{r}}$ is defined as:     $\Delta \vec{\mathbf{r}} = \vec{\mathbf{r}} - \vec{\mathbf{r}}_0$     The displacement vector points directly from the initial position of the object at time $t_0$ to its final position at time $t$.

• Average Velocity: Average velocity $\vec{\mathbf{v}}$ is defined as the displacement vector divided by the elapsed time interval:     $\vec{\mathbf{v}} = \frac{\vec{\mathbf{r}} - \vec{\mathbf{r}}_0}{t - t_0} = \frac{\Delta \vec{\mathbf{r}}}{\Delta t}$

• Instantaneous Velocity: Instantaneous velocity indicates how fast an object is moving and its direction of motion at a specific instant in time. It is mathematically defined as the limit of average velocity as the time interval approaches zero:     $\vec{\mathbf{v}} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{\mathbf{r}}}{\Delta t}$

  • Direction: The instantaneous velocity vector $\vec{\mathbf{v}}$ is always tangent to the trajectory path of the object at any point.
  • Vector Components: In a two-dimensional Cartesian coordinate system, the instantaneous velocity vector $\vec{\mathbf{v}}$ decomposes into two perpendicular scalar/vector components $\vec{\mathbf{v}}_x$ and $\vec{\mathbf{v}}_y$, parallel to the x-axis and y-axis, respectively.

• Average Acceleration: Average acceleration $\vec{\mathbf{a}}$ measures the rate of change of velocity over an elapsed time interval:     $\vec{\mathbf{a}} = \frac{\vec{\mathbf{v}} - \vec{\mathbf{v}}_0}{t - t_0} = \frac{\Delta \vec{\mathbf{v}}}{\Delta t}$

Kinematic Equations for Two-Dimensional Motion

• Independence of Vector Components: Motion in two dimensions can be analyzed as two completely independent one-dimensional motions along orthogonal axes (the x-axis and y-axis). The motion along the horizontal x-axis occurs exactly as if the y-axis motion were non-existent, and vice versa.

• Horizontal (x-component) Equations of Motion:

  • $v_x = v_{0x} + a_x t$
  • $\Delta x = \frac{1}{2}(v_{0x} + v_x) t$
  • $\Delta x = v_{0x} t + \frac{1}{2} a_x t^2$
  • $v_x^2 = v_{0x}^2 + 2 a_x \Delta x$
  • Variable Definitions:
    • $v_{0x}$: initial velocity component along the x-axis
    • $v_x$: final velocity component along the x-axis
    • $a_x$: acceleration component along the x-axis
    • $\Delta x$: displacement along the x-axis ($x_f - x_i$)
    • $t$: elapsed time

• Vertical (y-component) Equations of Motion:

  • $v_y = v_{0y} + a_y t$
  • $\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$
  • $\Delta y = \frac{1}{2}(v_{0y} + v_y) t$
  • $v_y^2 = v_{0y}^2 + 2 a_y \Delta y$
  • Variable Definitions:
    • $v_{0y}$: initial velocity component along the y-axis
    • $v_y$: final velocity component along the y-axis
    • $a_y$: acceleration component along the y-axis
    • $\Delta y$: displacement along the y-axis ($y_f - y_i$)
    • $t$: elapsed time

Worked Example: Kinematics of a Moving Spacecraft

• Problem Statement: In the x-direction, a spacecraft has an initial velocity component of $+22\,\text{m/s}$ and an acceleration of $+24\,\text{m/s}^2$. In the y-direction, the initial velocity component is $+14\,\text{m/s}$ and the acceleration is $+12\,\text{m/s}^2$. Assuming initial position $(x_0, y_0) = (0, 0)\,\text{m}$, calculate the following at time $t = 7.0\,\text{s}$:

  • (a) Final x-position $x_f$ and x-velocity component $v_x$
  • (b) Final y-position $y_f$ and y-velocity component $v_y$
  • (c) Final resultant velocity $\vec{\mathbf{v}}_f$ (magnitude and direction angle $\theta$)

• Part (a) Solution (x-axis Motion):

  • Given parameters: $v_{0x} = 22\,\text{m/s}$, $a_x = 24\,\text{m/s}^2$, $t = 7.0\,\text{s}$, $x_i = 0\,\text{m}$
  • Displacement:         $\Delta x = v_{0x} t + \frac{1}{2} a_x t^2 = (22\,\text{m/s})(7.0\,\text{s}) + \frac{1}{2}(24\,\text{m/s}^2)(7.0\,\text{s})^2 = 154\,\text{m} + 588\,\text{m} = 742\,\text{m}$$x_f = 742\,\text{m}$
  • Velocity component:         $v_x = v_{0x} + a_x t = 22\,\text{m/s} + (24\,\text{m/s}^2)(7.0\,\text{s}) = 22\,\text{m/s} + 168\,\text{m/s} = 190\,\text{m/s}$

• Part (b) Solution (y-axis Motion):

  • Given parameters: $v_{0y} = 14\,\text{m/s}$, $a_y = 12\,\text{m/s}^2$, $t = 7.0\,\text{s}$, $y_i = 0\,\text{m}$
  • Displacement:         $\Delta y = v_{0y} t + \frac{1}{2} a_y t^2 = (14\,\text{m/s})(7.0\,\text{s}) + \frac{1}{2}(12\,\text{m/s}^2)(7.0\,\text{s})^2 = 98\,\text{m} + 294\,\text{m} = 392\,\text{m}$$y_f = 392\,\text{m}$
  • Velocity component:         $v_y = v_{0y} + a_y t = 14\,\text{m/s} + (12\,\text{m/s}^2)(7.0\,\text{s}) = 14\,\text{m/s} + 84\,\text{m/s} = 98\,\text{m/s}$

• Part (c) Solution (Resultant Final Velocity Vector):

  • Magnitude:         $v_f = \sqrt{v_x^2 + v_y^2} = \sqrt{(190\,\text{m/s})^2 + (98\,\text{m/s})^2} = \sqrt{36100 + 9604} = \sqrt{45704} \approx 210\,\text{m/s}$
  • Direction angle $\theta$ above the positive x-axis:         $\tan(\theta) = \frac{v_y}{v_x} \implies \theta = \tan^{-1}\left(\frac{98\,\text{m/s}}{190\,\text{m/s}}\right) = \tan^{-1}(0.5158) \approx 27^\circ$

Fundamentals of Projectile Motion

• Definition of a Projectile: A projectile is any object launched with an initial velocity that subsequently moves under the sole influence of gravitational acceleration and air resistance (or gravity alone when air resistance is neglected). Examples include a batted baseball, a thrown football, and a bullet fired from a rifle.
• Trajectory: The curved path followed by a projectile is called its trajectory, which forms a parabolic arc.

• Simultaneous Fall Demonstration:
  • Setup: One ball is dropped from rest at the same instant another ball is fired horizontally to the right.
  • Finding: Both balls hit the ground simultaneously.
  • Explanation: Vertical motion is governed exclusively by gravitational acceleration $a_y = -g = -9.80\,\text{m/s}^2$. Because both balls start with zero initial vertical velocity ($v_{0y} = 0\,\text{m/s}$) and experience identical downward gravitational acceleration, their vertical positions match at every point in time regardless of horizontal speed.

• Key Acceleration and Velocity Properties:

  • Horizontal Acceleration: $a_x = 0\,\text{m/s}^2$ (assuming negligible air resistance). Thus, horizontal velocity remains strictly constant throughout flight ($v_x = v_{0x} = \text{constant}$).
  • Vertical Acceleration: $a_y = -g = -9.80\,\text{m/s}^2$ (directed vertically downward).
  • Apex / Peak of Trajectory: At the highest point of the flight, the vertical velocity component vanishes ($v_y = 0\,\text{m/s}$), but vertical acceleration remains non-zero and equals $-9.80\,\text{m/s}^2$. The overall speed at the peak equals the horizontal velocity component ($v = v_x = v_{0x}$).
  • Symmetry of Vertical Velocity: At equal vertical heights, the magnitude of the vertical velocity component during descent equals the magnitude of the vertical velocity component during ascent, but points in the opposite direction ($v_{y,\text{descent}} = -v_{y,\text{ascent}}$).

• Conceptual Question Assessment:

  • Question: A baseball is hit upward and travels along a parabolic arc before striking the ground. Which statement is necessarily true?
    1. The velocity of the ball is a maximum when the ball is at the highest point in the arc.
    2. The x-component of the velocity of the ball is the same throughout the ball's flight. [TRUE]
    3. The acceleration of the ball decreases as the ball moves upward.
    4. The velocity of the ball is zero m/s when the ball is at the highest point in the arc.
    5. The acceleration of the ball is zero m/s² when the ball is at the highest point in the arc.

Worked Example: Time of Flight and Impact Velocity of a Falling Care Package

• Problem Statement: An airplane is flying horizontally with a constant velocity of $+115\,\text{m/s}$ at an altitude of $1050\,\text{m}$. A care package is released from the aircraft.
  • (a) Determine the time required for the care package to hit the ground.
  • (b) Determine the magnitude and direction of the final velocity of the care package just before impact.

• Part (a) Solution (Time of Flight):

  • Given parameters: $v_{0x} = 115\,\text{m/s}$, $v_{0y} = 0\,\text{m/s}$, $\Delta y = -1050\,\text{m}$, $a_y = -9.80\,\text{m/s}^2$, $a_x = 0\,\text{m/s}^2$
  • Using vertical displacement kinematic equation:         $\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$
  • Substituting $v_{0y} = 0\,\text{m/s}$:         $-1050\,\text{m} = \frac{1}{2}(-9.80\,\text{m/s}^2) t^2$$-1050\,\text{m} = -4.90\,\text{m/s}^2 \times t^2$$t^2 = \frac{-1050\,\text{m}}{-4.90\,\text{m/s}^2} \approx 214.2857\,\text{s}^2$$t = \sqrt{214.2857\,\text{s}^2} \approx 14.6\,\text{s}$

• Part (b) Solution (Impact Velocity Vector):

  • Horizontal velocity component at impact:         $v_x = v_{0x} = 115\,\text{m/s}$
  • Vertical velocity component at impact:         $v_{fy} = v_{0y} + a_y t = 0\,\text{m/s} + (-9.80\,\text{m/s}^2)(14.6\,\text{s}) \approx -143\,\text{m/s}$
  • Magnitude of final velocity:         $v_f = \sqrt{v_x^2 + v_{fy}^2} = \sqrt{(115\,\text{m/s})^2 + (-143\,\text{m/s})^2} = \sqrt{13225 + 20449} = \sqrt{33674} \approx 184\,\text{m/s}$
  • Direction angle $\theta$ below horizontal:         $\tan(\theta) = \frac{v_{fy}}{v_x} \implies \theta = \tan^{-1}\left(\frac{-143\,\text{m/s}}{115\,\text{m/s}}\right) = \tan^{-1}(-1.2435) \approx -51^\circ$

Worked Example: Symmetrical Projectile Motion (Football Kickoff)

• Problem Statement: A placekicker kicks a football at an angle of $40.0^\circ$ above the horizontal with an initial speed of $22\,\text{m/s}$. Ignoring air resistance, calculate:
  • (a) The maximum height $H$ attained by the football.
  • (b) The total time of flight $t$ between kickoff and landing.
  • (c) The horizontal range $R$ of the football.

• Resolution of Initial Velocity Vectors:

  • $v_{0x} = v_0 \cos(\theta) = (22\,\text{m/s}) \cos(40.0^\circ) \approx 16.85\,\text{m/s}$
  • $v_{0y} = v_0 \sin(\theta) = (22\,\text{m/s}) \sin(40.0^\circ) \approx 14.14\,\text{m/s}$

• Part (a) Solution (Maximum Height $H$):

  • At peak height, $v_y = 0\,\text{m/s}$ and a_y = -9.80\,\text{m/s}^2$.\n * v_y^2 = v_{0y}^2 + 2 a_y \Delta y\n * 0 = (14.14\,\text{m/s})^2 + 2(-9.80\,\text{m/s}^2) \Delta y\n * 0 = 200 - 19.6 \Delta y\n * \Delta y = \frac{200}{19.6} \approx 10.2\,\text{m}\n\n* **Part (b) Solution (Total Time of Flight t):**\n * For landing at the same elevation as launch, net vertical displacement \Delta y = 0\,\text{m}.\n * \Delta y = v_{0y} t + \frac{1}{2} a_y t^2\n * 0 = v_{0y} t + \frac{1}{2} a_y t^2 \implies t = \frac{-2 v_{0y}}{a_y} = \frac{-2(14.14\,\text{m/s})}{-9.80\,\text{m/s}^2} \approx 2.9\,\text{s}\n\n* **Part (c) Solution (Horizontal Range R):**\n * \Delta x = v_{0x} t = (16.85\,\text{m/s})(2.9\,\text{s}) \approx 48.9\,\text{m}\n\n# Worked Example: Non-Symmetrical Projectile Motion (Stone Launched from a Building)\n\n* **Problem Statement:** A stone is thrown upward from the top of a building at an angle of 30.0^\circ$to the horizontal with an initial speed of$20.0\,\text{m/s}$. The launch point is$45.0\,\text{m} above level ground.\n * (a) Calculate the time required for the stone to hit the ground.\n * (b) Calculate the speed of the stone at impact.\n * (c) Calculate the horizontal range \Delta x of the stone.\n\n\n\n* **General Method for Non-Symmetrical Trajectories:**\n * Define launch position as origin (0,0)$, setting total vertical displacement$\Delta y = y_f - y_i = -45.0\,\text{m}.\n * Resolve initial velocity into components:\n * v_{0x} = v_0 \cos(30.0^\circ) = (20.0\,\text{m/s}) \cos(30.0^\circ) \approx 17.32\,\text{m/s}\n * v_{0y} = v_0 \sin(30.0^\circ) = (20.0\,\text{m/s}) \sin(30.0^\circ) = 10.0\,\text{m/s}\n\n* **Part (a) Solution (Time of Flight t):**\n * \Delta y = v_{0y} t + \frac{1}{2} a_y t^2\n * -45.0 = 10.0 t + \frac{1}{2}(-9.80) t^2\n * -45.0 = 10.0 t - 4.90 t^2 \implies 4.90 t^2 - 10.0 t - 45.0 = 0\n * Applying quadratic formula t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:\n        t = \frac{-(-10.0) \pm \sqrt{(-10.0)^2 - 4(4.90)(-45.0)}}{2(4.90)}\n        t = \frac{10.0 \pm \sqrt{100 + 882}}{9.80} = \frac{10.0 \pm \sqrt{982}}{9.80} \approx \frac{10.0 \pm 31.34}{9.80}\n * Selecting positive root:\n        t = \frac{41.34}{9.80} \approx 4.18\,\text{s}\n\n* **Part (b) Solution (Impact Speed v_f):**\n * Vertical velocity component at ground level:\n        v_{fy} = v_{0y} + a_y t = 10.0\,\text{m/s} + (-9.80\,\text{m/s}^2)(4.18\,\text{s}) = 10.0\,\text{m/s} - 40.96\,\text{m/s} \approx -31\,\text{m/s}\n * Horizontal velocity component remains constant:\n        v_x = v_{0x} \approx 17.32\,\text{m/s}\n * Resultant impact speed:\n        v_f = \sqrt{v_x^2 + v_{fy}^2} = \sqrt{(17.32\,\text{m/s})^2 + (-31\,\text{m/s})^2} = \sqrt{300 + 961} = \sqrt{1261} \approx 35.5\,\text{m/s}\n\n* **Part (c) Solution (Horizontal Distance \Delta x):**\n * \Delta x = v_{0x} t = (17.32\,\text{m/s})(4.18\,\text{s}) \approx 72.3\,\text{m}\n\n# Worked Example: Biological Application (Froghopper Record Jump)\n\n* **Problem Statement:** The froghopper (*Philaenus spumarius*) holds the world record for insect jumps. Leaping at an angle of 58.0^\circ$above the horizontal, some reach a maximum height of$58.7\,\text{cm}$($0.587\,\text{m}) above level ground. Neglecting air resistance, determine:\n * (a) The takeoff speed v_0 required for such a leap.\n * (b) The horizontal distance \Delta x covered during this record leap.\n\n* **Part (a) Solution (Takeoff Speed v_0):**\n * Given: \theta = 58.0^\circ$, maximum height$h = 0.587\,\text{m}$,$a_y = -9.80\,\text{m/s}^2$,$v_y = 0\,\text{m/s} at peak.\n * Using vertical kinematic equation:\n        v_y^2 = v_{0y}^2 + 2 a_y \Delta y\n        0 = v_{0y}^2 + 2(-9.80\,\text{m/s}^2)(0.587\,\text{m})\n        0 = v_{0y}^2 - 11.5052\,\text{m}^2/\text{s}^2 \implies v_{0y} = \sqrt{11.5052} \approx 3.39\,\text{m/s}\n * Connecting vertical velocity component to launch speed v_0:\n        v_{0y} = v_0 \sin(58.0^\circ)\n        v_0 = \frac{v_{0y}}{\sin(58.0^\circ)} = \frac{3.39\,\text{m/s}}{\sin(58.0^\circ)} = \frac{3.39}{0.84805} \approx 4.0\,\text{m/s}\n\n* **Part (b) Solution (Horizontal Range \Delta x):**\n * Total flight duration t for level jump:\n        \Delta y = v_{0y} t + \frac{1}{2} a_y t^2 = 0 \implies t = \frac{-2 v_{0y}}{a_y} = \frac{-2(3.39\,\text{m/s})}{-9.80\,\text{m/s}^2} \approx 0.692\,\text{s}\n * Horizontal velocity component v_{0x}:\n        v_{0x} = v_0 \cos(58.0^\circ) = (4.0\,\text{m/s}) \cos(58.0^\circ) \approx 2.12\,\text{m/s}\n * Horizontal distance covered:\n        \Delta x = v_{0x} t = (2.12\,\text{m/s})(0.692\,\text{s}) \approx 1.47\,\text{m}$$