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USAAAO 2026, I кръг — Задача 1

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Всички материали от USAAAO 2026

USAAAO 2026 - First Round · 9 февруари 2026 г.

Problem writers

Problem writers: Srihari (Hari) Balaji, Vincent Bian, Ferdinand, Hagan Hensley, Elizabeth Lee, Joe McCarty, Lucas Pinheiro, Feodor Yevtushenko

Условие

A planet moves in a circular orbit in a fixed plane with angular velocity $\omega$ , and let $\hat{n}$ be a unit vector perpendicular to the orbital plane. Let the planet’s velocity vector as a function of time be given by

\vec{v}(t) = v_0\big(\cos(\omega t)\,\hat{\imath} + \sin(\omega t)\,\hat{\jmath}\big)

Define $\vec{a}(t) = \frac{d\vec{v}}{dt}$ . Which of the following statements is correct? (Here, $\hat{\imath}$ and $\hat{\jmath}$ are orthogonal unit vectors in the orbital plane.)

(a) $\vec{a}$ is always parallel to $\vec{v}$

(b) $\vec{a}$ has a magnitude $v_0\omega$ and is perpendicular to $\vec{v}$

(d) $\vec{v}\cdot\vec{a} = v_0^2\omega$

(e) $\vec{a}$ is parallel to $\hat{n}$

Отговори

Решение

Solution: Given

\vec{v}(t) = v_0\big(\cos(\omega t)\,\hat{\imath} + \sin(\omega t)\,\hat{\jmath}\big),

which represents a velocity vector of constant magnitude $v_0$ rotating uniformly in the orbital plane.

Time derivative

The acceleration is defined as

\vec{a}(t) = \frac{d\vec{v}}{dt} = v_0\big(-\omega\sin(\omega t)\,\hat{\imath} + \omega\cos(\omega t)\,\hat{\jmath}\big).

Magnitude of the acceleration

|\vec{a}| = v_0\omega\sqrt{\sin^2(\omega t) + \cos^2(\omega t)} = v_0\omega.

Direction of the acceleration

The dot product of the velocity and acceleration is

\vec{v}\cdot\vec{a} = v_0^2\omega\big(\cos(\omega t)(-\sin(\omega t)) + \sin(\omega t)\cos(\omega t)\big) = 0.

Since the dot product vanishes, the acceleration vector is perpendicular to the velocity vector.

Equivalently, because the speed of the planet is constant,

\frac{d}{dt}(\vec{v}\cdot\vec{v}) = 0 \quad\Rightarrow\quad 2\vec{v}\cdot\vec{a} = 0,

which again implies that $\vec{a}$ is perpendicular to $\vec{v}$ .

Evaluation of answer choices

Option (a) is incorrect because the acceleration of uniform circular motion is perpendicular, not parallel, to the velocity.
Option (b) is correct: the acceleration has magnitude $v_0\omega$ and is perpendicular to the velocity.
Option (c) is incorrect because a constant speed does not imply zero acceleration; the direction of the velocity is changing.
Option (d) is incorrect because $\vec{v}\cdot\vec{a} = 0$ , not $v_0^2\omega$ .
Option (e) is incorrect because the acceleration lies in the orbital plane, whereas $\hat{n}$ is perpendicular to that plane.

Answer: B

Оригинал в Архива: usaaao_first_exam_2026.pdf · официални решения: usaaao_first_exam_solutions_2026.pdf