Statics and Dynamics Quiz: Motion Vectors

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Motion VectorsQuestion 1 of 5

An observer in reference frame S measures a particle's position as $\mathbf{r}_S(t) = (5t)\hat{i} + (10 - 4.9t^2)\hat{j}$ m. A second observer in frame S′ moves with constant velocity $\mathbf{V} = 5\hat{i}$ m/s relative to frame S (both frames coincide at $t = 0$ ).

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?

In S′, the particle has position

x′ = 10t

m (the observer's velocity adds to the horizontal component), and

y′ = 10 - 4.9t^2

, giving a wider parabola

y′ = 10 - 4.9(x′)^2/100

, because relative velocity between frames is subtracted from the particle's velocity, not added.

In S′, the trajectory is

y′ = 10 - 4.9(x′)^2/25

, which is a downward-opening parabola in

x′

y′

space, because the horizontal position in S′ is

x′ = 5t - 5t = 0

and the observer still measures a parabolic path due to the relative acceleration between the frames.

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.

In S′, the acceleration vector changes because the non-inertial transformation modifies the gravitational component, giving a trajectory that is neither a straight line nor a parabola but a cycloid, since the constant frame velocity introduces a Coriolis-type coupling in the

\hat{j}

direction.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Motion Vectors

Practice Motion Vectors in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Motion Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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?

In S′, the particle has position $x′ = 10t$ m (the observer's velocity adds to the horizontal component), and $y′ = 10 - 4.9t^2$ , giving a wider parabola $y′ = 10 - 4.9(x′)^2/100$ , because relative velocity between frames is subtracted from the particle's velocity, not added.
In S′, the trajectory is $y′ = 10 - 4.9(x′)^2/25$ , which is a downward-opening parabola in $x′$ - $y′$ space, because the horizontal position in S′ is $x′ = 5t - 5t = 0$ and the observer still measures a parabolic path due to the relative acceleration between the frames.
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. (correct answer)
In S′, the acceleration vector changes because the non-inertial transformation modifies the gravitational component, giving a trajectory that is neither a straight line nor a parabola but a cycloid, since the constant frame velocity introduces a Coriolis-type coupling in the $\hat{j}$ direction.

Explanation: 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.

Question 2

A drone flies such that its position vector (in meters) relative to a fixed origin is $\mathbf{r}(t) = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j} + ct\hat{k}$ , where $R = 5$ m, $\omega = 2$ rad/s, and $c = 3$ m/s. This describes a helical (corkscrew) path.

Which of the following correctly characterizes the relationship between the speed of the drone and the magnitude of its acceleration at any time $t$ ?

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. (correct answer)
The speed is constant at $R\omega$ m/s because the vertical component $c$ does not contribute to tangential speed, and the acceleration magnitude varies as $R\omega^2\cos(\omega t)$ m/s² due to the changing orientation of the centripetal vector.
The speed increases linearly with time because the helical geometry continuously redirects velocity, and the acceleration magnitude is constant at $\sqrt{R^2\omega^4 + c^2}$ m/s² since both circular and vertical accelerations are present.
The speed is constant at $\sqrt{R^2\omega^2 + c^2}$ m/s, and the acceleration magnitude is constant at $\sqrt{R^2\omega^4 + c^2\omega^2}$ m/s², indicating that the vertical velocity component contributes a time-varying acceleration alongside the centripetal term.

Explanation: 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.

Question 3

A particle's position in two dimensions is described by $\mathbf{r}(t) = A\sin(bt)\hat{i} + A\cos(bt)\hat{j}$ where $A > 0$ and $b > 0$ are constants.

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 $t$ ?

$\mathbf{a}$ is always perpendicular to $\mathbf{v}$ and parallel to $\mathbf{r}$ , the speed $|\mathbf{v}|$ varies sinusoidally with time, and $|\mathbf{r}|$ is constant only when $\sin(bt) = \cos(bt)$ , making this an example of non-uniform circular motion.
$\mathbf{v}$ is always parallel to $\mathbf{r}$ because both vectors rotate at rate $b$ , $\mathbf{a}$ is always perpendicular to $\mathbf{r}$ , and the speed $|\mathbf{v}| = Ab$ is constant, which is characteristic of spiral motion with constant angular speed.
$\mathbf{v}$ is always perpendicular to $\mathbf{r}$ , $\mathbf{a}$ is always parallel to $\mathbf{v}$ (in the same direction as $\mathbf{v}$ ), and the magnitude $|\mathbf{r}| = A$ is constant, so the particle moves on a circle with tangential acceleration $Ab^2$ .
$\mathbf{v}$ 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 $Ab^2$ . (correct answer)

Explanation: 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.

Question 4

A projectile is launched from the origin with initial speed $v_0$ at angle $\theta$ above the horizontal. Neglect air resistance. Consider the position vector $\mathbf{r}(t)$ , velocity vector $\mathbf{v}(t)$ , and acceleration vector $\mathbf{a}$ (constant) during flight.

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 $0 < \theta < 90°$ ?

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. (correct answer)
The velocity vector $\mathbf{v}$ is perpendicular to the acceleration vector $\mathbf{a}$ , and the position vector $\mathbf{r}$ is also perpendicular to $\mathbf{v}$ for all valid launch angles because the position at max height is always directed away from the trajectory's tangent.
The position vector $\mathbf{r}$ is parallel to the velocity vector $\mathbf{v}$ at maximum height because both vectors lie in the horizontal direction at that instant, with the vertical components of both being zero simultaneously.
The acceleration vector $\mathbf{a}$ is parallel to the position vector $\mathbf{r}$ at maximum height because gravity acts downward while the highest point of the trajectory is directly above the launch point, aligning the two vectors vertically.

Explanation: 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.

Question 5

A particle moves along a path defined in Cartesian coordinates. The path is constrained such that $y = x^2/4$ (a parabola). At a particular instant, the $x$ -coordinate is $x = 4$ m and is increasing at $\dot{x} = 3$ m/s. The rate $\dot{x}$ is itself increasing at $\ddot{x} = 1$ m/s².

At this instant, what is the magnitude of the particle's acceleration vector?

$|\mathbf{a}| = \sqrt{(\ddot{x})^2 + (\ddot{y})^2} = \sqrt{1 + (2)^2} = \sqrt{5} \approx 2.24$ m/s², where $\ddot{y} = x\ddot{x}/2 = 2$ m/s², obtained by differentiating $\dot{y} = x\dot{x}/2$ but omitting the $\dot{x}^2/2$ term that arises from differentiating the factor $\dot{x}$ via the product rule.
$|\mathbf{a}| = \sqrt{(\ddot{x})^2 + (\ddot{y})^2} \approx 6.58$ 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. (correct answer)
$|\mathbf{a}| = \sqrt{(\ddot{x})^2 + (\ddot{y})^2} = \sqrt{1 + (4.5)^2} = \sqrt{21.25} \approx 4.61$ m/s², where $\ddot{y} = \dot{x}^2/2 = 4.5$ m/s², obtained by differentiating $y = x^2/4$ twice but retaining only the $\dot{x}^2$ term and omitting the $x\ddot{x}/2$ contribution.
$|\mathbf{a}| = \sqrt{(\ddot{x})^2 + (\ddot{y})^2} = \sqrt{1 + (3)^2} = \sqrt{10} \approx 3.16$ m/s², where $\ddot{y} = (dy/dx)\cdot\ddot{x} = (x/2)\cdot\ddot{x} = 2\cdot 1 = 2$ m/s² is incorrectly doubled by also including the path slope evaluated at the initial position, conflating path geometry with the time-derivative of the constraint.

Explanation: 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.