1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    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.

  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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook