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

The Position, Velocity, and Acceleration Vectors

Two- and Three-Dimensional Motion with Constant AccelerationPrevious Section
Tangential and Radial AccelerationNext Section
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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:



Wize Tip
Reduce errors by re-writing the coordinates A(x, y, z) and B(x, y, z) BEFORE building the position vector. Don't pull the numbers right from the figure as you build it - easy to make mistakes!

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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.
r(t)⟶=x(t)ı^+y(t)ȷ^+z(t)k^\boxed{\stackrel{\longrightarrow}{r(t)}=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat k}







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

Δr⟶=rf→−rı^→=Δxı^+Δyȷ^+Δzk^\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

vave⟶=Δr⟶Δt\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
v(t)⟶=vxı^+vyȷ^+vzk^=lim⁡Δt→0Δr⟶Δt=dr(t)⟶dt\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}}
dr(t)⟶dt=dx(t)dtı^+dy(t)dtȷ^+dz(t)dtk^\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.
Wize Tip
Average speed and instantaneous speed could be found similarly if displacement vector is replaced by distance in above equations.



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Example: Position and Velocity

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

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

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

Integrate to find position:

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

=9.82 t2−5t+C\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)=0s(0)=0. Put this in the equation to get:

0=9.82 (0)2−(0)t+C\displaystyle 0=\dfrac{9.8}{2} \ (0)^2-(0)t+C

⇒C=0\Rightarrow C=0

Therefore the position function is:

s(t)=9.82 t2−5t\displaystyle s(t)=\dfrac{9.8}{2}\ t^2-5t

At t=3t=3 the position is:

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

=29.1=\boxed{29.1}

Practice: Position Vectors

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

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Example: Velocity and Acceleration in 3D


The position of a particle in 3D is given by
r⃗(t)=(10t+t2+7)i^+2j^+4tk^  m\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
v⃗=dr⃗dt\vec{v} = \dfrac{d\vec{r}}{dt}

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

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

Part b)

Average velocity:

v⃗average=Δr⃗Δt=r⃗(5)−r⃗(3)5−3\vec{v}_{average}=\dfrac{\Delta\vec{r}}{\Delta t}=\dfrac{\vec{r}(5)-\vec{r}(3)}{5-3}

v⃗average=[82i^+2j^+20k^]−[46i^+2j^+12k^]2=36i^+8k^2=18i^+4k^ m/s\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:
a⃗instant=d2r⃗dt2=2i^ ms2\vec{a}_{instan t}=\dfrac{d^2\vec{r}}{dt^2}=2\hat{i} \ \dfrac{m}{s^2}




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Q:\textbf{Q:} A particle moves back and forth along a line such that it has a position function s=f(t)=t3−9t2+24ts=f(t)=t^3-9t^2+24t, where tt is measured in seconds and ss in meters. Find the velocity, speed and acceleration at t=5t=5.