Vectors and Motion in Two Dimensions
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| component/kəmˈpəʊnənt/ | thành phần |
| resultant/rɪˈzʌltənt/ | hợp lực |
| projectile/prəˈdʒektaɪl/ | chuyển động ném |
A plane flies northeast — two motions in one
- One arrow points "northeast", but the plane is really moving east and north at once.
- Almost no real motion is purely left–right or up–down.
- The trick of two-dimensional kinematics: split each vector into perpendicular pieces.
- Handle the pieces separately, then put them back together.
Resolving into components
- Any vector can be replaced by two perpendicular components 分量 — one along $x$, one along $y$.
- Together the two components have exactly the same effect as the original vector.
- Choose axes that make the problem easy (often horizontal and vertical).
- The components carry signs, so "left" and "down" are just negative values.
The component formulas
- For a vector $\vec A$ of magnitude $A$ at angle $\theta$ above the $x$-axis:
- $A_x = A\cos\theta$ and $A_y = A\sin\theta$.
- These come straight from the right triangle the vector makes with its components.
- Going back: $A = \sqrt{A_x^2 + A_y^2}$ and $\theta = \tan^{-1}\!\big(\tfrac{A_y}{A_x}\big)$.
A velocity of $10\ \tfrac{\text{m}}{\text{s}}$ points $37^\circ$ above the horizontal. Its horizontal component is $10\cos 37^\circ$. With $\cos 37^\circ = 0.8$, what is $v_x$ in $\tfrac{\text{m}}{\text{s}}$?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$v_x = A\cos\theta = 10 \times 0.8 = 8\ \tfrac{\text{m}}{\text{s}}$.
For the same velocity, the vertical component is $10\sin 37^\circ$. With $\sin 37^\circ = 0.6$, what is $v_y$ in $\tfrac{\text{m}}{\text{s}}$?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$v_y = A\sin\theta = 10 \times 0.6 = 6\ \tfrac{\text{m}}{\text{s}}$.
A vector has components $A_x = 3$ and $A_y = 4$. What is its magnitude?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$A = \sqrt{A_x^2 + A_y^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Not $3+4=7$ — the components are perpendicular.
The horizontal component of a vector $A$ at angle $\theta$ is $A$ ____ $\theta$.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The adjacent side of the right triangle is $A\cos\theta$, so $A_x = A\cos\theta$.
Adding vectors by components
- To add vectors, add the $x$-components and separately add the $y$-components.
- Never add the magnitudes directly — that only works if they point the same way.
- The sums $(\Sigma A_x,\ \Sigma A_y)$ are the components of the resultant 合矢量.
- Recombine with $\sqrt{\;}$ and $\tan^{-1}$ to get its magnitude and direction.
Add two vectors by components
Drag two vectors and watch the resultant — its components are the sums of the parts.
Order the steps to add two vectors using components, first step at the top.
Độ khó 3/5 · 200 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Resolve, add component-wise, then recombine magnitude and direction.
Perpendicular motions are independent
- The $x$-motion and the $y$-motion evolve separately — neither affects the other.
- Gravity pulls only downward, so it changes $v_y$ but leaves $v_x$ alone.
- This independence is the whole secret of projectile 抛体 motion (Topic 2).
- Drop a ball and fire one horizontally at the same instant: they land together.
The horizontal and vertical parts of a two-dimensional motion are independent of each other.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Perpendicular components evolve separately — gravity changes $v_y$ but not $v_x$. This is the basis of projectile motion.
You cannot add two vectors by adding their sizes unless they point along the same line. A $3$-unit vector plus a perpendicular $4$-unit vector gives $5$, not $7$ — resolve into components (or use the triangle) first.
Resolve a velocity of $10\ \tfrac{\text{m}}{\text{s}}$ aimed at $37^\circ$ above the horizontal (take $\cos 37^\circ \approx 0.8$, $\sin 37^\circ \approx 0.6$).
- $v_x = 10\cos 37^\circ \approx 8\ \tfrac{\text{m}}{\text{s}}$.
- $v_y = 10\sin 37^\circ \approx 6\ \tfrac{\text{m}}{\text{s}}$.
Check: $\sqrt{8^2 + 6^2} = 10$. ✓
Split a 2-D vector into perpendicular components: $A_x = A\cos\theta$, $A_y = A\sin\theta$. Add vectors by adding $x$- and $y$-components separately, then recombine with $A = \sqrt{A_x^2+A_y^2}$. Perpendicular motions are independent — the key to projectiles.
Write from a magnitude of 15
- A vector has components 9 and 12.
- Find its magnitude.
A vector has perpendicular components 9 and 12. What is its magnitude?
81 plus 144 is 225, and the square root is 15.
What is the magnitude?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A 9-12-15 right triangle gives the magnitude. Other options — 21 (the components added); 5 (the lesson 3-4-5 magnitude).