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Math Help - acceleration and velocity

  1. #1
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    acceleration and velocity

    I don't understand the reasoning behind a few steps in this problem:

    What constant acceleration is required to increase the speed of a car from 30 mi to 50 mi in 5 seconds?


    a(t) = C = v'(t)

    v'(t) = Ct + K

    v(0)=30= C(0) + k

    k = 44 ?

    v(5) = (5/6)88 ft/s ?

    v(5) = 5c + 44

    5c = 88/3 = 88/15 ft/s^2


    I think I know where k = 44 comes from.

    50 mi - 30mi = 20mi

    20mi/5s = 4mi/s(30mi) = 120mi/s

    5280/120 = 44

    But I don't understand where the other number comes from.
    Last edited by zachb; April 25th 2007 at 12:17 PM.
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  2. #2
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    Quote Originally Posted by zachb View Post
    I don't understand the reasoning behind a few steps in this problem:

    What constant acceleration is required to increase the speed of a car from 30 mi to 50 mi in 5 seconds?
    What unit of speed is a "mi"?

    RonL
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  3. #3
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    Quote Originally Posted by zachb View Post
    I don't understand the reasoning behind a few steps in this problem:

    What constant acceleration is required to increase the speed of a car from 30 mi to 50 mi in 5 seconds?
    Lets assume the change in speed is from 30 mph to 50 mph, that is a change
    of 20 mph. Convert this to fps:

    dv=[20*5280]/[60*60] = 29.333.. fps

    So the speed changes by 29.333.. fps in 5 seconds so the avaerage
    acceleration is 29.333../5 ~= 5.866.. f/s^2

    Which is the constant acceleration that will change the speed by the
    required amount in 5 seconds.

    RonL
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  4. #4
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    Yeah, I don't follow any of that.

    I would appreciate it if somone can explain why v(5) = 5/6(88) or 440/6, and please use fractions, not decimals.

    Thanks.
    Last edited by zachb; April 25th 2007 at 12:50 PM.
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  5. #5
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    Quote Originally Posted by zachb View Post
    I don't understand the reasoning behind a few steps in this problem:

    What constant acceleration is required to increase the speed of a car from 30 mi to 50 mi in 5 seconds?


    a(t) = C = v'(t)

    v'(t) = Ct + K

    v(0)=30= C(0) + k

    k = 44 ?
    You seem to have v(0) = 30 = C.0 + k = k

    so in these units k=30.

    But watch your units, in you next section you are using fps, while here you
    have mi (whatever they are).

    Now if we convert 30 mph to fps we have 44 fps, and if we are working in fps
    we have:

    v(0) = 44 = C.0 + k = k

    so in fps k does equal 44 fps.

    RonL
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  6. #6
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    But watch your units, in you next section you are using fps, while here you
    have mi (whatever they are).
    I made a mistake it my first post, it should have been mi/h (mph). Anyway, I'm sure you know what it should be.

    Now if we convert 30 mph to fps we have 44 fps, and if we are working in fps
    we have:

    v(0) = 44 = C.0 + k = k

    so in fps k does equal 44 fps.
    Captin, I know that it should equal 44 but I don't know why. Also, you're not answering my other question as I still don't know why v(5) = 440/6 and it doesn't look like I'm going to get any help with that.

    Anyway, thanks . . .

    Can someone else explain how to convert from mph to fps, completely, and by using fractions only?
    Last edited by zachb; April 25th 2007 at 06:36 PM.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    I made a mistake it my first post, it should have been mi/h (mph). Anyway, I'm sure you know what it should be.


    Captin, I know that it should equal 44 but I don't know why. Also, I still don't know why v(5) = 440/6 and it doesn't look like I'm going to get any help with that.

    Anyway, thanks . . .

    Can someone else explain how to convert from mph to fps, completely, and by using fractions only?
    Ok, i hope you get this.

    first let's do the conversions:
    1 mile = 5280 ft
    1 hour = 3600 sec

    ...30 miles...........5280 ft...........1 hour..........30(5280) ft
    .-----------..x..-----------..x..----------..=..-------------..= 44 ft/s
    .....hour..............1 mile...........3600 sec..........3600 sec

    did you get that method? i used conversion factors that you know of, and i placed them in such a way, that the units i don't want cancel out, and the units i do want stay. (i hate this method, but people seem to understand it easier). similarly,

    ...50 miles...........5280 ft...........1 hour..........50(5280) ft
    .-----------..x..-----------..x..----------..=..-------------..= 220/3 ft/s
    .....hour..............1 mile...........3600 sec.........3600 sec

    Now let's get to the meat of the matter. we now know 30 mph = 44 ft/s, and 50 mph = 220/3 ft/s

    a(t) = C = v'(t)
    => v(t) = int{a(t)}dt = int{C}dt = Ct + D
    so v(t) = Ct + D

    at t = 0, v(t) = 44 ft/s
    => 44 = C(0) + D
    => 44 = D
    so v(t) = Ct + 44
    Now, when t = 5, we attain 220/3 ft/s
    => 220/3 = C(5) + 44
    => 5C = 220/3 - 44 = 88/5
    => C = (88/5)/5 = 88/15

    so v(t) = (88/15)t + 44
    thus the constant of acceleration is 88/15


    if you have problems with my conversion procedure i can show you another way, the one i usually use when i'm writing stuff out
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  8. #8
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    I understand that conversion method just fine. My professor absolutely refuses to show work like that, so when he did this problem I had no idea where he was getting those numbers from.

    Thank you. I really appreciate it.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    it won't hurt to show you the other method anyway. this method i think is a bit simpler. we simply replace what we don't want with what we do want.

    1 mile = 5280 ft
    1 hour = 3600 sec

    this means wherever i see mile i can replace it with 5280 ft, becuase they are equal, they're the same thing. and wherever i see hour i can replace that with 3600 sec since they're the same thing. so,

    30 miles/hour = 30 (5280 ft)/(3600 sec) = 30(5280)/3600 ft/sec = 44 ft/sec
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