Position and Velocity as Vectors

Where the object starts

Measure positions in meters from a fixed origin O, across and up. An object starts at the point (1, 2). Its position vector is the arrow from O to the object, written $r_0=\left(1,2\right)$r₀ = (1, 2): the 0 marks time zero.

Velocity is a vector too

The object moves with a constant velocity v = (3, 2) meters per second: every second it moves 3 m across and 2 m up. A velocity has a direction and a size, just like a position vector.

Its size is the speed. The object moves 3 across and 2 up each second, along the diagonal of that step, so by Pythagoras' theorem the speed is $|v|=\sqrt{3^2+2^2}=\sqrt{13}$|v| = √(3² + 2²) = √13, about 3.606 meters per second.

The position vector $r_0=\left(1,2\right)$r₀ = (1, 2) from the origin, then one arrow v = (3, 2) for each second. After 2 seconds the object is at (1, 2) + 2(3, 2) = (7, 6), and every position lies on the gold line.

$r=r_0+vt$r = r₀ + vt

After 1 second the object is at (1, 2) + (3, 2) = (4, 4). After 2 seconds it is at (1, 2) + 2(3, 2) = (7, 6). After t seconds it has moved t copies of v, so its position is $r=r_0+vt$r = r₀ + vt.

Written out, r = (1, 2) + t(3, 2) = (1 + 3t, 2 + 2t). Each coordinate is its own equation in t: x = 1 + 3t and y = 2 + 2t. Neither needs the other, and the time t can be any number, not only whole seconds. At t = 2.5 the object is at (1 + 7.5, 2 + 5) = (8.5, 7).

These are parametric equations with time as the parameter, and eliminating t gives the path. From x = 1 + 3t, $t=\frac{x-1}{3}$t = (x − 1)/3, so $y=2+\frac{2\left(x-1\right)}{3}=\frac{2x+4}{3}$y = 2 + 2(x − 1)/3 = (2x + 4)/3. That is the straight line of gradient $\frac{2}{3}$2/3 through (1, 2) in the figure, the same direction as v, which rises 2 for every 3 across.

Distance and time

In t seconds the object covers t × |v| meters. After 2 seconds that is $2\sqrt{13}$2√13, about 7.211 m. Check: the displacement is 2(3, 2) = (6, 4), and $|\left(6,4\right)|=\sqrt{52}=2\sqrt{13}$|(6, 4)| = √52 = 2√13.

To find when the object reaches a place, use one coordinate to find t and test the other. When is it at x = 10? 1 + 3t = 10 gives t = 3, and then y = 2 + 2 × 3 = 8, so it is at (10, 8) after 3 seconds. Is (13, 10) on its path? x = 13 needs t = 4, and y = 2 + 8 = 10 agrees, so yes, after 4 seconds. Is (13, 9)? The same t = 4 gives y = 10, not 9, so the object never passes through (13, 9).

A velocity can be found from two sightings. An object is at (2, 1) at t = 0 and at (14, 17) at t = 4. In 4 seconds it moved (12, 16), so each second it moved a quarter of that: v = (3, 4), and its speed is $\sqrt{9+16}=5$√(9 + 16) = 5.

An object starts at (1, 1) with velocity (4, 3). At t = 1 it is at (1, 1) + (4, 3) = (5, 4), and its speed is $\sqrt{4^2+3^2}=5$√(4² + 3²) = 5 meters per second. Drag t to move it along its path; drag the arrow to change v.

The usual mistakes

Adding v once. $r_0+v$r₀ + v is the position after 1 second. After t seconds, v is added t times: $r_0+vt$r₀ + vt.

Multiplying the start by t as well. The start is where the object already was; only the velocity is multiplied by the time. With $r_0=\left(1,2\right)$r₀ = (1, 2) and v = (3, 2), the position at t = 3 is (1, 2) + 3(3, 2) = (10, 8), not 3(1, 2) + (3, 2) = (6, 8).

Adding the components for the speed. With v = (3, 4), the object goes straight along the diagonal, so the speed is $\sqrt{3^2+4^2}=5$√(3² + 4²) = 5, not 3 + 4 = 7. Multiplying them, 3 × 4 = 12, gives the area of the rectangle, not its diagonal.

Velocities that add

A wind or a current adds its own velocity vector to a craft's velocity through the air or the water. In the first application below, a crosswind adds an east component to an aircraft flying north. In the second, a ferry must be steered upstream so that its velocity and the current's add to a velocity straight across.