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Braking Distance and Speed

  1. Jun 3, 2009 #1

    MxS

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    A car was involved in an accident. There are skid marks of 40m and the car has a constant deceleration of 7m/(8^2) when braking. How fast was the car travelling when it started braking? (Not actual question I'm trying to solve, but very similar)

    Hint:
    f(x)= location
    f′(x) = speed
    f′′(x) = acceleration

    How would you work something like this out? Probably simple, but I don't do physics and have no idea what to do.

    Would greatly appreciate any help or guidance.
     
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2

    cepheid

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    Re: Breaking Distance and Speed

    I think there are probably two ways. Since the acceleration is constant and you know the total distance d over which the motion occured, you can just use the equations of kinematics (which are derived using calculus) to deduce the initial velocity.

    Or, since you seem to know calculus, and you know how position, velocity and acceleration are related, you could just integrate the acceleration (which is given) in order to determine the velocity as a function of time, and then integrate that to determine the position as a function of time.
     
  4. Jun 3, 2009 #3

    MxS

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    thanks for the quick reply. :smile:

    i think the lecturer wants us to go for the 2nd option out of the 2 (which would explain why he gave the hints about how the above are related).

    sorry if i'm being stupid but i'm really confused about how to do it.

    if i integrate the acceleration i get -7m^2/128
    then what do i do?

    sorry, i really don't get this. :(
     
  5. Jun 3, 2009 #4

    cepheid

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    Sorry, but 7m/8^2 makes no sense, so I assumed it was a typo. Acceleration is the rate at which the velocity is changing in (metres per second) PER second. Hence, acceleration is measured in units of m/s^2. Is it possible that that is what your problem actually says (i.e. the 8 is an s)?
     
  6. Jun 3, 2009 #5

    MxS

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    how stupid of me. it is actually an s yes. looks like an 8 on the printed sheet but i got the pdf off the website and zoomed in. :redface:

    ok so i have:
    7m/s^2

    what do i integrate that in respect to?
     
  7. Jun 3, 2009 #6

    cepheid

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    Oh yeah, that's the other thing. The derivatives in your first post should be with respect to time, not x. This is because velocity is the rate of change of position with time. In other words, it tells you by how much your position changes per unit time. Similarly, acceleration is the rate of change of velocity with time. It tells you how much your velocity changes per unit time.
     
  8. Jun 3, 2009 #7

    MxS

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    thanks again for your help.

    i'm looking up how to integrate acceleration to find the velocity, i've came across this:
    http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity_14.gif

    is this what i need to be doing? i'm not sure which values i substitute in where from my problem. :confused:
     
    Last edited by a moderator: Apr 24, 2017
  9. Jun 3, 2009 #8

    cepheid

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    Hmm, okay, so you don't know calculus?

    You're on the right track. I would use indefinite integrals, just to get the velocity and position as functions of time.

    You know that:

    [tex] v(t) = \int a(t) dt = \int a dt = at + C [/tex]

    The constant of integration is determined by the initial conditions. At t = 0, we have

    [tex]v_0 = a(0) + C = C [/tex]

    Therefore, C = v(0), and we get:

    [tex] v(t) = v_0 + at [/tex]

    In addition to v(0), you have another unknown, which is the time required to decelerate. For this you need another equation, which you can get by integrating v(t) to get the position as a function of time.
     
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