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Homework Help: A few problems regarding motion

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    2.15

    A turtle crawls a straight line, which we call the x-axis with positive direction to the right.Equation for the turtle's position as a function of time

    is x(t)=50,0cm + (2,0cm/s)t - (0,625cm/s^2)t^2


    a.) find the turtle's initial velocity, initial position, initial acceleration

    b.) at what time t is the velocity of the turtle zero (derivation?)

    c.) how long after does it take the turtle to return to it's starting point

    d.)At what time t is the turtle a distance of 10,0cm from it's starting poin



    3. The attempt at a solution

    a.) Initial position at x(0)
    50cm +(2cm/s)(0) -(0,065cm/s)(0)
    =50cm

    Initial velocity
    x'(0) =0 + (2cm/s)(1) -(0,065cm/s^2)(2*0)
    =2cm/s

    initial acceleration
    x''(0) = 0 + (-0,065cm/s^2)(2)
    = -0,125cm/s^2

    b.)
    at what time t is the velocity of the turtle zero
    x'(t)= 0 + (2cm/s)(1) - (0,065cm/s^2)(2t)

    then x'(t) = 0

    2cm/s - (0,0625cm/s^2)(2t) = 0
    1cm/s =(0,0625cm/s^2)(t)
    1cm/s / -0,0625cm/s^2=t
    = -16s


    c.) how long after does it take the turtle to return to it's starting point

    I tried

    (-0,065cm/s^2)(t^2) + (2cm/s)(t) + 50cm = 0

    QUADRATIC FORMULA

    got either -16,49..... or 48,3...
    but answer should be 32s


    same with d.)
     
    Last edited: Jan 18, 2009
  2. jcsd
  3. Jan 18, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    Maybe there was a typo in the problem?

    Maybe it should have been .0625 and not .065 in the statement?
     
  4. Jan 18, 2009 #3
    Hello and thank you.

    I double checked, and yes it was indeed a typo. No wonder... :)
     
  5. Jan 18, 2009 #4
    a.) and b.) are now correct, but am I doing something wrong in c?
     
  6. Jan 18, 2009 #5

    LowlyPion

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    Homework Helper

    The starting point is at 50 cm.
     
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