Math Core

When $\mathbf{r}(t)$ is the position of an object at time $t$ , its derivatives have physical meaning: the first derivative is velocity and the second is acceleration. This lesson turns the calculus of vector functions into the language of motion, from thrown balls to cars rounding curves.

Velocity, speed and acceleration

Definition

Velocity, speed and acceleration

If $\mathbf{r}(t)$ is the position of a particle at time $t$ , then

\mathbf{v}(t) = \mathbf{r}'(t), \qquad \text{speed} = |\mathbf{v}(t)|, \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t).

The velocity $\mathbf{v}(t)$ is tangent to the path and points in the direction of motion. The speed is a scalar; it equals $\dfrac{ds}{dt}$ , the rate at which distance along the path is covered. The acceleration $\mathbf{a}(t)$ describes how the velocity is changing, in length, in direction, or both.

Because speed is $\dfrac{ds}{dt}$ , the distance traveled from $t = a$ to $t = b$ is $\displaystyle\int_a^b |\mathbf{v}(t)|\,dt$ , exactly the arc length formula from the previous lesson.

Worked example: Circular motion with a climb

A particle has position $\mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; t^2 \rangle$ . Find its velocity, speed and acceleration at $t = \pi$ .

\mathbf{v}(t) = \langle -3\sin t,\; 3\cos t,\; 2t \rangle, \qquad \mathbf{a}(t) = \langle -3\cos t,\; -3\sin t,\; 2 \rangle.

At $t = \pi$ : $\mathbf{v}(\pi) = \langle 0, -3, 2\pi \rangle$ , the speed is $\sqrt{9 + 4\pi^2}$ , and $\mathbf{a}(\pi) = \langle 3, 0, 2 \rangle$ . The horizontal part of the acceleration, $\langle 3, 0 \rangle$ , points from the particle at $(-3, 0)$ back toward the $z$ -axis: that's what keeps it going around in a circle.

From acceleration back to position

Newton's second law, $\mathbf{F} = m\mathbf{a}$ , tells you the acceleration when you know the forces. Integrating twice recovers the position, with one constant vector fixed by the initial velocity $\mathbf{v}(0) = \mathbf{v}_0$ and another by the initial position $\mathbf{r}(0) = \mathbf{r}_0$ .

The classic case is a projectile near Earth's surface with air resistance ignored. Gravity gives constant acceleration $\mathbf{a} = \langle 0, 0, -g \rangle$ , where $g \approx 9.8\ \text{m/s}^2$ or $32\ \text{ft/s}^2$ . Integrating,

\mathbf{v}(t) = \mathbf{v}_0 + \langle 0, 0, -gt \rangle, \qquad \mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v}_0 + \left\langle 0, 0, -\tfrac{1}{2}gt^2 \right\rangle.

The horizontal motion has constant velocity; only the vertical motion feels gravity. So a projectile always travels in a vertical plane, the one containing its initial velocity, and inside that plane its path is a parabola. If the launch speed is $v_0$ and the launch angle above the ground is $\alpha$ , the initial velocity can be written as $v_0\langle \cos\alpha, 0, \sin\alpha \rangle$ after rotating the axes so the motion is in the $xz$ -plane. The flight time from level ground is then $\dfrac{2v_0\sin\alpha}{g}$ , and the range is $\dfrac{v_0^2\sin 2\alpha}{g}$ , which is largest at $\alpha = 45^\circ$ . You don't need to memorize these; they come straight out of setting $z(t) = 0$ .

Worked example: A launched ball

A ball is launched from the origin with initial velocity $\mathbf{v}_0 = \langle 48, 0, 64 \rangle$ ft/s. Using $g = 32\ \text{ft/s}^2$ , find when it lands, how far away it lands and its maximum height.

The position is $\mathbf{r}(t) = \langle 48t,\; 0,\; 64t - 16t^2 \rangle$ . It lands when $z = 16t(4 - t) = 0$ with $t > 0$ , so at $t = 4$ s, at the point $(192, 0, 0)$ : 192 feet away.

The height is largest when $z'(t) = 64 - 32t = 0$ , at $t = 2$ , where $z = 128 - 64 = 64$ feet. As expected for a symmetric flight, the peak is at half the flight time.

The ball's path in the xz-plane (vertical axis is height z): a parabola from (0, 0) to (192, 0) with peak height 64.Open in grapher →

Tangential and normal components of acceleration

Acceleration does two jobs: it changes how fast you go and it changes which way you go. Splitting $\mathbf{a}$ along the unit tangent $\mathbf{T}$ and the principal normal $\mathbf{N}$ separates those jobs. Write $v = |\mathbf{v}|$ for the speed. Differentiating $\mathbf{v} = v\,\mathbf{T}$ and using $\mathbf{T}' = |\mathbf{T}'|\,\mathbf{N} = \kappa v\,\mathbf{N}$ gives the following decomposition.

Components of acceleration

\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, \qquad a_T = \frac{dv}{dt} = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|}, \qquad a_N = \kappa v^2 = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|}.

The tangential component $a_T$ is the rate of change of speed. The normal component $a_N \ge 0$ measures how hard the path is turning. Also $|\mathbf{a}|^2 = a_T^2 + a_N^2$ .

Notice what each formula needs. The dot product version of $a_T$ and the cross product version of $a_N$ use only $\mathbf{v}$ and $\mathbf{a}$ , so you never have to compute $\mathbf{T}$ , $\mathbf{N}$ or $\kappa$ first. Once you know $|\mathbf{a}|$ and one of the two components, the identity $|\mathbf{a}|^2 = a_T^2 + a_N^2$ hands you the other.

The formula $a_N = \kappa v^2$ explains why you feel pushed sideways in a car: taking the same curve twice as fast quadruples the sideways acceleration. It also explains why highways use gentle curves (small $\kappa$ ) where speeds are high.

Worked example: Splitting an acceleration

For $\mathbf{r}(t) = \langle t^2,\; 2t,\; \ln t \rangle$ , find $a_T$ and $a_N$ at $t = 1$ .

$\mathbf{v}(t) = \left\langle 2t, 2, \dfrac{1}{t} \right\rangle$ and $\mathbf{a}(t) = \left\langle 2, 0, -\dfrac{1}{t^2} \right\rangle$ . At $t = 1$ : $\mathbf{v} = \langle 2, 2, 1 \rangle$ with $|\mathbf{v}| = 3$ , and $\mathbf{a} = \langle 2, 0, -1 \rangle$ with $|\mathbf{a}| = \sqrt{5}$ .

a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|} = \frac{4 + 0 - 1}{3} = 1, \qquad a_N = \sqrt{|\mathbf{a}|^2 - a_T^2} = \sqrt{5 - 1} = 2.

As a check, $\mathbf{v} \times \mathbf{a} = \langle -2, 4, -4 \rangle$ has length $6$ , and $\dfrac{6}{3} = 2$ .

Common mistake

Don't confuse the speed $|\mathbf{v}|$ with the magnitude of acceleration $|\mathbf{a}|$ , and don't assume zero $a_T$ means zero acceleration. A particle moving at constant speed around a circle has $a_T = 0$ but a nonzero $a_N$ , because its direction keeps changing.

Tip

If the speed is constant, then $\mathbf{v} \cdot \mathbf{v}$ is constant, and the same argument as for constant-length vectors shows $\mathbf{v} \cdot \mathbf{a} = 0$ . Constant speed means the acceleration is perpendicular to the velocity.

Practice

Practice 1

A particle has position $\mathbf{r}(t) = \langle t^3,\; 2t,\; t^2 \rangle$ . Find its speed at $t = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the acceleration at $t = \tfrac{\pi}{2}$ for $\mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; t^2 \rangle$ .

Enter a point like (2, -3)

Practice 3

An object has acceleration $\mathbf{a}(t) = \langle 0, 0, -10 \rangle$ , initial velocity $\mathbf{v}(0) = \langle 2, 1, 15 \rangle$ and initial position $\mathbf{r}(0) = \langle 0, 0, 0 \rangle$ . Find its position at $t = 1$ .

Enter a point like (2, -3)

Practice 4

A ball is launched from the origin with initial velocity $\langle 30, 40, 64 \rangle$ ft/s. With $g = 32\ \text{ft/s}^2$ , how far from the launch point (measured along the ground) does it land?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

For the ball in the previous problem, what is its maximum height in feet?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A 2 kg object moves with position $\mathbf{r}(t) = \langle t^3,\; 4t,\; 5 - t^2 \rangle$ (meters, seconds). Find the net force on it at $t = 2$ , in newtons.

Enter a point like (2, -3)

Practice 7

For $\mathbf{r}(t) = \langle 3t,\; 4t,\; t^2 \rangle$ , find the normal component of acceleration $a_N$ at $t = 6$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A particle moves so that its speed is constant but its path is curved. Which statement must be true?