Motion Motion
Introduction
When analyzing motion in space, we use the following:
\begin{align*}
\text{Position: } &&\mathbf{r}(t)=\,\,& \langle x(t),\, y(t),\, z(t) \rangle \\
\text{Velocity: } &&\mathbf{v}(t)=\,\,& \mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle \\
\text{Speed: } &&\|\mathbf{v}(t)\|=\,\,& \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \\
\text{Acceleration: } &&\mathbf{a}(t)=\,\,& \mathbf{v}'(t) = \mathbf{r}''(t) = \langle x''(t),\, y''(t),\, z''(t) \rangle \\
\text{Direction: } &&\mathbf{T}(t)=\,\,& \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}
\end{align*}
Motion Under Gravity
For projectile motion problems under gravity (g = -9.8 m/s²):
\begin{align*}
r(t) &= \text{Initial Position} + \text{Initial velocity} + \text{Gravity} \\
&= \langle x_0, y_0, z_0 \rangle + t\langle v_{0x}, v_{0y}, v_{0z} \rangle + \frac{1}{2}t^2\langle 0, g, 0 \rangle \\
&= \langle x_0 + v_{0x}t,\, y_0 + v_{0y}t - \frac{1}{2}gt^2,\, z_0 + v_{0z}t \rangle \\
\end{align*}
Where
(x_0, y_0, z_0)
is the initial position and
(v_{0x}, v_{0y}, v_{0z})
is the initial velocity.
Circular Motion
Circular motion in the xy-plane with
r
- The radius of the circle
\omega
- The speed of rotation
\begin{align*}
\text{Position: } &&\mathbf{r}(t)=\,\,& \langle r\cos(\omega t),\, r\sin(\omega t),\, 0 \rangle \\
\text{Velocity: } &&\mathbf{v}(t)=\,\,& \mathbf{r}'(t) = \langle -r\omega\sin(\omega t),\, r\omega\cos(\omega t),\, 0 \rangle \\
\text{Speed: } &&\|\mathbf{v}(t)\|=\,\,& r\omega \\
\text{Acceleration: } &&\mathbf{a}(t)=\,\,& \mathbf{v}'(t) = \langle -r\omega^2\cos(\omega t),\, -r\omega^2\sin(\omega t),\, 0 \rangle \\
\end{align*}
Note that the speed is a constant value, equal to the product of the radius of the circle and the speed of rotating around the circle.
Examples
Example 1: Projectile Motion
A ball is thrown from initial position
(0, 0, 0)
with initial velocity
\langle 2, 3, 0 \rangle
m/s. Find its position, velocity, and maximum height.
\begin{align*}
\mathbf{r}(t) &= \langle 0, 0, 0 \rangle + t\langle 2, 3, 0 \rangle + \frac{1}{2}t^2\langle 0, -9.8, 0 \rangle \\
&= \langle 2t, 3t - 4.9t^2, 0 \rangle \\
\mathbf{v}(t) &= \langle 2, 3 - 9.8t, 0 \rangle
\end{align*}
For maximum height, set y-component of velocity to zero:
\begin{align*}
3 - 9.8t &= 0 \\
t &= 0.306 \text{ seconds} \\
y_{max} &= 3(0.306) - 4.9(0.306)^2 = 0.459 \text{ meters}
\end{align*}
Example 2: Circular Motion Speed
A particle moves in a circle of radius 5 meters with angular velocity
\omega = 2
rad/s. Find its position, velocity, and speed at
t = \pi/4
.
\begin{align*}
\mathbf{r}(\pi/4) &= \langle 5\cos(\pi/2), 5\sin(\pi/2), 0 \rangle \\
&= \langle 0, 5, 0 \rangle \\
\mathbf{v}(\pi/4) &= \langle -10\sin(\pi/2), 10\cos(\pi/2), 0 \rangle \\
&= \langle -10, 0, 0 \rangle \\
\text{Speed} &= \|\mathbf{v}(\pi/4)\| = r\omega = 5(2) = 10 \text{ m/s}
\end{align*}
Example 3: Direction of Motion
A particle moves according to
\mathbf{r}(t) = \langle t^2, t^3, t \rangle
. Find its velocity and direction at
t = 1
.
\begin{align*}
\mathbf{v}(t) &= \langle 2t, 3t^2, 1 \rangle \\
\mathbf{v}(1) &= \langle 2, 3, 1 \rangle \\
\|\mathbf{v}(1)\| &= \sqrt{4 + 9 + 1} = \sqrt{14} \\
\mathbf{T}(1) &= \frac{\mathbf{v}(1)}{\|\mathbf{v}(1)\|} = \left\langle \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right\rangle
\end{align*}
The direction vector
\mathbf{T}(1)
is a unit vector pointing in the direction of motion at
t = 1
.