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Calculus intuitive and physical approach - please help me understand.

  1. Sep 12, 2014 #1
    Hi if i have a function that relates time and distance travelled, is this correctly understood: (please tell me what is correct and what is wrong - it would help me understand)

    The function is
    f(t)=16(t^2)

    1. f(4) = 256 This tells me that at the END of 4 seconds (or is it at the beginning?) i travelled a total distance of 256 ft so the number 256 has the units feet

    2. d(f(5))/dt = 256 tells me that at 5 seconds my instantaneous speed is 256 feet per seconds, so the units are feet/seconds

    My final question is this: How can an object have an instantaneous speed of 256 ft/seconds at 5 seconds, if the object has only travelled 256 feet by the END of 4 seconds (which is close to 5 seconds)??

    The idea that f(4) means "At the end of 4 seconds" is something i got from "Calculus an intuitive and physical approach" says on page 24 at the top.
     
  2. jcsd
  3. Sep 12, 2014 #2

    Mentallic

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    There is no such thing as "at the end of 4 seconds" versus "at the start of 4 seconds". That's like saying 4.0 and 4.9 are at the start and end of 4.

    On a stopwatch, you'll start at t=0.00, then after 1/10 of a sec, it's t=0.10. After 1 second you're at t=1.00. After 1 second and 990 milliseconds, you're at t=1.99 secs. Don't think of this as being "at the end of 1 second". It's almost 2 seconds.
    When you're at t=4, this means 4 seconds have passed. A moment before 4 seconds the stopwatch would've been at 3.99 secs. A moment after 4 seconds it's at 4.01 secs.

    And just a little precaution: Unless f(t) is said to be measured in feet, or if its implied in the class, then you should stick to unspecified units.
    If you're studying physics then you should start using the metric system also ;)

    Check your numbers again. What is df/dt?

    Well, besides the fact that you made a little slip up with your instantaneous speed calculation, it's perfectly possible to have such a situation that you described.

    Let's say you start from a stationary position and begin accelerating at 1 ft/s2 which in words is "1 foot per second per second", this means that after 1 second, you've increased your speed by 1 foot per second, so you've gone from 0 ft/s to 1 ft/s. In this time, you must have travelled more than 0 feet but less than 1 foot. The reason for this is because if you were travelling 1 ft/s for the entire second, then you'd move 1 foot in distance, but you were travelling slower than 1 ft/s for the entire second (until you reached that speed at t=1).

    Your study of calculus and motion will help you answer questions such as how far you've travelled in that 1 second, etc.
     
  4. Sep 12, 2014 #3
    Hi, Thank you for the end of your reply

    I just don't understand what the author means by "at the end of 5 seconds" so let me show you his example:

    He uses that increment method to find the derivative of s=16t^2 + 96t

    he states "We first calculate the distance fallen at the end of 5 +h seconds. this distance is"

    s5 +k = 16(5+h)^2+96(5+h) = 880+256h +h^2

    He then states "The distance falled at the end of 5 seconds is"
    s5 = 16*5^2 +96*5=880

    I don't understand what is meant by "at the end of 5 seconds".
     
  5. Sep 12, 2014 #4

    Mentallic

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    I think I can see what he's trying to get across to the audience, but it was clearly a swing and a miss.

    So the distance travelled is

    [tex]s(t)=16t^2+96t[/tex]

    so at 5 seconds (after 5 seconds is valid terminology too) we have travelled

    [tex]s(5)=16(5)^2+96(5)=880[/tex]

    What he's done is converted the function into something that starts at t=5 as opposed to t=0.

    If we compute s(5+t) then we get

    [tex]s(5+t)=16(5+t)^2+96(5+t)=880+256t+t^2[/tex]

    So to now find what distance we've travelled after 5 seconds, we would usually need to calculate s(5) but now we have a function that starts at 5 seconds since t=0 gives us s(5) as its initial value, and in order to try and avoid confusion, we'll give this a new function named S and rather than the variable t we'll use h.

    [tex]S(h)=880+256h+h^2[/tex]

    hence

    [tex]S(0)=880[/tex]

    this is the same distance travelled as s(5) because S(t)=s(5+t) so using t=0 we get S(0)=s(5).


    If all of this confuses you, then you can just ignore it and solve for s(5) like you usually do.
     
  6. Sep 12, 2014 #5
    thank you metallic, that makes more sense. I'm interested in learning the language of calculus but also in learning how use it in physics, and how to set up relationships between different variables in real life since i want to be a scientist (so i want to be good at interpreting math and understanding how i can use to set up formulaes from real world data)

    Would you recommend any other books for this purpose?
     
  7. Sep 12, 2014 #6

    Mentallic

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