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Math Help - Beginner Velocity/Acceleration question

  1. #1
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    Beginner Velocity/Acceleration question

    Well, 2 actually.

    "In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

    Okay, so what I (think) I've got here is:

    Initial Velocity = 35 m/s
    Acceleration = -20 m/s^2
    D= ?
    Final Velocity = 0 m/s

    What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.


    I feel silly asking, but any help would be tremendously appreciated.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Caturdayz View Post
    Well, 2 actually.

    "In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

    Okay, so what I (think) I've got here is:

    Initial Velocity = 35 m/s
    Acceleration = -20 m/s^2
    D= ?
    Final Velocity = 0 m/s

    What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.


    I feel silly asking, but any help would be tremendously appreciated.
    Maybe this'll help

    t=\frac{v}{a} and d=\frac{1}{2}at^2
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  3. #3
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    Quote Originally Posted by Caturdayz View Post
    Well, 2 actually.

    "In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

    Okay, so what I (think) I've got here is:

    Initial Velocity = 35 m/s
    Acceleration = -20 m/s^2
    D= ?
    Final Velocity = 0 m/s

    What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.


    I feel silly asking, but any help would be tremendously appreciated.
    v_f^2 = v_0^2 + 2a(\Delta x)

    solve for \Delta x ... don't forget to convert to km.
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  4. #4
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    Okay, so I took 35/-20 to get -1.75

    I then took (1/2)(-20*-1.75)^(2)

    Which equals 612.5

    I then took the square root of 612.5, which came out to 24.75 km.
    Last edited by Caturdayz; September 13th 2009 at 04:25 PM.
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  5. #5
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    Using Skeeter's formula I got 0^2 = 35^2 + 2(-20)(deltaX)

    Solving for (deltaX), where i got .0306 kilometers
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Caturdayz View Post
    Okay, so I took 35/-20 to get -1.75

    I then took (1/2)(-20*-1.75)^(2)

    Which equals 612.5

    I then took the square root of 612.5, which came out to 24.75 km.
    hmmm.... We've got a negative time here, and that's not good. Maybe...

    V=\int-20dt=-20t+v_0 where v_0 is the initial velocity which we are given. So, we have

    V=-20t+35

    Now s=\int(-20t+35)dt=-10t^2+35t+s_0 where s_0 is the initial displacement.

    At t=0,s=0\Rightarrow{s}_0=0

    So we have s=-10t^2+35t

    Now set V=0 to determine the time taken to stop, then sub this time into the displacement function to determine the distance taken to stop.
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  7. #7
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    Quote Originally Posted by Caturdayz View Post
    Using Skeeter's formula I got 0^2 = 35^2 + 2(-20)(deltaX)

    Solving for (deltaX), where i got .0306 kilometers
    and that would be correct.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeeter View Post
    and that would be correct.
    Wow, I've really got to read things better...

    "Beginner Velocity/Acceleration question"

    I'm slow sometimes...
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