4.2 — Velocity and acceleration vectors

The time-derivative of \vec r(t) is the velocity vector. Second derivative is the acceleration vector. All Chapter 2 kinematic equations become vector equations.

Average velocity

\[ \vec v_\text{avg} = \frac{\Delta \vec r}{\Delta t} = \frac{\vec r_2 - \vec r_1}{t_2 - t_1} \]

Interpretation: vector — magnitude AND direction. Its direction is the direction of the displacement.

Instantaneous velocity

\[ \vec v = \frac{d\vec r}{dt} = \frac{dx}{dt}\hat i + \frac{dy}{dt}\hat j + \frac{dz}{dt}\hat k = v_x \hat i + v_y \hat j + v_z \hat k \]

Note: each component is differentiated independently. Instantaneous velocity is tangent to the trajectory at every point.

Geometric intuition: as \Delta t → 0, \Delta \vec r/\Delta t rotates and approaches the tangent to the path.

Speed vs velocity

• Speed: |\vec v| = \sqrt{v_x^2 + v_y^2 + v_z^2} (scalar, always ≥ 0)
• Velocity: \vec v (vector, magnitude + direction)

Average speed: total distance / time. Average velocity: displacement vector / time. They differ (§2.1).

Average acceleration

\[ \vec a_\text{avg} = \frac{\Delta \vec v}{\Delta t} \]

Instantaneous acceleration

\[ \vec a = \frac{d\vec v}{dt} = \frac{d^2 \vec r}{dt^2} \]

Componentwise:

\[ \vec a = a_x \hat i + a_y \hat j + a_z \hat k \]

Key point: acceleration can arise for two reasons:

1. Changing magnitude of velocity (speeding up or slowing down)
2. Changing direction of velocity (even if speed is constant!)

Example: a car in a curve at constant speed. The velocity vector is changing, so there is acceleration.

Vector kinematics under constant acceleration

If \vec a = const, the Chapter 2 equations hold in vector form:

\[ \vec v(t) = \vec v_0 + \vec a\, t \]

\[ \vec r(t) = \vec r_0 + \vec v_0 t + \frac{1}{2}\vec a\, t^2 \]

Power of this form: each component (x, y, z) gives an independent equation — each just like Chapter 2.

The principle of component independence

If acceleration only points along one axis, motion along the other axes is uniform.

Classic example: a projectile. Acceleration is downward (\vec a = -g\hat j). So:

• Horizontal (x): a_x = 0 → uniform — v_x = v_{0x} constant — x = v_{0x} t
• Vertical (y): a_y = -g → free-fall equations — y = v_{0y} t - \frac{1}{2}g t^2

Two independent motions, simultaneous. This principle helped Galileo discover free fall.

Example — throwing a ball

Ball launched from (0, 0) with initial velocity \vec v_0 = 15\hat i + 20\hat j\ \mathrm{m/s}. \vec a = -9.8\hat j\ \mathrm{m/s^2}.

Velocity at t = 2 s:

\[ \vec v(2) = (15\hat i + 20\hat j) + (-9.8\hat j)(2) = 15\hat i + 0.4\hat j \]

Position at t = 2 s:

\[ \vec r(2) = 0 + (15\hat i + 20\hat j)(2) + \frac{1}{2}(-9.8\hat j)(4) = 30\hat i + 20.4\hat j \]

Ball is at (30 m, 20.4 m) with nearly horizontal velocity (v_y ≈ 0 — near peak).

A few notes and common mistakes

1. Velocity is tangent to the trajectory, not necessarily horizontal. On a parabola, velocity is tangent to the parabola at each point.

2. Acceleration can be perpendicular to velocity. In uniform circular motion, \vec a \perp \vec v. Constant speed, but changing direction.

3. Constant speed does not mean zero acceleration. If direction is changing, acceleration is nonzero — always toward the center of curvature.

What you should be able to do

• Use \vec v = d\vec r/dt and \vec a = d\vec v/dt in component form
• Distinguish speed vs velocity
• Recognize vector kinematic equations for constant acceleration
• Apply the component-independence principle to projectile problems
• Identify the two sources of acceleration (magnitude or direction change)

Preview of §4.3

The classical 2D-motion example: the projectile. §4.3 focuses on this — constant gravitational acceleration downward, uniform horizontal motion, all needed equations.

📚 See also: Halliday Vol 1, Ch 4, §4.2 — Velocity and Acceleration.

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