Velocity That Varies with Time

A position that changes with time

In Position and Velocity as Vectors the position was $r=r_0+vt$r = r₀ + vt: a fixed start plus one copy of a fixed velocity for each second. Most motion is not like that. The position is still a vector whose coordinates depend on the time t, but the coordinates need not be straight-line functions of t.

A ball is thrown from the origin, and t seconds later its position is $r\left(t\right)=\left(2t,5t-5t^2\right)$r(t) = (2t, 5t − 5t²) meters, the first coordinate across and the second up. The first coordinate grows steadily, 2 meters each second. The second rises and then falls, because the $-5t^2$−5t² term grows faster than 5t.

At t = 0 the ball is at (0, 0). At t = 0.5 it is at (1, 1.25), since 5 × 0.5 − 5 × 0.25 = 2.5 − 1.25 = 1.25. At t = 1 it is at (2, 0), back at the height it started from, 2 meters away.

The path

Plotting the second coordinate against the first draws the path. From x = 2t, $t=\frac{x}{2}$t = x/2, so $y=5\left(\frac{x}{2}\right)-5{\left(\frac{x}{2}\right)}^2=2.5x-1.25x^2$y = 5(x/2) − 5(x/2)² = 2.5x − 1.25x². The path is a parabola, and the ball moves along it from (0, 0) to (2, 0).

The path alone has lost the time. The point (1, 1.25) is on it, but nothing in $y=2.5x-1.25x^2$y = 2.5x − 1.25x² says the ball is there at t = 0.5. The position vector r(t) keeps both: where the ball is, and when.

The gold curve is the path $y=2.5x-1.25x^2$y = 2.5x − 1.25x². The ball is at (0, 0) at t = 0, at the top (1, 1.25) at t = 0.5, and at (2, 0) at t = 1.

Velocity: differentiate each coordinate

The velocity is the rate of change of the position, $v=\frac{dr}{dt}$v = dr/dt. A vector changes coordinate by coordinate, so differentiate each coordinate on its own: v(t) = (2, 5 − 10t). For the line $r=r_0+vt$r = r₀ + vt this gives back v, because $r_0$r₀ is a constant and each coordinate of vt is a constant times t.

The first coordinate of the velocity is always 2: the ball moves across at 2 meters per second the whole time. The second is 5 − 10t. At t = 0 the velocity is (2, 5), rising. At t = 0.5 it is (2, 0): the ball is moving straight across, at the top of its flight. At t = 1 it is (2, −5), falling.

The speed is the length of the velocity, $|v|=\sqrt{4+{\left(5-10t\right)}^2}$|v| = √(4 + (5 − 10t)²). At t = 0 it is $\sqrt{29}$√29, about 5.39 meters per second. It is least at the top, where the upward part is 0 and the speed is $\sqrt{4}=2$√4 = 2 meters per second.

The velocity points along the path. Its second coordinate divided by its first is the gradient of the path, $\frac{5-10t}{2}$(5 − 10t)/2, as in Parametric Differentiation. At t = 0 that is 2.5, and the gradient of $y=2.5x-1.25x^2$y = 2.5x − 1.25x² at x = 0 is 2.5 − 2.5 × 0 = 2.5.

The velocity at t = 0, 0.5 and 1, each arrow drawn at a fifth of its length from the ball's position: (2, 5) rising, (2, 0) straight across at the top, (2, −5) falling. Each arrow lies along the path.

Acceleration: differentiate again

The acceleration is the rate of change of the velocity, $a=\frac{dv}{dt}$a = dv/dt. Differentiating (2, 5 − 10t) coordinate by coordinate gives a = (0, −10), the same at every moment.

The first coordinate is 0, so the speed across never changes. The second is −10, so the upward velocity falls by 10 meters per second every second: from 5 to −5 over the one second of flight. This is gravity, taken as 10 meters per second per second, pulling straight down.

Moving round a circle

An object has position r(t) = (cos t, sin t). Since ${\cos }^{2}t+{\sin }^{2}t=1$cos²t + sin²t = 1, it is always 1 unit from the origin: it moves round the circle of radius 1, once every $2\pi$2π seconds.

Differentiating each coordinate gives v = (−sin t, cos t). Its length is $\sqrt{{\sin }^{2}t+{\cos }^{2}t}=1$√(sin²t + cos²t) = 1, so the speed is constant. It is also perpendicular to r, because r · v = −cos t sin t + sin t cos t = 0. The velocity is along the tangent to the circle, at right angles to the radius.

Differentiating again gives a = (−cos t, −sin t), which is −r. The acceleration has length 1 and points from the object straight back at the center. The speed is constant, and the acceleration is not zero: the velocity keeps changing direction, and acceleration measures any change in the velocity, not only a change in speed.

The object at $t=\frac{\pi }{3}$t = π/3, at (0.5, 0.87) on the circle of radius 1. The velocity v = (−0.87, 0.5) runs along the tangent; the acceleration a = (−0.5, −0.87) runs from the object to the center.

Any circle, any rate

For a circle of radius R swept at $\omega$ω radians per second, $r=\left(R\cos \omega t,R\sin \omega t\right)$r = (R cos ωt, R sin ωt). By the chain rule, $v=\left(-R\omega \sin \omega t,R\omega \cos \omega t\right)$v = (−Rω sin ωt, Rω cos ωt), of length $R\omega$Rω, and $a=\left(-R{\omega }^2\cos \omega t,-R{\omega }^2\sin \omega t\right)=-{\omega }^{2}r$a = (−Rω² cos ωt, −Rω² sin ωt) = −ω²r, of length $R{\omega }^2$Rω², pointing at the center.

A point 3 meters from the center, turning at 2 radians per second, moves at 3 × 2 = 6 meters per second, and its acceleration toward the center is $3\times 2^2=12$3 × 2² = 12 meters per second per second.

The usual mistakes

Differentiating $5t^2$5t² as 5t. The power comes down: the derivative is 10t, so the vertical velocity is 5 − 10t.

Giving the position as the velocity. At t = 1 the ball is at (2, 0) and its velocity is (2, −5): one is where it is, the other how it is moving.

Dividing the position by the time. That is the average velocity since t = 0, and it matches the velocity only when the velocity is constant. At t = 1, $\frac{r}{t}=\left(2,0\right)$r/t = (2, 0), while v = (2, −5).

Giving the velocity as the acceleration on the circle, or taking constant speed to mean no acceleration. On the circle the speed is 1 throughout, and a = −r has length 1.