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Velocity rate question

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A coin is dropped from a height of 800 feet. The height, s(measured in feet), at time, t(measured in seconds is given by:

    s(t)= Vit + 1/2 a t^2 + So

    a) find the average velocity on the interval [1,3]
    b) find the instantaneous velocity when t= 3s.
    c) how long does it take the coin to hit the ground?
    d) find the velocity on the interval [1,3]


    2. Relevant equations
    a= -32 ft/s^2
    h=800 ft


    3. The attempt at a solution
    a) I understand that you have to derive the position function to get the velocity function but I'm having trouble doing it. To find the average though i can do f(3)-f(1)/3-1
    b) I can't find it because I'm having trouble with the derivative
    c &d) I do not understand it because I missed the class that my teacher went over it when i was sick
     
  2. jcsd
  3. Nov 10, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    For (a): the average is not f(3)-f(1)/3-1 (which means f(3) - [f(1)/3] - 1 exactly as you have written it). If you mean (f(3)-f(1))/(3-1) you should use parentheses.

    Anyway, your expression for s(t) does not have an 'h' in it; maybe you really meant to say S0 = 800. You see, these errors start to add up and can reduce your marks!

    For (b): you say you are having trouble with the derivative. Please show your work, so we can see exactly what is your source of trouble.

    For (c): you need to know when s(t) = 0.

    Question (d) does not make much sense as written, but maybe they want you to write a formula for velocity v(t) that applies on the interval 1 ≤ t ≤ 3.

    RGV
     
  4. Nov 10, 2012 #3
    Letter d, i typed up b again. so letter D actually is below
    D) find the velocity when the coin hits the ground.

    My work so far

    I used the 800 ft from the original statement as So instead of height


    S(t)=Vi(0)t + 1/2 (-32ft/s^2)t^2 +So
    S(t)= 0 -16t^2 + 800
    v(t) = -32t

    a) V(3)-V(1)/(3-1)
    V(3) = -32(3)=-96
    V(1) = -32(1)= -32

    -96-(-32)/2 = -64ft/sec is my final answer

    b) v(3)= -32(3)
    v(3)= -96 ft/Sec is my final answer
    C) I just don't understand
    D)I know i need C but don't understand it
     
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