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 * Vector Kinematics**

Recall that a parametric vector equation **__r__(t)**, can be used to give the position of a particle at time, t.

The __**distance**__ from the origin of the particle at time, t, will therefore be the magnitude of position. math . \qquad \textbf{distance } = \big| \underline{r}(t) \big| math

Recall that: math . \qquad \underline{r} \centerdot \underline{r} = \big| \underline{r} \big|^2 math

Hence math . \qquad \big| \underline{r} \big| = \sqrt{ \underline{r} \centerdot \underline{r} } = \sqrt{x^2 + y^2} math

The __**displacement**__ between when time(t) is a and time is b will be
 * **__r__(b) – __r__(a)**

The __**average velocity**__ will be displacement divided by time. math . \qquad \textbf{Average Velocity} \big( \text{ between t = a and t = b } \big) = \dfrac{ \underline{r}(b) - \underline{r}(a) }{b - a} math

Refer to Chapter 8: Kinematics for definitions of these terms

Example 1

The path of a particle is described by the equation (distances in metres, time in seconds): math . \qquad \underline{r}(t)=\big( 2t^2-4 \big)\,\underline{i}+\big( t-2 \big)\,\underline{j} math

a) Find the position of the particle after 3 seconds math \\ . \qquad \underline{r}(3)=\big( 2\times 3^2-4 \big)\,\underline{i}+\big( 3-2 \big)\,\underline{j} \\ . \\ . \qquad \underline{r}(3)=14\,\underline{i}+\underline{j} math

b) Find the distance from the origin after 3 seconds math \\ . \qquad |\underline{r}(3)| =|14\,\underline{i}+\underline{j}| \\ . \\ . \qquad |\underline{r}(3)| =\sqrt{14^2+1^2} \\ . \\ . \qquad |\underline{r}(3)| =\sqrt{197} \\ . \\ . \qquad |\underline{r}(3)| \approx 14 m math

c) Find the initial position of the particle math \\ . \qquad \underline{r}(0)=\big( 2\times 0^2-4 \big)\,\underline{i}+\big( 0-2 \big)\,\underline{j} \\ . \\ . \qquad \underline{r}(0)=-4\,\underline{i}-2\,\underline{j} math

d) Hence find the displacement of the particle after 3 seconds math \\ . \qquad \text{displacement} = \underline{r}(3)-\underline{r}(0) \\ . \\ . \qquad = \big( 14\,\underline{i}+\underline{j} \big) - \big( -4\,\underline{i}-2\,\underline{j} \big) \\ . \\ . \qquad = 18\,\underline{i}+3\,\underline{j} math

e) Hence find the Average Velocity in the first 3 seconds math . \qquad \textbf{Ave Velocity } = \dfrac{ 18\,\underline{i}+3\,\underline{j} }{3} = 6\,\underline{i}+ \underline{j} \; \textit{ m/s} math

Differentiating Vectors

The **__instantaneous velocity__** of a particle is the derivative of the position vector math \\ \text{Given } \\ . \qquad \underline{r}(t)=x(t)\,\underline{i}+y(t)\,\underline{j} \\. \\ \text{Velocity is} \\ . \qquad \underline{v}(t)=\underline{r}'(t)=x'(t)\,\underline{i}+y'(t)\,\underline{j} math

{the velocity vector gives the __direction of travel__}

The **__speed__** of a particle is the magnitude of velocity, **|__v__(t)|**

Similarly, the **__acceleration__** of a particle is the derivative of the velocity vector.

To differentiate a vector, simply differentiate the **__i__** coeeficient and differentiate the **__j__** coefficient seperately.

Example 2

The path of a particle is described by the equation (distances in metres, time in seconds): math . \qquad \underline{r}(t)=\big( 2t^2-4 \big)\,\underline{i}+\big( t^3-6t+3 \big)\,\underline{j} math

a) Find the velocity vector math . \qquad \underline{v}(t)=4t\,\underline{i}+ \big( 3t^2-6 \big)\,\underline{j} math

b) Find the acceleration vector math . \qquad \underline{a}(t)=4\,\underline{i}+6t\,\underline{j} math

c) Find the speed at time, 3 math \\ . \qquad \underline{v}(3) = 12\,\underline{i}+ 21\, \underline{j} \\ . \\ . \qquad |\underline{v}(3)| =\sqrt{12^2+21^2} \\ . \\ . \qquad |\underline{v}(3)| =\sqrt{585} \\ . \\ . \qquad |\underline{v}(3)| \approx 24 \; \textit{ m/s} math

d) Find the time at which the particle is travelling parallel to the x-axis.
 * {__Direction of travel__ is given by the __velocity__ vector, so this occurs when the coefficient of j is 0 in the velocity vector} **

math \\ . \qquad 3t^2-6=0 \\. \\ . \qquad t^2=2 \\. \\ . \qquad t=\sqrt{2} \; seconds math


 * {discard negative because t is always positive by definition} **

Alternate Notation

When doing kinematics with vectors, we sometimes use dots above the vector names to indicate derivatives with respect to t:

Let position be __x__(t)

math \\ . \qquad \underline{v}(t)=\underline{\dot{x}}(t) \\. \\ . \qquad \underline{a}(t)=\underline{\dot{v}}(t)=\underline{\ddot{x}}(t) math

Vector Calculus in 3 Dimensions

Using __k__ as the unit vector in the direction of the //z//-axis (usually pictured coming up out of the page perpendicular to the surface), it is easy to do these types of calculations in 3D

Example Find the acceleration vector **__a__(t)** for the object whose position vector is: math . \qquad \underline{r}(t)=e^{2t}\underline{i}-(t^3+t)\underline{j}+\big( t\,\log_e(t)\big)\underline{k} math

math . \qquad \underline{v}(t)=2e^{2t}\underline{i}-(3t^2+1)\underline{j}+\big( 1+\log_e(t) \big) \underline{k} math
 * Differentiate once to get velocity **

math . \qquad \underline{a}(t)=4e^{2t}\underline{i}-6t\underline{j}+\big( \frac{1}{t} \big) \underline{k} math .
 * Differentiate again to get acceleration **