Displacement and velocity vectors

Position and displacement vectors, motion with constant velocity, speed as the magnitude of velocity, finding when and where a moving object reaches a line, and closest approach using the dot product.

Constant velocity

An object that starts at position $\underset{\sim}{r}_0$ and moves with constant velocity $\underset{\sim}{v}$ has position $$\underset{\sim}{r}(t) = \underset{\sim}{r}_0 + t\underset{\sim}{v}$$ at time $t$. Its speed is $|\underset{\sim}{v}|$, a scalar. Its path is a straight line in the direction of $\underset{\sim}{v}$.

Example 1●

A particle starts at $(2, -1)$ with velocity $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ m/s. Find its position at time $t$, its speed, and where it crosses the $x$-axis.

$\underset{\sim}{r} = \begin{pmatrix} 2 + t \\ -1 + 2t \end{pmatrix}$.

Speed $= \sqrt{1 + 4} = \sqrt5$ m/s.

$-1 + 2t = 0$ when $t = \tfrac12$, at the point $(2.5, 0)$.

Closest approach

The object is closest to a point $P$ when its displacement from $P$ is perpendicular to its velocity, so $(\underset{\sim}{r} - \underset{\sim}{p}) \cdot \underset{\sim}{v} = 0$.

Example 2●●●

A ship's position in km after $t$ hours is $\underset{\sim}{r} = \begin{pmatrix} -10 + 2t \\ 5 + t \end{pmatrix}$. When is it closest to a lighthouse at the origin, and how close does it get?

$\underset{\sim}{r} \cdot \underset{\sim}{v} = 2(-10 + 2t) + (5 + t) = 5t - 15 = 0$, so $t = 3$ hours.

Then $\underset{\sim}{r} = \begin{pmatrix} -4 \\ 8 \end{pmatrix}$, at distance $\sqrt{16 + 64} = 4\sqrt5 \approx 8.94$ km.

Flashcards

Position with constant velocity

$\underset{\sim}{r} = \underset{\sim}{r}_0 + t\underset{\sim}{v}$

Speed

$|\underset{\sim}{v}|$

Condition for closest approach to $P$

$(\underset{\sim}{r} - \underset{\sim}{p}) \cdot \underset{\sim}{v} = 0$

Practice

A particle starts at $(2, -1)$ with velocity $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Find its position when $t = 3$. Enter it as $x, y$.
Find the speed of that particle.
When does that particle cross the $x$-axis?
A ship's position is $\begin{pmatrix} -10 + 2t \\ 5 + t \end{pmatrix}$. At what time is it closest to the origin?
Find that closest distance.
A car has velocity $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$ m/s. Find its speed.
A drone starts at the origin and flies with constant velocity $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ m/s. Find its position after $10$ s. Enter it as $x, y$.
With constant velocity, the path of an object is:
At the moment of closest approach to $P$, the displacement from $P$ is: