Comprehensive Two-Dimensional and Three-Dimensional Particle Kinematics

Two-Dimensional Relative Velocity: Airplane Navigation in Wind

• Galilean Relative Velocity Principle:

  • The velocity of a body relative to a reference frame is given by the vector sum of its velocity relative to an intermediate frame and the velocity of that intermediate frame relative to the primary frame.
  • For an airplane ($P$) flying through moving wind/air ($W$) over the ground ($G$):     $\vec{v}_{PG} = \vec{v}_{PW} + \vec{v}_{WG}$

• Given Parameters:

  • Plane airspeed relative to the wind: $v_{PW} = 215\,km/h$
  • Wind velocity relative to the ground: $v_{WG} = 65\,km/h$
  • Wind direction: $20^\circ$ East of North
  • Desired ground track: Due East (along the positive $x$-axis, requiring $v_{PG, y} = 0$

• Vector Decomposition:

  • Coordinate axes: $+x$ directed due East; $+y$ directed due North.
  • Ground velocity components:     $v_{PG, x} = v_{PG}$$v_{PG, y} = 0$
  • Wind velocity components (directed $20^\circ$ East of North):     $v_{WG, x} = v_{WG} \sin(20^\circ) = 65 \sin(20^\circ)$$v_{WG, y} = v_{WG} \cos(20^\circ) = 65 \cos(20^\circ)$
  • Plane-to-wind velocity components (heading angle $\theta$ South of East):     $v_{PW, x} = v_{PW} \cos(\theta) = 215 \cos(\theta)$$v_{PW, y} = -v_{PW} \sin(\theta) = -215 \sin(\theta)$

• Determination of Heading Angle $\theta$:

  • Balancing the vertical ($y$) components:     $v_{PG, y} = v_{PW, y} + v_{WG, y}$$0 = -215 \sin(\theta) + 65 \cos(20^\circ)$
  • Solving for $\sin(\theta)$$\sin(\theta) = \frac{65 \cos(20^\circ)}{215} = \frac{65 \times 0.9397}{215} \approx 0.2841$
  • Calculating the angle:     $\theta = \arcsin(0.2841) \approx 16.5^\circ$
  • The plane must be directed at an angle of $16.5^\circ$ South of East (or $\theta = -16.5^\circ$ relative to the $+x$-axis).

• Determination of Ground Speed $v_{PG}$:

  • Combining the horizontal ($x$) components:     $v_{PG} = v_{PW, x} + v_{WG, x}$$v_{PG} = 215 \cos(16.5^\circ) + 65 \sin(20^\circ)$
  • Evaluating terms:     $215 \cos(16.5^\circ) \approx 215 \times 0.9588 = 206.15\,km/h$$65 \sin(20^\circ) \approx 65 \times 0.3420 = 22.23\,km/h$$v_{PG} = 206.15 + 22.23 = 228.38\,km/h \approx 228\,km/h$

Position and Displacement in Three Dimensions

• Proton Position Vectors:

  • Initial position vector:     $\vec{r}_i = 5.0\hat{i} - 6.0\hat{j} + 2.0\hat{k}\,\text{m}$
  • Final position vector:     $\vec{r}_f = -2.0\hat{i} + 6.0\hat{j} + 2.0\hat{k}\,\text{m}$

• Displacement Vector Calculation:

  • Displacement definition:     $\Delta\vec{r} = \vec{r}_f - \vec{r}_i$
  • Evaluating individual cartesian components:     $\Delta x = -2.0\,\text{m} - 5.0\,\text{m} = -7.0\,\text{m}$$\Delta y = 6.0\,\text{m} - (-6.0\,\text{m}) = 12.0\,\text{m}$$\Delta z = 2.0\,\text{m} - 2.0\,\text{m} = 0.0\,\text{m}$
  • Result in unit-vector notation:     $\Delta\vec{r} = -7.0\hat{i} + 12.0\hat{j} + 0.0\hat{k}\,\text{m} = -7.0\hat{i} + 12.0\hat{j}\,\text{m}$

• Plane Parallel to Displacement:

  • Because the displacement along the $z$-axis is zero ($\Delta z = 0$), the displacement has no perpendicular component to the $xy$-plane.
  • Consequently, the displacement vector is parallel to the $xy$-plane.

Parametric Harmonic Motion and Circular Trajectories

• Parametric Position Vector:

  • Position as a function of time:     $\vec{r}(t) = 4\cos\left(\frac{\pi t}{T}\right)\hat{i} - 4\sin\left(\frac{\pi t}{T}\right)\hat{j}$
  • Baseline position at $t = 0$:     $\vec{r}(0) = 4\cos(0)\hat{i} - 4\sin(0)\hat{j} = 4\hat{i}$

• Displacements and Distances from Origin at Specific Times:

  • Case 1: $t = \frac{T}{3}$
  • Position:       $\vec{r}\left(\frac{T}{3}\right) = 4\cos\left(\frac{\pi}{3}\right)\hat{i} - 4\sin\left(\frac{\pi}{3}\right)\hat{j} = 4\left(\frac{1}{2}\right)\hat{i} - 4\left(\frac{\sqrt{3}}{2}\right)\hat{j} = 2\hat{i} - 2\sqrt{3}\hat{j}$
  • Displacement vector:       $\Delta\vec{r} = \vec{r}\left(\frac{T}{3}\right) - \vec{r}(0) = (2 - 4)\hat{i} - 2\sqrt{3}\hat{j} = -2\hat{i} - 2\sqrt{3}\hat{j}$
  • Magnitude of displacement:       $|\Delta\vec{r}| = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$
  • Distance to the origin:       $|\vec{r}(T/3)| = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$
  • Case 2: $t = \frac{T}{2}$
  • Position:       $\vec{r}\left(\frac{T}{2}\right) = 4\cos\left(\frac{\pi}{2}\right)\hat{i} - 4\sin\left(\frac{\pi}{2}\right)\hat{j} = 0\hat{i} - 4\hat{j} = -4\hat{j}$
  • Displacement vector:       $\Delta\vec{r} = \vec{r}\left(\frac{T}{2}\right) - \vec{r}(0) = -4\hat{i} - 4\hat{j}$
  • Magnitude of displacement:       $|\Delta\vec{r}| = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2} \approx 5.66$
  • Distance to the origin:       $|\vec{r}(T/2)| = \sqrt{0^2 + (-4)^2} = 4$
  • Case 3: $t = 2T$
  • Position:       $\vec{r}(2T) = 4\cos(2\pi)\hat{i} - 4\sin(2\pi)\hat{j} = 4\hat{i} - 0\hat{j} = 4\hat{i}$
  • Displacement vector:       $\Delta\vec{r} = \vec{r}(2T) - \vec{r}(0) = 4\hat{i} - 4\hat{i} = \vec{0} = 0\hat{i} + 0\hat{j}$
  • Magnitude of displacement:       $|\Delta\vec{r}| = 0$
  • Distance to the origin:       $|\vec{r}(2T)| = \sqrt{4^2 + 0^2} = 4$
  • General Distance to Origin for all $t$:     $r(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{16\cos^2\left(\frac{\pi t}{T}\right) + 16\sin^2\left(\frac{\pi t}{T}\right)} = \sqrt{16} = 4$
  • The particle undergoes uniform circular motion of constant radius $R = 4$ around the origin.

• Angle with the Positive $x$-Axis for Arbitrary $t$:

  • Tangent of the directional angle:     $\tan(\theta) = \frac{y(t)}{x(t)} = \frac{-4\sin\left(\frac{\pi t}{T}\right)}{4\cos\left(\frac{\pi t}{T}\right)} = -\tan\left(\frac{\pi t}{T}\right)$
  • Evaluating angle:     $\theta(t) = -\frac{\pi t}{T}$
  • Measured in radians clockwise from the positive $x$-axis, or as standard counterclockwise angle $2\pi - \frac{\pi t}{T}\,\text{rad}$.

Differential Kinematics in Two and Three Dimensions

• Kinematics of an Electron:

  • Given position vector ($t$ in seconds, $\vec{r}$ in meters):     $\vec{r}(t) = 3.00t\hat{i} - 4.00t^2\hat{j} + 2.00\hat{k}$
  • Velocity Function by Differentiation:     $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.00t)\hat{i} - \frac{d}{dt}(4.00t^2)\hat{j} + \frac{d}{dt}(2.00)\hat{k}$$\vec{v}(t) = 3.00\hat{i} - 8.00t\hat{j} + 0\hat{k}\,\text{m/s}$
  • Unit-Vector Notation at $t = 2.00\,\text{s}$:     $\vec{v}(2.00) = 3.00\hat{i} - 8.00(2.00)\hat{j} = 3.00\hat{i} - 16.0\hat{j}\,\text{m/s}$
  • Velocity Magnitude at $t = 2.00\,\text{s}$:     $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.00)^2 + (-16.0)^2} = \sqrt{9.00 + 256.0} = \sqrt{265.0} \approx 16.3\,\text{m/s}$
  • Orientation Angle Relative to Positive $x$-Axis:     $\theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{-16.0}{3.00}\right) \approx \arctan(-5.333) \approx -79.4^\circ$
  • Corresponds to $79.4^\circ$ below the positive $x$-axis (or $280.6^\circ$ standard counterclockwise).

• Two-Dimensional Kinematics of Polynomial Position:

  • Given position:     $\vec{r}(t) = (2t^3 - 5t)\hat{i} + (6 - 7t^4)\hat{j}$
  • Position at $t = 2.0\,\text{s}$:     $x(2.0) = 2(2.0)^3 - 5(2.0) = 16 - 10 = 6\,\text{m}$$y(2.0) = 6 - 7(2.0)^4 = 6 - 7(16) = 6 - 112 = -106\,\text{m}$$\vec{r}(2.0) = 6\hat{i} - 106\hat{j}\,\text{m}$
  • Velocity at $t = 2.0\,\text{s}$:     $v_x(t) = \frac{d}{dt}(2t^3 - 5t) = 6t^2 - 5$$v_y(t) = \frac{d}{dt}(6 - 7t^4) = -28t^3$$v_x(2.0) = 6(2.0)^2 - 5 = 24 - 5 = 19\,\text{m/s}$$v_y(2.0) = -28(2.0)^3 = -28(8) = -224\,\text{m/s}$$\vec{v}(2.0) = 19\hat{i} - 224\hat{j}\,\text{m/s}$
  • Acceleration at $t = 2.0\,\text{s}$:     $a_x(t) = \frac{d}{dt}(6t^2 - 5) = 12t$$a_y(t) = \frac{d}{dt}(-28t^3) = -84t^2$$a_x(2.0) = 12(2.0) = 24\,\text{m/s}^2$$a_y(2.0) = -84(2.0)^2 = -84(4) = -336\,\text{m/s}^2$$\vec{a}(2.0) = 24\hat{i} - 336\hat{j}\,\text{m/s}^2$
  • Orientation of Path Tangent at $t = 2.0\,\text{s}$:
  • The tangent vector is aligned with the velocity vector $\vec{v}(2.0)$:       $\theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{-224}{19}\right) \approx \arctan(-11.789) \approx -85.2^\circ$
  • Angle is $-85.2^\circ$ relative to the positive $x$-axis (or $274.8^\circ$ counterclockwise).

Integral Kinematics: Reconstructing Position from Rates of Change

• Motion Governed by Time-Varying Velocity:

  • Velocity function:     $\vec{v}(t) = (4 + 2t^2)\hat{i} + (-10 - 4t)\hat{j}$
  • Boundary condition:     $\vec{r}(0) = 0\hat{i} + 0\hat{j}$
  • Velocity at $t = 3\,\text{s}$:     $v_x(3) = 4 + 2(3)^2 = 4 + 18 = 22\,\text{m/s}$$v_y(3) = -10 - 4(3) = -10 - 12 = -22\,\text{m/s}$$\vec{v}(3) = 22\hat{i} - 22\hat{j}\,\text{m/s}$
  • Integration for Position:     $x(t) = \int_0^t (4 + 2\tau^2)\,d\tau = 4t + \frac{2}{3}t^3$$y(t) = \int_0^t (-10 - 4\tau)\,d\tau = -10t - 2t^2$
  • Position at $t = 3\,\text{s}$:     $x(3) = 4(3) + \frac{2}{3}(27) = 12 + 18 = 30\,\text{m}$$y(3) = -10(3) - 2(9) = -30 - 18 = -48\,\text{m}$$\vec{r}(3) = 30\hat{i} - 48\hat{j}\,\text{m}$
  • Moving Direction with Respect to the $x$-Axis:     $\theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{-22}{22}\right) = \arctan(-1) = -45^\circ$
  • Directed at $45^\circ$ below the positive $x$-axis (or $315^\circ$).

• Motion Governed by Time-Varying Acceleration:

  • Given acceleration and initial states:     $\vec{a}(t) = (2 + 4t^3)\hat{i} + (6 - 3t^2)\hat{j}$$\vec{v}(0) = 10\hat{i}\,\text{m/s}$$\vec{r}(0) = 0\hat{i} + 0\hat{j}\,\text{m}$
  • Velocity Vector by Integration:     $v_x(t) = v_{0x} + \int_0^t (2 + 4\tau^3)\,d\tau = 10 + 2t + t^4$$v_y(t) = v_{0y} + \int_0^t (6 - 3\tau^2)\,d\tau = 0 + 6t - t^3$$\vec{v}(t) = (10 + 2t + t^4)\hat{i} + (6t - t^3)\hat{j}$
  • Velocity at $t = 2\,\text{s}$:     $v_x(2) = 10 + 2(2) + (2)^4 = 10 + 4 + 16 = 30\,\text{m/s}$$v_y(2) = 6(2) - (2)^3 = 12 - 8 = 4\,\text{m/s}$$\vec{v}(2) = 30\hat{i} + 4\hat{j}\,\text{m/s}$
  • Position Vector by Integration:     $x(t) = x_0 + \int_0^t (10 + 2\tau + \tau^4)\,d\tau = 10t + t^2 + \frac{t^5}{5}$$y(t) = y_0 + \int_0^t (6\tau - \tau^3)\,d\tau = 3t^2 - \frac{t^4}{4}$$\vec{r}(t) = \left(10t + t^2 + \frac{t^5}{5}\right)\hat{i} + \left(3t^2 - \frac{t^4}{4}\right)\hat{j}$
  • Position at $t = 2\,\text{s}$:     $x(2) = 10(2) + (2)^2 + \frac{(2)^5}{5} = 20 + 4 + 6.4 = 30.4\,\text{m}$$y(2) = 3(2)^2 - \frac{(2)^4}{4} = 12 - 4 = 8.0\,\text{m}$$\vec{r}(2) = 30.4\hat{i} + 8.0\hat{j}\,\text{m}$

Projectile Acceleration Components and Launch Optimization

• Tangential and Radial Acceleration for a Horizontal Projectile:

  • Physical conditions:
  • Initial velocity: $\vec{v}(0) = v_0\hat{i}$
  • Constant gravitational acceleration: $\vec{a} = -g\hat{j}$
  • Air resistance: neglected
  • Velocity Components:     $v_x(t) = v_0$$v_y(t) = -gt$$v(t) = \sqrt{v_0^2 + g^2t^2}$
  • Derivation of Tangential Acceleration ($a_t$):
  • Rate of change of scalar speed:       $a_t = \frac{dv}{dt} = \frac{d}{dt}\left(\sqrt{v_0^2 + g^2t^2}\right) = \frac{2g^2t}{2\sqrt{v_0^2 + g^2t^2}} = \frac{g^2 t}{\sqrt{v_0^2 + g^2t^2}}$
  • Equivalent vector projection:       $a_t = \frac{\vec{a} \cdot \vec{v}}{v} = \frac{(0)(v_0) + (-g)(-gt)}{\sqrt{v_0^2 + g^2t^2}} = \frac{g^2 t}{\sqrt{v_0^2 + g^2t^2}}$
  • Derivation of Radial (Normal) Acceleration ($a_r$):
  • Orthogonal acceleration relationship:       $a^2 = a_t^2 + a_r^2 = g^2$
  • Solving for radial component:       $a_r = \sqrt{g^2 - a_t^2} = \sqrt{g^2 - \frac{g^4 t^2}{v_0^2 + g^2t^2}} = \sqrt{\frac{g^2 v_0^2 + g^4 t^2 - g^4 t^2}{v_0^2 + g^2t^2}} = \frac{g v_0}{\sqrt{v_0^2 + g^2t^2}}$
  • Boundary Limits:
  • At launch ($t = 0$): $a_t = 0$, $a_r = g$
  • At asymptotic limit ($t \rightarrow \infty$): $a_t \rightarrow g$, $a_r \rightarrow 0$

• Launch Angle for Range Equal to Twice Maximum Height:

  • Target condition:     $R = 2H$
  • Analytic expressions:
  • Maximum height: $H = \frac{v_0^2 \sin^2(\theta)}{2g}$
  • Range on flat ground: $R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{2v_0^2 \sin(\theta)\cos(\theta)}{g}$
  • Algebraic reduction:     $\frac{2v_0^2 \sin(\theta)\cos(\theta)}{g} = 2\left(\frac{v_0^2 \sin^2(\theta)}{2g}\right)$$2\sin(\theta)\cos(\theta) = \sin^2(\theta)$
  • Dividing by $\sin(\theta)\cos(\theta)$ ($\sin(\theta) \neq 0$, $\cos(\theta) \neq 0$):     $\tan(\theta) = 2$
  • Angle calculation:     $\theta = \arctan(2) \approx 63.4^\circ$

Circular Kinematics: Uniform and Non-Uniform Motion

• Revolution Period of a Merry-Go-Round:

  • Given values:
  • Distance from axis (radius): $r = 7.5\,\text{m}$
  • Centripetal acceleration: $a_c = 3.3\,\text{m/s}^2$
  • Formula relating centripetal acceleration and period $T$:     $a_c = \omega^2 r = \left(\frac{2\pi}{T}\right)^2 r = \frac{4\pi^2 r}{T^2}$
  • Solving for period $T$:     $T = 2\pi \sqrt{\frac{r}{a_c}} = 2\pi \sqrt{\frac{7.5}{3.3}} \approx 2\pi \sqrt{2.2727} \approx 2\pi \times 1.5076 \approx 9.47\,\text{s}$
  • Equivalent linear speed check:     $v = \sqrt{a_c r} = \sqrt{3.3 \times 7.5} \approx 4.975\,\text{m/s}$$T = \frac{2\pi r}{v} = \frac{2\pi (7.5)}{4.975} \approx 9.47\,\text{s}$

• Acceleration of a Decelerating Vehicle on a Circular Track:

  • Given values:
  • Track radius: $R = 400\,\text{m}$
  • Instantaneous speed: $v = 10\,\text{m/s}$
  • Tangential acceleration: $a_t = -0.2\,\text{m/s}^2$
  • Radial (Normal) Acceleration:     $a_r = \frac{v^2}{R} = \frac{(10)^2}{400} = \frac{100}{400} = 0.25\,\text{m/s}^2$
  • Total Acceleration Magnitude:
  • By orthogonality of radial and tangential components:       $a = \sqrt{a_r^2 + a_t^2} = \sqrt{(0.25)^2 + (-0.2)^2} = \sqrt{0.0625 + 0.0400} = \sqrt{0.1025} \approx 0.320\,\text{m/s}^2$
  • Direction Relative to Inward Radial Direction:     $\phi = \arctan\left(\frac{|a_t|}{a_r}\right) = \arctan\left(\frac{0.2}{0.25}\right) = \arctan(0.8) \approx 38.7^\circ$
  • Total acceleration points inwards and backwards at an angle of $38.7^\circ$ behind the center-pointing normal.

Relative Motion: Observer Reference Frames and Windblown Rain

• Motion Decomposition and Frame Notation:

  • Subscript notation:
  • $R$: Raindrops
  • $G$: Stationary Ground observer
  • $T$: Moving Train observer
  • Vector addition equation:     $\vec{v}_{RG} = \vec{v}_{RT} + \vec{v}_{TG}$

• Kinematic Observations and Coordinates:

  • Let horizontal South be $+x$, and vertical downward be $+y$.
  • Train velocity relative to ground:     $\vec{v}_{TG} = 30\hat{i}\,\text{m/s}$
  • Rain path observed on train:
  • Drops fall perfectly vertically, implying no horizontal component relative to the train:       $v_{RT, x} = 0\,\text{m/s}$$v_{RT, y} = v_y$
  • Rain path observed from ground:
  • Vector sum:       $v_{RG, x} = v_{RT, x} + v_{TG, x} = 0 + 30\,\text{m/s} = 30\,\text{m/s}$$v_{RG, y} = v_y$
  • Angle with vertical: $\theta = 70^\circ$

• Geometric Relationship and Speed Determination:

  • Right triangle formed by components:     $\sin(70^\circ) = \frac{v_{RG, x}}{v_{RG}} = \frac{30}{v_{RG}}$
  • Solving directly for the raindrop speed relative to the ground ($v_{RG}$):     $v_{RG} = \frac{30}{\sin(70^\circ)} = \frac{30}{0.9397} \approx 31.9\,\text{m/s}$
  • Vertical Speed Component ($v_{RG, y}$):     $v_{RG, y} = \frac{v_{RG, x}}{\tan(70^\circ)} = \frac{30}{\tan(70^\circ)} = \frac{30}{2.7475} \approx 10.9\,\text{m/s}$
  • Verification via Pythagorean Theorem:     $v_{RG} = \sqrt{(30)^2 + (10.92)^2} = \sqrt{900 + 119.2} = \sqrt{1019.2} \approx 31.9\,\text{m/s}$