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Kinematics in Two Dimensions: Vectors, Projectiles, and Relative Motion

Capítulo 3

Estimated reading time: 10 minutes

1) Vectors for Motion in a Plane

Why vectors appear in 2D kinematics

In two-dimensional motion, position, velocity, and acceleration must describe both “how much” and “which way.” That is exactly what vectors do. We will represent vectors using components along chosen axes (usually horizontal x and vertical y), then combine them using vector addition.

Components and unit vectors

A 2D vector \vec{A} can be written in component form as \vec{A} = A_x\,\hat{i} + A_y\,\hat{j}, where \hat{i} is the unit vector in the +x direction and \hat{j} is the unit vector in the +y direction. The numbers A_x and A_y are the signed components (they can be negative).

Vector addition: \vec{C}=\vec{A}+\vec{B} means C_x=A_x+B_x and C_y=A_y+B_y.
Vector subtraction: \vec{A}-\vec{B} = \vec{A} + (-\vec{B}), so subtract components.

Magnitude and angle relationships

If \vec{A} = (A_x, A_y), its magnitude is |\vec{A}| = \sqrt{A_x^2 + A_y^2}. If the angle \theta is measured from the +x axis toward the vector, then (when using that convention):

A_x = |\vec{A}|\cos\theta
A_y = |\vec{A}|\sin\theta
\theta = \tan^{-1}(A_y/A_x) (use a quadrant-aware calculator function like atan2(A_y, A_x) to get the correct quadrant)

Checkpoint: signs, angles, and units

Signs: A negative component means the vector points partly in the negative axis direction. For example, v_x<0 means motion toward −x.
Angles: Always state your angle reference (from +x? from horizontal? above the horizon?). Keep it consistent.
Units: Components have the same units as the vector. If \vec{v} is in m/s, then v_x and v_y are also in m/s.

Mini-practice: component extraction

A runner’s velocity has magnitude 6.0 m/s at 30° above +x. Find components.

Use v_x = v\cos\theta, v_y = v\sin\theta.
v_x = 6.0\cos30^\circ \approx 5.20 m/s.
v_y = 6.0\sin30^\circ = 3.00 m/s.

2) Constant-Acceleration Motion in 2D: Split into x and y

The key idea: independence of perpendicular components

If acceleration is constant, you can treat motion in x and y as two separate constant-acceleration problems that share the same time t. The vector equations are:

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\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \tfrac{1}{2}\vec{a}\,t^2  \n\vec{v}(t) = \vec{v}_0 + \vec{a}\,t

In components:

x(t) = x_0 + v_{0x} t + \tfrac{1}{2} a_x t^2  \ny(t) = y_0 + v_{0y} t + \tfrac{1}{2} a_y t^2  \n\n v_x(t) = v_{0x} + a_x t  \n v_y(t) = v_{0y} + a_y t

This “splitting” works because x and y directions are perpendicular: acceleration in y does not directly change the x-component of velocity, and vice versa.

Step-by-step workflow for 2D kinematics problems

Choose axes (often x horizontal, y vertical; but you can rotate axes to simplify).
Write knowns separately for x and y: x_0, v_{0x}, a_x and y_0, v_{0y}, a_y.
Pick target unknowns (time, final position, final velocity components, etc.).
Use one component to solve for time if needed, then substitute that same t into the other component.
Recombine if asked for speed or direction: v = \sqrt{v_x^2+v_y^2}, \theta = \tan^{-1}(v_y/v_x).

Checkpoint: consistent sign conventions

Pick a sign convention and stick to it. A common choice for projectile problems is:

+x: to the right
+y: upward
Gravity: a_y = -g (negative because it points downward)

If you instead choose +y downward (sometimes convenient), then gravity is positive, but all y-positions and velocities must follow that choice consistently.

3) Projectile Motion (No Air Drag, Constant g)

Model assumptions (state them explicitly)

Air resistance is neglected.
Acceleration is constant and downward: \vec{a} = (0, -g), with g \approx 9.8\,\text{m/s}^2.
Earth’s curvature and rotation are ignored (valid for typical everyday ranges).

Under these assumptions, horizontal acceleration is zero (a_x=0), so v_x stays constant. Vertical motion has constant acceleration -g.

Launch speed and angle: convert to components

If a projectile is launched with speed v_0 at angle \theta above the horizontal:

v_{0x} = v_0\cos\theta
v_{0y} = v_0\sin\theta

Core equations for projectile motion

With launch point (x_0, y_0):

x(t) = x_0 + (v_0\cos\theta) t  \n\n y(t) = y_0 + (v_0\sin\theta) t - \tfrac{1}{2} g t^2  \n\n v_x(t) = v_0\cos\theta  \n v_y(t) = v_0\sin\theta - g t

Time to peak and peak height (physical interpretation)

At the highest point, the vertical velocity becomes zero (v_y=0) while horizontal velocity remains unchanged. Solve:

0 = v_0\sin\theta - g t_{\text{peak}} so t_{\text{peak}} = \dfrac{v_0\sin\theta}{g}.

Peak height relative to launch height (\Delta y_{\max}) comes from plugging t_{\text{peak}} into y(t) or using a vertical kinematics relation:

\Delta y_{\max} = \dfrac{(v_0\sin\theta)^2}{2g}.

Interpretation: A larger vertical launch component increases time to peak and peak height; the horizontal component does not affect peak height in this model.

Time of flight and range (same launch and landing height)

If the projectile lands at the same height it was launched (y_f = y_0), then the vertical displacement is zero:

0 = (v_0\sin\theta)t - \tfrac{1}{2} g t^2.

Factor out t to get two solutions: t=0 (launch) and the nonzero flight time:

T = \dfrac{2 v_0\sin\theta}{g}.

The horizontal range is then:

R = (v_0\cos\theta)T = \dfrac{v_0^2 \sin(2\theta)}{g}.

Interpretation: For fixed v_0 and same-height landing, the range depends on \sin(2\theta), which is maximized at 2\theta=90^\circ → \theta=45^\circ. Angles \theta and 90^\circ-\theta give the same range (but different peak heights and flight times).

Projectile launched from a height (general method)

If the projectile lands at a different height, do not memorize a special formula—solve systematically:

Use y(t)=y_f to solve for t. This is usually a quadratic in t.
Choose the physically meaningful root (typically the positive time after launch).
Compute horizontal distance with x(t).

Example workflow: A ball is thrown from a balcony at y_0=12 m with v_0=10 m/s at 30^\circ. Find time to hit the ground (y_f=0).

Compute v_{0y}=10\sin30^\circ=5 m/s.
Write 0 = 12 + 5t - \tfrac{1}{2}(9.8)t^2.
Solve the quadratic for t; keep the positive root.

Checkpoint: units and “sanity checks” for projectile results

Time of flight should be in seconds; if you get meters, you mixed formulas.
Range should increase with v_0^2 and decrease with larger g.
Peak time should be half of total time only when launch and landing heights match.
Horizontal velocity stays constant only if air drag is neglected; if a problem mentions drag, this model does not apply.

Problem set: choose axes and solve

P1 (axis choice): A puck slides down a straight ramp that makes a 20° angle below the horizontal. You track its motion in the air after it leaves the ramp edge. Choose axes to simplify the launch components and write x(t), y(t) for the flight. (Hint: use standard horizontal/vertical axes during flight.)
P2 (same-height range): A ball is kicked at v_0=18 m/s and \theta=40^\circ. Find T and R assuming it lands at the same height.
P3 (different heights): A stone is thrown horizontally from a cliff of height 45 m with speed 12 m/s. Find impact time and horizontal distance. (Hint: v_{0y}=0.)
P4 (sign convention): Redo P3 using +y downward. Write y(t) and show that the final time is the same number.

4) Relative Motion: Velocities Add as Vectors

What “relative” means

Relative motion compares measurements made in different frames (different observers). The most common quantity is relative velocity. In vector form:

\vec{v}_{A/B} = \vec{v}_{A/G} - \vec{v}_{B/G}

Read this as: “velocity of A relative to B equals velocity of A relative to ground minus velocity of B relative to ground.” Rearranging gives a useful addition form:

\vec{v}_{A/G} = \vec{v}_{A/B} + \vec{v}_{B/G}

Everyday example 1: walking on a moving bus

A person walks forward inside a bus. Let:

\vec{v}_{P/B} = person relative to bus (what you’d measure inside)
\vec{v}_{B/G} = bus relative to ground (what a roadside observer measures)
\vec{v}_{P/G} = person relative to ground

Step-by-step:

Choose +x along the road.
Use v_{P/G} = v_{P/B} + v_{B/G} in the x-direction.
Example numbers: if the bus moves at +12 m/s and the person walks at +1.5 m/s relative to the bus, then v_{P/G}=13.5 m/s.
If the person walks toward the back at −1.5 m/s, then v_{P/G}=10.5 m/s.

Everyday example 2: wind and airplanes (2D vector addition)

An airplane’s velocity relative to the air (airspeed) combines with the wind’s velocity relative to the ground to produce the plane’s velocity relative to the ground (ground speed):

\vec{v}_{P/G} = \vec{v}_{P/A} + \vec{v}_{A/G}

Here A stands for “air.” This is a 2D vector problem because wind can blow sideways relative to the desired travel direction.

Step-by-step crosswind example (axis selection matters):

Choose +x east, +y north.
Suppose the plane’s airspeed is 80 m/s aimed due north relative to the air: \vec{v}_{P/A}=(0,80) m/s.
Wind blows east at 20 m/s: \vec{v}_{A/G}=(20,0) m/s.
Add components: \vec{v}_{P/G}=(20,80) m/s.
Ground speed magnitude: v=\sqrt{20^2+80^2}\approx 82.5 m/s.
Drift angle east of north: \theta=\tan^{-1}(20/80)\approx 14.0^\circ.

Heading correction (to go due north): If the pilot wants v_{P/G,x}=0, they must aim slightly into the wind so that the plane’s air-velocity has a westward x-component that cancels the wind’s eastward component.

Checkpoint: common mistakes in relative motion

Mixing frames: Do not add a ground-measured velocity to an air-measured velocity unless the equation explicitly relates those frames.
Forgetting vector nature: In 2D, add components, not magnitudes.
Sign errors: Decide which direction is positive before plugging numbers.

Problems: relative motion with smart axis choices

R1 (bus, 2D): A bus moves east at 15 m/s. A passenger walks at 1.2 m/s relative to the bus toward the front, but also 0.5 m/s toward the left aisle (north). Find the passenger’s velocity relative to the ground (components and magnitude).
R2 (wind correction): A plane’s airspeed is 90 m/s. Wind is 30 m/s from west to east. What heading (angle west of north) must the plane choose to travel due north? (Hint: set the ground x-component to zero.)
R3 (river crossing axis choice): A swimmer can swim at 1.5 m/s relative to the water. The river flows east at 0.8 m/s. Choose axes and find (a) the swimmer’s ground velocity if they aim due north relative to the water, and (b) the angle they must aim upstream to go straight north relative to the ground.

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