Motion Vectors - Statics and Dynamics

1. At the instant of maximum height, which of the following statements about the vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ is necessarily true for all valid launch angles $0 < \theta < 90°$?
   Answer: The velocity vector $\mathbf{v}$ is perpendicular to the acceleration vector $\mathbf{a}$, and the position vector $\mathbf{r}$ is not necessarily perpendicular to $\mathbf{v}$ unless a specific relationship between $v_0$, $\theta$, and $g$ is satisfied. When analyzing projectile motion vectors, always start by writing out what you know at each key instant — don't rely on intuition alone. At maximum height, the vertical velocity component is zero, so $\mathbf{v} = v_x\hat{i} = v_0\cos\theta\,\hat{i}$, pointing purely horizontal. Since gravity gives $\mathbf{a} = -g\hat{j}$, pointing purely downward, $\mathbf{v} \cdot \mathbf{a} = 0$ — they are always perpendicular at this instant. That part holds for every valid launch angle. Now check the position vector. At maximum height, $\mathbf{r} = \left(\frac{v_0^2 \sin\theta\cos\theta}{g}\right)\hat{i} + \left(\frac{v_0^2\sin^2\theta}{2g}\right)\hat{j}$. This vector has both horizontal and vertical components, so it generally points diagonally — not horizontally like $\mathbf{v}$. For $\mathbf{r} \perp \mathbf{v}$, you'd need $\mathbf{r} \cdot \mathbf{v} = 0$, which requires $\frac{v_0^2\sin\theta\cos^2\theta}{g} = 0$ — only satisfied under specific conditions, not universally. This confirms answer A is correct. Choice B fails because it claims $\mathbf{r} \perp \mathbf{v}$ for all launch angles — the math above disproves this. Choice C is wrong because $\mathbf{r}$ retains a vertical component (the height!) even when $\mathbf{v}$ becomes horizontal; "both having zero vertical components simultaneously" is false for $\mathbf{r}$. Choice D is wrong because the highest point is not directly above the launch origin — the projectile has traveled horizontally — so $\mathbf{a}$ and $\mathbf{r}$ are never parallel. Your study tip: when a question says "necessarily true for all angles," test extreme cases (like $\theta = 45°$ vs. $\theta = 80°$) to quickly eliminate answers that only work sometimes.
2. In frame S′, what is the particle's trajectory equation $y′$ as a function of $x′$, and what type of motion does observer S′ measure?
   Answer: In S′, the particle undergoes purely vertical free-fall: $x′ = 0$ for all $t$, and $y′ = 10 - 4.9t^2$, so the trajectory is a vertical line at $x′ = 0$. Observer S′ sees only vertical acceleration with no horizontal motion, consistent with a particle thrown straight up from S′'s origin. When you see a question involving two reference frames with a constant relative velocity, your first instinct should be to apply the Galilean transformation: $\mathbf{r}' = \mathbf{r}_S - \mathbf{V}t$, where $\mathbf{V}$ is the velocity of frame S′ relative to S. Here, the particle in S has $x_S = 5t$ and $y_S = 10 - 4.9t^2$. Since $\mathbf{V} = 5\hat{i}$ m/s, the transformed coordinates are $x' = x_S - Vt = 5t - 5t = 0$ and $y' = y_S = 10 - 4.9t^2$. The horizontal components exactly cancel, meaning observer S′ sees the particle sitting directly overhead at $x' = 0$ the entire time, rising and falling vertically — pure free-fall. Answer C correctly captures this: the trajectory is a vertical line at $x' = 0$, and the physics is equivalent to a ball thrown straight up. Answer A incorrectly adds the frame velocity to the particle's horizontal component rather than subtracting it, producing $x' = 10t$ — a sign and concept error in the Galilean transformation. Answer B arrives at $x' = 0$ correctly but then contradicts itself by claiming the trajectory is still a parabola in $x'$-$y'$ space; if $x' = 0$ always, you cannot form a parabolic relationship between $x'$ and $y'$. Answer D introduces a Coriolis effect, which only appears in rotating (non-inertial) frames — not in frames moving at constant velocity, which remain inertial. Your key takeaway: in Galilean relativity between inertial frames, subtract $\mathbf{V}t$ from position component-by-component and remember that accelerations are unchanged — only velocities and positions transform.
3. Which of the following correctly characterizes the relationship between the speed of the drone and the magnitude of its acceleration at any time $t$?
   Answer: The speed is constant at $\sqrt{R^2\omega^2 + c^2}$ m/s and the acceleration magnitude is constant at $R\omega^2$ m/s², confirming that the acceleration is entirely centripetal (directed toward the helix axis) with no tangential component. When a position vector traces a helix, your first move should always be to differentiate to find velocity and acceleration, then compute their magnitudes directly — don't let the 3D geometry intimidate you. Differentiating $\mathbf{r}(t) = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j} + ct\hat{k}$ gives velocity $\mathbf{v}(t) = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j} + c\hat{k}$. The speed is $|\mathbf{v}| = \sqrt{R^2\omega^2\sin^2(\omega t) + R^2\omega^2\cos^2(\omega t) + c^2} = \sqrt{R^2\omega^2 + c^2}$, which is constant because $\sin^2 + \cos^2 = 1$. Differentiating again gives $\mathbf{a}(t) = -R\omega^2\cos(\omega t)\hat{i} - R\omega^2\sin(\omega t)\hat{j}$, with magnitude $|\mathbf{a}| = R\omega^2$ — purely radial (centripetal), pointing toward the helix axis, with no vertical component at all. This confirms answer A. Answer B is wrong on two counts: it ignores the vertical component $c$ when computing speed (the full Pythagorean sum is required), and the acceleration magnitude doesn't vary with time — it's constant. Answer C incorrectly claims speed increases with time. Constant component magnitudes in velocity produce constant speed. It also invents a $\sqrt{R^2\omega^4 + c^2}$ acceleration by wrongly adding a $c^2$ term — differentiating the constant $c\hat{k}$ term yields zero, not $c$. Answer D fabricates a $\sqrt{R^2\omega^4 + c^2\omega^2}$ acceleration. Since $c$ is constant, its derivative vanishes — there is no vertical acceleration contribution. Your study tip: on helical motion problems, always differentiate fully before drawing conclusions. A constant vertical velocity contributes to speed but produces zero acceleration — a distinction many distractors exploit.
4. At this instant, what is the magnitude of the particle's acceleration vector?
   Answer: $|\mathbf{a}| = \sqrt{(\ddot{x})^2 + (\ddot{y})^2} \approx 6.58$ m/s², where $\ddot{y} = \frac{\dot{x}^2 + x\ddot{x}}{2} = \frac{9 + 4}{2} = 6.5$ m/s², obtained by differentiating the velocity constraint $\dot{y} = x\dot{x}/2$ with respect to time using the product rule. When a particle is constrained to a curved path, finding its acceleration requires careful application of the chain rule and product rule — not just geometric intuition. The key is differentiating the position constraint twice with respect to time to extract the true Cartesian acceleration components. Starting with $y = x^2/4$, differentiate once to get the velocity constraint: $\dot{y} = \frac{2x\dot{x}}{4} = \frac{x\dot{x}}{2}$. Now differentiate again using the product rule on $x\dot{x}$: $\ddot{y} = \frac{\dot{x}\cdot\dot{x} + x\cdot\ddot{x}}{2} = \frac{\dot{x}^2 + x\ddot{x}}{2}$. Substituting $x = 4$, $\dot{x} = 3$, $\ddot{x} = 1$: $\ddot{y} = \frac{9 + 4}{2} = 6.5$ m/s². The magnitude is then $|\mathbf{a}| = \sqrt{(1)^2 + (6.5)^2} \approx 6.58$ m/s², confirming B is correct. A drops the $\dot{x}^2/2$ term when differentiating $x\dot{x}$ — a classic product rule error. It treats $\dot{x}$ as a constant during differentiation, yielding only $x\ddot{x}/2 = 2$ instead of the full 6.5. C makes the opposite mistake: it retains only the $\dot{x}^2/2$ term and ignores the $x\ddot{x}/2$ contribution, essentially forgetting that $x$ itself is changing. D conflates the slope of the path ($dy/dx$) with the time derivative of the constraint, incorrectly treating the geometric slope as if it directly gives the acceleration relationship. Study tip: Whenever a constraint links two coordinates, always differentiate with respect to time using the full product rule — never substitute path-slope formulas directly for time derivatives.
5. Which of the following correctly describes the relationship between the position vector $\mathbf{r}$, velocity vector $\mathbf{v}$, and acceleration vector $\mathbf{a}$ for all times $t$?
   Answer: $\mathbf{v}$ is always perpendicular to $\mathbf{r}$, $\mathbf{a}$ is always antiparallel to $\mathbf{r}$ (directed opposite to $\mathbf{r}$), and the speed $|\mathbf{v}| = Ab$ is constant, confirming uniform circular motion with centripetal acceleration of magnitude $Ab^2$. When you see a position vector defined as sinusoidal components, your first instinct should be to differentiate to find velocity and acceleration, then test key geometric relationships using the dot product. Starting with $\mathbf{r}(t) = A\sin(bt)\hat{i} + A\cos(bt)\hat{j}$, differentiate once to get $\mathbf{v}(t) = Ab\cos(bt)\hat{i} - Ab\sin(bt)\hat{j}$, and again to get $\mathbf{a}(t) = -Ab^2\sin(bt)\hat{i} - Ab^2\cos(bt)\hat{j} = -b^2\mathbf{r}$. Notice that last step: acceleration equals $-b^2$ times the position vector, meaning a always points opposite to r — that's the definition of antiparallel. Now check the dot product $\mathbf{r} \cdot \mathbf{v} = A\sin(bt)(Ab\cos(bt)) + A\cos(bt)(-Ab\sin(bt)) = 0$ for all $t$, confirming v is always perpendicular to r. The speed is $|\mathbf{v}| = Ab$ (constant), and $|\mathbf{r}| = A$ (constant), confirming uniform circular motion. This makes D correct. Choice A fails because the speed is constant — not sinusoidally varying — and the particle traces a perfect circle, not non-uniform circular motion. Choice B incorrectly claims v is parallel to r; the dot product proves they are perpendicular, not parallel. It also misidentifies the path as a spiral. Choice C gets the r–v perpendicularity right but then claims a is parallel (same direction) to v — in reality, a is antiparallel to r, not aligned with v at all. A reliable shortcut: if you spot $\mathbf{a} = -b^2\mathbf{r}$, you immediately know the motion is uniform circular — the negative sign signals centripetal (inward) acceleration, and the constant prefactor guarantees constant speed.

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