CLASSICAL-MECHANICS

Motion in Two Dimensions: Splitting the Plane

IKey Formulas

Formula	Name	Description
$\vec{r}(t) = x(t)\,\hat{\imath} + y(t)\,\hat{\jmath}$	Position vector	Horizontal coordinate $x$ and vertical coordinate $y$ , bundled with direction
$\vec{v}(t) = \dfrac{dx}{dt}\,\hat{\imath} + \dfrac{dy}{dt}\,\hat{\jmath} = v_x\,\hat{\imath} + v_y\,\hat{\jmath}$	Velocity vector	Derivative of position, component by component
$\vec{a}(t) = \dfrac{dv_x}{dt}\,\hat{\imath} + \dfrac{dv_y}{dt}\,\hat{\jmath} = a_x\,\hat{\imath} + a_y\,\hat{\jmath}$	Acceleration vector	Derivative of velocity, component by component
$v_x \equiv \dfrac{dx}{dt}, \quad v_y \equiv \dfrac{dy}{dt}$	Velocity components	Each is just the ordinary 1-D derivative of its coordinate

IIInteractive Visualizations

Visualization 1 — Resolving a Vector into Components

The position vector is the sum of a horizontal leg of length $x$ and a vertical leg of length $y$ .

x (horizontal) 3 y (vertical) 2

Visualization 2 — Two Motions, One Trajectory

Two independent 1-D motions, sharing only the clock $t$ , combine point by point into one curved path.

vx 3 vy₀ 8 ay -3 t 2.00

💡 Challenge: raise $v_x$ and watch the orange height-versus-time curve on the left stay put — the vertical story never feels the horizontal one.

Visualization 3 — The Derivative Ladder

At each instant the slope of $y(t)$ is the velocity $v_y$ , and the slope of $v_y$ is the constant acceleration $a_y$ .

time marker t 1.50

💡 This is the ladder $\vec{r} \xrightarrow{d/dt} \vec{v} \xrightarrow{d/dt} \vec{a}$ , run on a single component.

IIIQuiz Questions

Question 1

A particle has position vector $\vec{r}(t) = (4t)\,\hat{\imath} + (3t^2)\,\hat{\jmath}$ (with $x$ and $y$ in meters, $t$ in seconds).

What is the horizontal component of velocity, $v_x(t)$ ?

✅ Correct! $\frac{d}{dt}(4t) = 4$ , a constant horizontal velocity.

❌ Not quite. Differentiate only the horizontal coordinate $x(t) = 4t$ with respect to $t$ . The $3t^2$ term is vertical and is irrelevant here.

Show solution

Solution:

The horizontal component is $x(t) = 4t$ . Velocity is the time derivative of position, taken component by component:

$v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(4t) = 4 \text{ m/s}$

The $\hat{\jmath}$ term ( $3t^2$ ) belongs to the vertical motion and does not affect the horizontal component — the two directions are independent. So $v_x = 4$ , a constant.

Question 2

For the same particle, $\vec{r}(t) = (4t)\,\hat{\imath} + (3t^2)\,\hat{\jmath}$ .

What is the acceleration vector $\vec{a}(t)$ ?

✅ Correct! $a_x = 0$ and $a_y = 6$ , so $\vec{a} = 6\,\hat{\jmath}$ .

❌ Not quite. Acceleration is the second derivative of position. Differentiate each component twice: $x = 4t \to v_x = 4 \to a_x = 0$ , and $y = 3t^2 \to v_y = 6t \to a_y = 6$ .

Show solution

Solution:

Differentiate position twice, component by component.

First, velocity: $v_x = \frac{d}{dt}(4t) = 4, \qquad v_y = \frac{d}{dt}(3t^2) = 6t$

Then, acceleration: $a_x = \frac{d}{dt}(4) = 0, \qquad a_y = \frac{d}{dt}(6t) = 6$

So $\vec{a}(t) = 0\,\hat{\imath} + 6\,\hat{\jmath} \text{ m/s}^2$

The horizontal motion has constant velocity (zero acceleration); the vertical motion has constant acceleration $6\ \text{m/s}^2$ .

Question 3

True or False: For a projectile moving under gravity, changing its horizontal launch speed $v_x$ will change how its vertical position $y(t)$ evolves in time.

✅ Correct! The vertical motion depends only on the vertical quantities; $v_x$ never enters the equation for $y(t)$ .

❌ Not quite. The components are independent. The equation for $y(t)$ contains no horizontal quantity, so $v_x$ cannot affect it.

Show solution

Solution: False.

The horizontal and vertical motions are independent — they share only the clock $t$ . The vertical motion is governed entirely by the vertical equations ( $v_y = v_{y0} + a_y t$ , $y = v_{y0}t + \tfrac{1}{2}a_y t^2$ ), which contain no $v_x$ at all.

Changing $v_x$ stretches or compresses the path horizontally but leaves the height-versus- time story untouched. This is exactly the independence highlighted in Visualization 2.

Question 4

A particle moves with $x(t) = 5t$ and $y(t) = 20t - 5t^2$ (SI units). At what time $t > 0$ is the vertical component of velocity $v_y$ equal to zero (the top of the arc)?

✅ Correct! $v_y = 20 - 10t = 0$ gives $t = 2$ s.

❌ Not quite. Differentiate only the vertical coordinate: $v_y = \frac{d}{dt}(20t - 5t^2) = 20 - 10t$ , then solve $v_y = 0$ .

Show solution

Solution:

The vertical velocity is the derivative of the vertical coordinate: $v_y(t) = \frac{dy}{dt} = \frac{d}{dt}\left(20t - 5t^2\right) = 20 - 10t$

Set it to zero: $20 - 10t = 0 \;\Longrightarrow\; t = 2 \text{ s}$

At $t = 2\ \text{s}$ the vertical velocity vanishes — the particle is momentarily moving purely horizontally, i.e. at the top of its arc. (Note $x(t) = 5t$ plays no role here; the horizontal and vertical pieces are independent.)

Solved: 0 / 4