Wize University Physics Textbook (Master) > Motion in Two Dimensions

The Position, Velocity, and Acceleration Vectors

Position Vector (r) = fixed vector that locates a point in space relative to another point. Some call it the "distance" vector, as its magnitude is the distance between the endpoints.

can be used to form a unit vector
can be used to determine the length or distance between 2 points

IF USE THE ORIGIN AS THE STARTING POINT:

ex: so two points are O (0,0,0) and A (Ax, Ay, Az)

IF the ORIGIN is NOT the STARTING POINT:

ex: so two points are A (Ax, Ay, Az) and B (Bx, By, Bz), then rAB can be written as:

Two- and Three-Dimensional Kinematics: Part 1

In general, the position of a particle is represented by a 3D vector that is changing with time. The components of this 3D vector are a function of time too.

\boxed{\stackrel{\longrightarrow}{r(t)}=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat k}

Image illustrating Two- and Three-Dimensional Kinematics : Part 1

Displacement is a vector that shows the change in position of the particle. The unit of displacement is m.

\stackrel{\longrightarrow}{\Delta r}=\stackrel{\rightarrow}{r_f}-\stackrel{\rightarrow}{r_{\hat{\imath}}}=\Delta x\hat{\imath}+\Delta y\hat{\jmath}+\Delta z\hat k

Distance is the length of the path travelled by the particle

Average velocity is defined as

\boxed{\stackrel{\longrightarrow}{v_{ave}}=\frac{\stackrel{\longrightarrow}{\Delta r}}{{\Delta t}}}

Instantaneous velocity is the velocity of an object in a specific time or position which in general might be different from the average velocity

\stackrel{\longrightarrow}{v(t)}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat k=\lim_{\Delta t\to0}\frac{\stackrel{\longrightarrow}{\Delta r}}{{\Delta t}}=\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}

\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}=\frac{dx(t)}{dt}\hat{\imath}+\frac{dy(t)}{dt}\hat{\jmath}+\frac{dz(t)}{dt}\hat k

So, basically, the instantaneous velocity is obtained by taking the derivative of position with respect to time.

Example: Position and Velocity

Given that the velocity of a body, starting from the origin, and moving along the x-axis, is $\displaystyle v(t)=9.8t-5$ , find the position of the moving body at $t=3$ .

Since the velocity is the derivative of position, position will be the integral of the velocity:

$\displaystyle v\left(t\right)=\frac{ds}{dt}=9.8t-5$

Integrate to find position:

$\displaystyle s(t)=\int{(9.8t-5)}\ dt$

$\displaystyle =\dfrac{9.8}{2}\ t^2-5t+C$

Find the constant using the fact that the object starts at the origin, that is, position is zero when time is zero: $s(0)=0$ . Put this in the equation to get:

$\displaystyle 0=\dfrac{9.8}{2} \ (0)^2-(0)t+C$

$\Rightarrow C=0$

Therefore the position function is:

$\displaystyle s(t)=\dfrac{9.8}{2}\ t^2-5t$

At $t=3$ the position is:

$s(3)=\dfrac{(9.8)(9)}{2}-5(3)$

$=\boxed{29.1}$

Practice: Position Vectors

Given the points $𝑃(1,2),\space 𝑄(0,1),$ and $𝑅(1,−1),$ find $\overrightarrow{𝑃𝑄}+ \overrightarrow{𝑅𝑄} .$

a. $\left(-2,1\right)$

b. $\left(2,\ -1\right)$

c. $\left(2,1\right)$

d. $\left(-1,\ 2\right)$

e. $\left(1,2\right)$

I don't know

Example: Velocity and Acceleration in 3D

The position of a particle in 3D is given by

$\vec{r}(t)=(10t+t^2+7)\hat{i}+2\hat{j}+4t\hat{k} \ \ m$

a) Calculate instantaneous velocity at t =5 sec

b) Calculate the average velocity for an interval (3-5 sec).

c) Calculate the instantaneous acceleration at t =1 sec.

Solution:

Part a)

We know that Instantaneous velocity is

$\vec{v} = \dfrac{d\vec{r}}{dt}$

$\frac{d\vec{r}}{dt}=(10+2t)\hat{i}+0\hat{j}+4\hat{k}$

$\vec{v}(t=5)= (10 + 2 (5))\hat{i}+4\hat{k} = 20\hat{i}+4\hat{k} \ m/s$

Part b)

Average velocity:

$\vec{v}_{average}=\dfrac{\Delta\vec{r}}{\Delta t}=\dfrac{\vec{r}(5)-\vec{r}(3)}{5-3}$

$\vec{v}_{average} = \dfrac{[82\hat{i}+2\hat{j}+20\hat{k}]-[46\hat{i}+2\hat{j}+12\hat{k}]}{2}=\dfrac{36\hat{i}+8\hat{k}}{2}= 18\hat{i}+4\hat{k}\ m/s$

Part c)

Instantaneous acceleration:

$\vec{a}_{instan t}=\dfrac{d^2\vec{r}}{dt^2}=2\hat{i} \ \dfrac{m}{s^2}$

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$\textbf{Q:}$ A particle moves back and forth along a line such that it has a position function $s=f(t)=t^3-9t^2+24t$ , where $t$ is measured in seconds and $s$ in meters. Find the velocity, speed and acceleration at $t=5$ .

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The Position, Velocity, and Acceleration Vectors

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Two- and Three-Dimensional Kinematics: Part 1

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Example: Position and Velocity

Practice: Position Vectors

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Example: Velocity and Acceleration in 3D

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