In short

The velocity vector \vec{v} = d\vec{r}/dt tells you how fast and in which direction an object moves. Its magnitude |\vec{v}| = \sqrt{v_x^2 + v_y^2} is the speed. The acceleration vector \vec{a} = d\vec{v}/dt has two roles: its tangential component changes the speed, and its normal component changes the direction. Horizontal and vertical motions are independent — gravity does not affect horizontal velocity, and horizontal velocity does not affect vertical fall.

A cricket ball leaves the bat at an angle, climbing steeply at first, then curving over and dropping to the ground. At every instant of that flight, the ball has a velocity — but that velocity is not a single number. It has a magnitude (how fast) and a direction (which way). The direction keeps changing as the ball arcs through the sky. At the top of its flight, the ball moves purely horizontally. A moment later, gravity has bent the velocity downward. The velocity is a vector, and it lives in two dimensions.

This article is about the machinery you need to describe that velocity precisely — and the acceleration that reshapes it. By the end, you will know how to take a position vector, differentiate it to get velocity, differentiate again to get acceleration, and decompose that acceleration into the part that speeds you up and the part that turns you.

The velocity vector — derivative of the position vector

Suppose an object moves in a plane. At any time t, its position is described by a position vector:

\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j}

Here x(t) and y(t) are the coordinates of the object at time t, and \hat{i} and \hat{j} are unit vectors along the x-axis and y-axis.

How do you find the velocity? The same way you find velocity in one dimension — take the derivative of position with respect to time. In one dimension, v = dx/dt. In two dimensions, you differentiate the entire position vector:

\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j}

Why: the unit vectors \hat{i} and \hat{j} are constant — they do not change with time. So differentiating \vec{r} means differentiating each component separately. The x-component of velocity is the rate of change of x, and the y-component is the rate of change of y.

Write the components as:

v_x = \frac{dx}{dt}, \qquad v_y = \frac{dy}{dt}

So \vec{v} = v_x\,\hat{i} + v_y\,\hat{j}.

This is the central idea: the velocity vector is the position vector's derivative, component by component. There is no mixing — v_x depends only on x(t), and v_y depends only on y(t). The two directions are independent.

Position vector and velocity vector of a moving object A curved path in 2D with a position vector from the origin to a point P, and a velocity vector tangent to the path at P, decomposed into vx and vy components. x y path r(t) P v vₓ vᵧ θ
The position vector $\vec{r}$ points from the origin to point P. The velocity vector $\vec{v}$ is tangent to the path and decomposes into $v_x$ (horizontal) and $v_y$ (vertical) components. The angle $\theta$ gives the direction of motion.

Speed — the magnitude of the velocity vector

The velocity vector has both magnitude and direction. The magnitude of \vec{v} is the speed — a scalar that tells you how fast the object is moving, without caring about direction:

\text{speed} = |\vec{v}| = \sqrt{v_x^2 + v_y^2}

Why: this is the Pythagorean theorem applied to the velocity components. Since v_x and v_y are perpendicular, the magnitude of the vector is the hypotenuse of a right triangle with legs v_x and v_y.

The direction of the velocity at any instant is given by:

\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)

Why: in the right triangle formed by v_x and v_y, the tangent of the angle with the x-axis is opposite over adjacent, which is v_y/v_x.

Notice the distinction: a car's speedometer reads speed (a number), not velocity (a number plus a direction). An autorickshaw whipping around a corner in Bangalore at a constant 30 km/h has constant speed but changing velocity — the direction changes at every instant of the turn.

The acceleration vector — derivative of velocity

Just as velocity is the rate of change of position, acceleration is the rate of change of velocity:

\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{dv_x}{dt}\,\hat{i} + \frac{dv_y}{dt}\,\hat{j}

Why: the same logic as before — \hat{i} and \hat{j} are constant, so you differentiate component by component. The x-component of acceleration is a_x = dv_x/dt, and the y-component is a_y = dv_y/dt.

Since \vec{v} = d\vec{r}/dt, you can also write:

\vec{a} = \frac{d^2\vec{r}}{dt^2} = \frac{d^2 x}{dt^2}\,\hat{i} + \frac{d^2 y}{dt^2}\,\hat{j}

Why: differentiating position once gives velocity, differentiating again gives acceleration. This is the second derivative of position.

The magnitude of the acceleration is:

|\vec{a}| = \sqrt{a_x^2 + a_y^2}

Acceleration in two dimensions is richer than acceleration in one dimension. In one dimension, acceleration can only speed you up or slow you down. In two dimensions, acceleration can do something new: change your direction without changing your speed. That distinction leads to the idea of tangential and normal components.

Tangential and normal components of acceleration

When a car drives around a curve, it experiences two kinds of acceleration at once. The engine pushes it forward along the road — that changes the speed. The curve bends the road — that changes the direction. These are the two components of acceleration.

Tangential acceleration — changing the speed

The tangential acceleration a_T is the component of \vec{a} along the direction of motion — along the velocity vector. It tells you how quickly the speed is changing:

a_T = \frac{d|\vec{v}|}{dt} = \frac{d(\text{speed})}{dt}

Why: the tangential component acts along the direction you are already moving. If it is positive, you speed up. If it is negative, you slow down. It cannot change your direction — only your speed.

You can compute a_T directly from the components:

a_T = \frac{v_x\,a_x + v_y\,a_y}{|\vec{v}|} = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|}

Why: the dot product \vec{v} \cdot \vec{a} projects \vec{a} onto the direction of \vec{v}. Dividing by |\vec{v}| gives the scalar component of acceleration along the velocity direction.

Normal acceleration — changing the direction

The normal acceleration a_N is the component of \vec{a} perpendicular to the velocity. It acts sideways — toward the centre of curvature of the path — and changes only the direction, not the speed:

a_N = \sqrt{|\vec{a}|^2 - a_T^2}

Why: the total acceleration vector is the vector sum of the tangential and normal parts. Since they are perpendicular, the Pythagorean theorem applies: |\vec{a}|^2 = a_T^2 + a_N^2. Rearranging gives a_N.

For circular motion at constant speed, a_T = 0 (the speed is not changing), and all the acceleration is normal — pointing toward the centre of the circle. That is the centripetal acceleration you will study in Uniform Circular Motion.

Tangential and normal acceleration components A curved path with a point P. The velocity vector is tangent. Acceleration is decomposed into tangential (along velocity) and normal (perpendicular, toward centre of curvature) components. P v a aᵀ aₙ toward centre of curvature
The acceleration $\vec{a}$ at point P is decomposed into a tangential component $a_T$ (along the velocity, changes speed) and a normal component $a_N$ (perpendicular to velocity, changes direction). On a curve, both are generally nonzero.

Here is the physical picture to carry in your head:

Component Direction What it changes Example
a_T (tangential) Along \vec{v} Speed (faster or slower) A car accelerating on a straight road
a_N (normal) Perpendicular to \vec{v} Direction (turning) A car on a circular roundabout at constant speed
Both nonzero At an angle to \vec{v} Speed and direction A car speeding up while rounding a curve

The independence of horizontal and vertical motions

This is the single most powerful idea in two-dimensional kinematics, and it follows directly from how you decompose vectors.

Consider a ball thrown horizontally from the top of a building. Gravity acts downward — it is an acceleration in the y-direction only. There is no horizontal force (neglect air resistance). So:

a_x = 0, \qquad a_y = g = 9.8 \text{ m/s}^2 \text{ (downward)}

Why: gravity is a purely vertical force. It pulls the ball down but has no opinion about horizontal motion. Newton's second law applied to each direction: F_x = 0 means a_x = 0; F_y = mg means a_y = g.

This means the horizontal velocity v_x never changes — the ball keeps moving horizontally at whatever speed it was thrown. Meanwhile, the vertical velocity v_y increases at the rate g, exactly as if the ball were simply dropped. The two motions are independent: each runs on its own clock, obeying its own equation, unaware of the other.

Take two balls at the same height. Drop one straight down. Throw the other horizontally at 20 m/s. Which one hits the ground first?

They hit at the same time.

The thrown ball covers horizontal distance while falling, but the falling itself — y = \frac{1}{2}gt^2 — is identical for both. The horizontal velocity of 20 m/s does not slow the fall or speed it up. Gravity does not care that the ball is also moving sideways.

This is the key to projectile motion. Every projectile problem decomposes into two independent one-dimensional problems: constant velocity horizontally, constant acceleration vertically. Solve each separately, combine the results.

Animated: dropped ball and horizontally thrown ball fall at the same rate Two balls start at the same height. One is dropped vertically. The other is thrown horizontally at 20 m/s. Both fall with y = 4.9 t squared. Horizontal connectors show they are always at the same height.
The red ball is dropped. The dark ball is thrown horizontally at 20 m/s. The dashed connectors prove they are always at the same height — both fall with $y = \frac{1}{2}gt^2$. Horizontal motion does not affect vertical fall. Click replay to watch again.

The simulation above is worth studying carefully. At t = 0.5 s, both balls have fallen y = 4.9 \times 0.25 = 1.225 m. At t = 1 s, both have fallen 4.9 m. At t = 1.5 s, both have fallen 11.025 m. The dark ball has also travelled 20 \times 1.5 = 30 m horizontally — but its vertical position is identical to the red ball's. The horizontal and vertical motions run independently.

Why independence works

The mathematical reason is that the equations decouple. For the thrown ball:

Horizontal: x = v_0 t (constant velocity, no force)

Vertical: y = \frac{1}{2}gt^2 (starting from rest vertically, constant acceleration g)

The equation for x contains no y. The equation for y contains no x. Each direction is its own one-dimensional problem. This decoupling happens whenever the forces act along only one coordinate direction — which, for projectile motion near Earth's surface (neglecting air resistance), is always the case.

Worked examples

Example 1: Finding velocity and acceleration from position

A particle moves in the xy-plane with position vector:

\vec{r}(t) = (3t^2)\,\hat{i} + (4t - t^2)\,\hat{j} \quad \text{(metres, with } t \text{ in seconds)}

Find the velocity vector, acceleration vector, speed, and direction of velocity at t = 1 s.

Step 1. Find the velocity vector by differentiating \vec{r}.

\vec{v} = \frac{d\vec{r}}{dt} = \frac{d(3t^2)}{dt}\,\hat{i} + \frac{d(4t - t^2)}{dt}\,\hat{j}
\vec{v} = 6t\,\hat{i} + (4 - 2t)\,\hat{j}

Why: differentiate each component separately. The derivative of 3t^2 is 6t. The derivative of 4t - t^2 is 4 - 2t.

Step 2. Find the acceleration vector by differentiating \vec{v}.

\vec{a} = \frac{d\vec{v}}{dt} = \frac{d(6t)}{dt}\,\hat{i} + \frac{d(4 - 2t)}{dt}\,\hat{j}
\vec{a} = 6\,\hat{i} + (-2)\,\hat{j} = 6\,\hat{i} - 2\,\hat{j}

Why: the derivative of 6t is 6 and the derivative of 4 - 2t is -2. The acceleration is constant — it does not depend on time. The particle accelerates at 6 m/s^2 in the x-direction and -2 m/s^2 (downward) in the y-direction.

Step 3. Evaluate \vec{v} at t = 1 s.

\vec{v}(1) = 6(1)\,\hat{i} + (4 - 2(1))\,\hat{j} = 6\,\hat{i} + 2\,\hat{j} \text{ m/s}

So v_x = 6 m/s and v_y = 2 m/s.

Step 4. Compute the speed.

|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ m/s}

Why: speed is the magnitude of the velocity vector — the length of the hypotenuse formed by v_x and v_y.

Step 5. Find the direction.

\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{2}{6}\right) = \tan^{-1}(0.333) \approx 18.4°

Why: the angle is measured from the positive x-axis. Since both v_x and v_y are positive, the velocity points into the first quadrant — upward and to the right.

Velocity vector at t = 1 s decomposed into components A vector from the origin with vx = 6 along the x-axis and vy = 2 along the y-axis. The resultant makes 18.4 degrees with the x-axis. Speed is 6.32 m/s. x y 1 2 3 4 5 6 1 2 vₓ = 6 m/s vᵧ = 2 m/s |v| = 6.32 m/s 18.4°
At $t = 1$ s: the velocity has components $v_x = 6$ m/s and $v_y = 2$ m/s. The speed $|\vec{v}| = \sqrt{40} \approx 6.32$ m/s. The direction is $18.4°$ above the positive $x$-axis.

Result: At t = 1 s, \vec{v} = 6\,\hat{i} + 2\,\hat{j} m/s (speed \approx 6.32 m/s, direction 18.4° above horizontal). The acceleration is constant: \vec{a} = 6\,\hat{i} - 2\,\hat{j} m/s^2.

What this shows: Differentiating the position vector component by component gives you the velocity. Differentiating again gives acceleration. The speed and direction come from the Pythagorean theorem and the inverse tangent — nothing more.

Example 2: Ball thrown horizontally from a building

A ball is thrown horizontally at 15 m/s from the top of a 20 m building. Neglect air resistance. Find the velocity vector at t = 1 s, and determine the speed and direction of the velocity at that instant.

Assumptions: Air resistance is neglected. The ball starts at the top of the building with v_x = 15 m/s and v_y = 0. Take downward as positive y.

Step 1. Write the position components.

x(t) = 15t, \qquad y(t) = \frac{1}{2}(9.8)t^2 = 4.9t^2

Why: horizontally, there is no force, so x increases at a constant rate of 15 m/s. Vertically, the ball starts with zero vertical velocity and accelerates downward at g = 9.8 m/s^2, giving y = \frac{1}{2}gt^2.

Step 2. Find the velocity components.

v_x = \frac{dx}{dt} = 15 \text{ m/s}, \qquad v_y = \frac{dy}{dt} = 9.8t

At t = 1 s:

v_x = 15 \text{ m/s}, \qquad v_y = 9.8 \text{ m/s}

Why: the horizontal velocity is constant (no horizontal acceleration). The vertical velocity grows linearly with time because the vertical acceleration is constant at 9.8 m/s^2.

Step 3. Write the velocity vector and compute speed.

\vec{v}(1) = 15\,\hat{i} + 9.8\,\hat{j} \text{ m/s}
|\vec{v}| = \sqrt{15^2 + 9.8^2} = \sqrt{225 + 96.04} = \sqrt{321.04} \approx 17.9 \text{ m/s}

Step 4. Find the direction.

\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{9.8}{15}\right) = \tan^{-1}(0.653) \approx 33.1°

Why: the velocity makes 33.1° below the horizontal. The ball was thrown horizontally, but after 1 second gravity has bent the velocity noticeably downward.

Animated: horizontal throw trajectory with velocity vector A ball thrown horizontally at 15 m/s from a 20 m building traces a parabolic path. The velocity vector tilts downward as gravity adds vertical speed. 20 m
The ball is thrown horizontally at 15 m/s. The trail shows the parabolic path — the ball moves steadily to the right while falling faster and faster. At $t = 1$ s, it has travelled 15 m horizontally and fallen 4.9 m. Click replay to watch again.

Result: At t = 1 s, the velocity vector is \vec{v} = 15\,\hat{i} + 9.8\,\hat{j} m/s. The speed is approximately 17.9 m/s, and the velocity makes 33.1° below the horizontal.

What this shows: Even though the ball was thrown horizontally, after just 1 second gravity has given it a significant downward velocity component (9.8 m/s). The speed has increased from 15 m/s to 17.9 m/s, and the direction has tilted 33° below horizontal. This is the independence of horizontal and vertical motions at work — v_x stays at 15, while v_y grows independently.

Common confusions

If you are comfortable computing velocity and acceleration vectors, finding speed and direction, and understanding the independence of horizontal and vertical motions, you have the tools for projectile motion. What follows is for readers who want the formal treatment of tangential and normal acceleration and the connection to curvature.

The tangential-normal decomposition — full derivation

The velocity vector can be written as:

\vec{v} = v\,\hat{T}

where v = |\vec{v}| is the speed and \hat{T} is the unit tangent vector (the direction of motion). Differentiate using the product rule:

\vec{a} = \frac{d\vec{v}}{dt} = \frac{dv}{dt}\,\hat{T} + v\,\frac{d\hat{T}}{dt}

Why: the product rule applies because both the speed v and the direction \hat{T} can change with time. The first term changes the speed (tangential). The second term changes the direction (normal).

The derivative d\hat{T}/dt is perpendicular to \hat{T} (since \hat{T} has constant magnitude 1, any change in \hat{T} must be sideways). Define the unit normal vector \hat{N} as the direction of d\hat{T}/dt, and write:

\frac{d\hat{T}}{dt} = \frac{v}{\rho}\,\hat{N}

Why: the rate at which the direction changes depends on both the speed v (faster motion means faster turning) and the curvature 1/\rho of the path (\rho is the radius of curvature — smaller radius means sharper turning). This can be proven rigorously from the arc-length parameterisation of the curve.

Substituting:

\vec{a} = \frac{dv}{dt}\,\hat{T} + \frac{v^2}{\rho}\,\hat{N}

This is the full tangential-normal decomposition:

\boxed{a_T = \frac{dv}{dt}, \qquad a_N = \frac{v^2}{\rho}}

The normal acceleration v^2/\rho is exactly the centripetal acceleration formula you will see in Uniform Circular Motion, where \rho is the radius of the circle.

Curvature and the radius of curvature

The radius of curvature \rho at a point on a curve y = f(x) is given by:

\rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}

Why: this formula comes from differential geometry. The curvature \kappa = 1/\rho measures how sharply the curve bends. A straight line has \kappa = 0 (infinite radius of curvature). A tight circle has large \kappa (small \rho).

For a projectile with x = v_0 t and y = \frac{1}{2}gt^2 (so y = gx^2/(2v_0^2)):

\frac{dy}{dx} = \frac{gx}{v_0^2}, \qquad \frac{d^2y}{dx^2} = \frac{g}{v_0^2}

At the launch point (x = 0):

\rho = \frac{[1 + 0]^{3/2}}{g/v_0^2} = \frac{v_0^2}{g}

Why: at the launch point the path is horizontal, so the first derivative is zero. The radius of curvature equals v_0^2/g — and the normal acceleration v^2/\rho = v_0^2/(v_0^2/g) = g, which is exactly the gravitational acceleration. This confirms that at the launch point, all of gravity acts as centripetal (normal) acceleration, bending the path downward.

Connection to polar coordinates

In polar coordinates (r, \theta), the velocity has radial and transverse components:

v_r = \frac{dr}{dt}, \qquad v_\theta = r\frac{d\theta}{dt}

The acceleration has four terms (two from each component):

a_r = \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2, \qquad a_\theta = r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}

The -r\dot{\theta}^2 term in a_r is the centripetal acceleration, and the 2\dot{r}\dot{\theta} term in a_\theta is the Coriolis term. These arise naturally when you differentiate the position vector in polar coordinates, where the unit vectors \hat{r} and \hat{\theta} themselves change direction with time. The full derivation belongs in the article on polar coordinates in kinematics, but the takeaway is this: the tangential-normal decomposition and the polar decomposition are two different ways of splitting the same acceleration vector, useful in different situations.

Where this leads next