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
[B] The velocity vector
Key Point 1.5
The velocity is the time rate of change of the position vector; it is a vector:![\[ \vec{v} = \frac{d \vec{r}}{dt} \hspace*{1cm}\mbox{{\rm In component form:}}\hspace*{0.3cm} v_x = \frac{dx}{dt} \hspace*{0.5cm} v_y = \frac{dy}{dt} \]](mastermathpng-3.png)
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:
![\[ \vec{r}= x\hat{i} + y\hat{j} \]](mastermathpng-4.png)
![\[ \frac{d \vec{r}}{dt} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{d x}{dt}\hat{i} + \frac{dy}{dt} \hat{j}} \]](mastermathpng-5.png)

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
- 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.)
- Doing the same with the
term gives the result.
Equate to:
![\[ \vec{v}= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{v_x\hat{i} + v_y\hat{j}} \]](mastermathpng-11.png)
![\[ v_x\hat{i} + v_y\hat{j} = \frac{d x}{dt}\hat{i} + \frac{dy}{dt} \hat{j} \]](mastermathpng-12.png)

Dotting with gives the vy equation.
Visualization
- 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:![\[ \vec{a} = \frac{d \vec{v}}{dt} \hspace*{1cm}\mbox{{\rm In component form:}}\hspace*{0.3cm} a_x = \frac{dv_x}{dt} \hspace*{0.5cm} a_y = \frac{dv_y}{dt} \]](mastermathpng-18.png)
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
Visualization
Learning Resources
![]() | HRW Chapter 4.1-4 |
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