S1.2 Kinematics in two (or three) dimensions
[A] Position and displacement vectors
- In space dimension d = 2 (or 3...) the position of a particle is specified by:
d coordinates (x, y …)
or
- a d-dimensional vector

Visualization
The position vector can be written as The displacement vector is![\[ \vec{r}= x\hat{i} + y\hat{j} \]](mastermathpng-1.png)
![\[ \Delta\vec{r}= \Delta x\,\hat{i} + \Delta y\,\hat{j} \]](mastermathpng-2.png)
[B] The velocity vector
Key Point 1.5
The velocity is the time rate of change of the position vector; it is a vector:Commentary
Reveal
Hide

Differentiate:![\[ \frac{d \vec{r}}{dt} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{d x}{dt}\hat{i} + \frac{dy}{dt} \hat{j}} \]](mastermathpng-5.png)
Then
Take the dot product of each side with î:
Commentary
Write:
Help?
Combining ‘vectors’ and ‘differentiation’ may feel like a ‘bridge too far’. But just take it step by step.Realise that
is simply short for![\[ x\hat{i} +y \hat{j} \]](mastermathpng-7.png)
- Recognise that the things that depend on time are the scalar
quantities
x and y; the unit vectors î and
are ‘constant’ (independent of time)...they are the signposts that define our coordinate axes. - Using the fact that î is constant we can write
since the unit vector can be pulled out from the differentation (which is only interested in things that do depend on time.)
![\[ \frac{d}{dt} (x\hat{i}) = \hat{i} \frac{dx}{dt} \]](mastermathpng-9.png)
- Doing the same with the
term gives the result.
Equate to:
Dotting with
gives the vy equation.
Visualization
![\[ \vec{v}= \lim_{\Delta t \rightarrow 0} \frac {\Delta \vec{r}}{\Delta t} \]](mastermathpng-16.png)
- As Δt→0, Δr→0
becomes tangent- And so …
Key Point 1.6
The velocity vector is always tangential to the particle path.[C] The acceleration vector
Key Point 1.7
The acceleration is the time rate of change of the velocity; it is a vector:The acceleration is non-zero if
the velocity vector is changing in direction
[ the path is not straight! ]
the velocity vector is changing in magnitude
[ the speed is changing! ]
[D] Constant acceleration equations
- If the acceleration vector is constant, the 1D kinematic
equations (Key Point 1.4) can be applied
to the motion associated with each of the
axes.

Example
Equation [EQ-constaccn in Node onedkinematics](a) applied to the y-motion gives
vy=vy0+aytEquation [EQ-constaccn in Node onedkinematics](b) applied to the x-motion gives
![\[ \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{x-x_0= v_{x0} t + \frac{1}{2} a_xt^2} \]](mastermathpng-19.png)
Visualization
Learning Resources
| HRW Chapter 4.1-4 | |


