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More theoretical questions about acceleration and velocity

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    This question is from a theory stand point.
    I understand the definitions of velocity and acceleration separately. For some reason, I can not conceptualize them in equation form. How do you really know a*t= velocity or d/t=v ? Is there a proof for this or something? I need to understand the reasoning better. The professor for my class is like it just is. Any suggestions on how to make understanding the equations and relationships in physics more concretely in my head? I need more background info then my book is giving to help me to better understand how to derive the constant acceleration equations.

    2. Relevant equations

    constant acceleration of motion
    3. The attempt at a solution
    read text/google search
     
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2

    PeterO

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    Homework Helper

    If a crowd of 60 000 people entered a sports stadium in 40 minutes, then they entered at an average rate of 90 000 per hour.
    Is that description OK
    Rate of Entry = number of people / time

    If a car covers 60 km in 40 minutes, then it was travelling at a rate of 90 km each hour. Rate of covering distance = distance covered / time.

    We call THAT rate speed, symbol V, the distance covered is given symbol d and time is given symbol t; so v = d/t

    So rather than a proof - it is a definition.

    Try this: Prove that the four legged piece of furniture people sit on at a restaurant is a chair. Of course the long legged chair at the bar is actually a stool !!!

    Also if a car increases its speed from 20 km/h to 80 km/h in 40 seconds, it was increasing its speed at the rate of 60 km/h per minute.
    We call the rate of increase of speed acceleration (a) so a = Δv/t or Δv = a*t

    Now if the initial speed was 0, then the change in speed (Δv) will equal the final speed v so that would look like a*t = v
     
  4. Sep 27, 2012 #3

    I understand this already. My question is more about constant acceleration of motion equations. Is there a good way to visualize this?
     
  5. Sep 27, 2012 #4

    PeterO

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    Homework Helper

    The equations of motion are mathematical summaries of a simple velocity-time graph.

    If a body is accelerating, its velocity-time graph is a sloping, straight line. The gradient of the line is the rate of change of the quantity graphed, defined as acceleration "a".

    If the initial velocity is called "u", then that us the intercept on the vertical axis.
    After a period of time "t", the velocity will have increased to a final velocity "v".
    Thus there is a co-ordinate on the line (t,v).

    If you draw such a graph, you add a pair of reference lines: a horizontal line at the initial velocity, and also at the final velocity.
    The gradient of the line [rise / run] tells us that the increase in velocity over this time interval is given by a*t.
    Thus the final velocity is given by:

    v = u + at

    The displacement during this time is given by the area under the graph.
    Those horizontal lines show that area in 3 ways.

    #1 - a lower rectangle plus a triangle [length x width plus 1/2 x base x height]
    The rectangle has length t, and width u --> area ut
    The triangle has base t, and height a*t so are 0.5 * t * a*t = 0.5at2

    using symbol s for displacement we have

    s = ut + 0.5at2

    #2 - a large rectangle, with a triangle removed from the top left.
    The large rectangle has length, t, and width, v, so area vt

    as above we get

    s = vt - 0.5at2

    #3 - before the horizontal lines were drawn, the graph represented a trapezium on its side - the parallel sides are the vertical lines along the vertical axis, and the ordinate t.

    The area of a trapezium is given by area = height * (base + top)/2
    using our symbols that means

    s = t(v+u)/2

    That gives us 4 equations of motion - all mathematical summaries of one simple graph.

    We can do one more algebraic manipulation using the first and last of them.

    s = t(v+u)/2 and v = u + at

    The second can be transformed to show t = (v-u)/a

    substituting for t in the first we get

    s = (v-u)/a * (v+u)/2

    multiply both sides by 2a to eliminate the "fractions"

    2as = (v-u)(v+u)

    The right hand side is the classic "difference of two squares" expression
    so

    2as = v2 - u2

    re-arranging so that there are no minus signs we get the more familiar

    v2 = u2 + 2as
     
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