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S1.1 One dimensional particle kinematics

[A] Context

IIs time simple?

[B] Displacement

Consider a particle whose x coordinate varies smoothly but arbitrarily with t.
Graphic - No title - xversustA

Commentary

Commentary

The displacement
  • is a vector –sense given by sign in 1D
  • is to be distinguished from ‘distance’
 

Example

Example

Graphic - No title - onedwalk
  • distance travelled: 7 units
  • displacement: –1 unit
 

[C] Velocity

From the variation of x with t we define the average velocity over a time interval as
Graphic - No title - xversustB

Key Point 1.1

The instantaneous velocity is defined as the average velocity over the next infinitesimally small time interval:
\[ v= \lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \]
Geometrical significance: v is gradient of x-t graph.

Commentary

Commentary

  • v is a 1D vector
  • the magnitude of v is the speed
 

[D] Acceleration

Now consider the variation of v with t.

The average acceleration over a time interval is defined as

Graphic - No title - vversust

Key Point 1.2

The instantaneous acceleration is the average acceleration over the next infinitesimally small time interval:
\[ a= \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \]
Geometrical significance: a is gradient of v-t graph.

Commentary

Commentary

  • acceleration is a vector –sense given by sign in 1D
  • beware confusion with colloquial usage
 

Example

Example

Graphic - No title - decellern
  • consider particle on spring, launched to left
  • plain-man-speak:

    particle slows down or decelerates

  • physics-speak:

    particle has a positive acceleration

 

[E] Integral forms of key equations

Key Point 1.3

The integral forms of the x-t and v-t relationships are:
\[ \Delta x= \int_{t_1}^{t_2} v dt \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \Delta v= \int_{t_1}^{t_2} a dt \]
In particular the displacement is the area under the v-t curve.

Analysis

Analysis

  • Consider the equation
    \[ \frac{dx}{dt}=v \]
  • Rewrite as relationship between infinitesimals:

    dx=vdt

  • Integrate both sides:
    \[ \Delta x= x_2-x_1 = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\int_{x_1}^{x_2} dx = \int_{t_1}^{t_2} v dt} \]
  • The integral is the area under the v-t curve
  • In the same way the equation
    \[ \frac{dv}{dt}=a \]
    can be integrated to give
    \[ \Delta v= v_2-v_1 = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\int_{t_1}^{t_2} a dt} \]
 

[F] Constant acceleration equations

Key Point 1.4

For 1D motion at constant acceleration a, the position and velocity (x and v) at the end of a time interval (t) are related to those at the beginning of the interval (x0 and v0) by
\begin{eqnarray*} v&=&v_0 + at \hspace*{1cm} (a)\\ x-x_0& =& v_0 t +\frac{1}{2}at^2 \hspace*{1cm} (b)\\ v^2&=&v_0^2 +2a(x-x_0)\hspace*{1cm} (c) \end{eqnarray*}

Analysis

Analysis

  • Let the times t1 and t2 be 0 and t respectively.

    Denote the associated positions and velocities by x0, v0 and x, v

  • Recall definition

    \[ a_{\mbox{\rm av}} = \frac{\Delta v}{\Delta t} =\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{ \frac{ v_2 -v_1}{t_2 -t_1}= \frac{ v -v_0}{t}} \]

    But for constant acceleration

    \[ a_{\mbox{\rm av}} = a \hspace*{1cm}\mbox{\rm (constant)} \]
    Thus

    \[ a = \frac{ v -v_0}{t} \]

    Rearranging gives

    v=v0+at

  • Next recall the definition

    \[ v_{\mbox{\rm av}} = \frac{\Delta x}{\Delta t}= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{ x_2 -x_1}{t_2 -t_1} =\frac{ x -x_0}{t}} \]
    But for constant acceleration
    \[ v_{\mbox{\rm av}} = \frac{v_0 +v}{2} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{v_0+\left[ v_0 +at \right]}{2} = v_0+\frac{1}{2}at} \]

    Combining these two equations gives

    \[ \frac{x -x_0}{t} = v_0+\frac{1}{2}at \]

    or

    $$x-x_0 = v_0 t +\frac{1}{2}at^2$$

  • Finally eliminating t between

    \[ v=v_0 + at \hspace*{1cm}\mbox{\rm and}\hspace*{1cm} x-x_0 = v_0 t +\frac{1}{2}at^2 \]

    gives DIY!

    $$v^2=v_0^2 +2a(x-x_0)$$

 
MGoing a different route

Visualization

Graphic - No title - constaccn
TAcceleration of a Ball Thrown Vertically
TVelocity of a Ball Thrown Vertically

Learning Resources

Textbook: HRW Chapter 2
Course Questions:
Self-Test Questions: