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!

Basic 1-d dynamics with a bit of calculus.

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A 500 kg boat moves with an initial speed of 20 m/s turns off it engine and slows down due to the drag force. If this force has magnitude F= 15.3v^2 find the boat speed after 12 seconds.


    2. Relevant equations
    F = ma
    dv = adt

    3. The attempt at a solution

    taking the side opposite to the drag force to be positive in our force equation we get:
    -15.3v2 = m(-a) = m(-dv/dt).

    so,
    ∫-15.3/500 dt = ∫ (1/v^2) dv with the correct limits i get V = -3.4m/s which is incorrect.

    however, when i use the force equation
    -15.3v2 = m(a) = m(dv/dt) { no negative sign for dv/dt} i get the right answer v = 2.4 m/s .

    I dont understand why this works. isnt acceleration in the same direction as the net drag force, so by how i set up my coordinate system they both should have a negative sign? Please help me understand. i looked online and in my textbook and i found nothing.
     
  2. jcsd
  3. Apr 2, 2016 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You can multiply this equation by -1, so it is identical with 15.3v2 = m(a) = m(dv/dt). That means positive acceleration and increasing speed.
    The force of drag is opposite to the velocity, it decreases the speed. Take the direction of velocity positive, then the force and acceleration are negative. .So you have to write m dv/dt= -15.3 v2.
     
  4. Apr 2, 2016 #3
    I understand that the velocity should decrease, but isnt acceleration in the same direction as the drag force so they should have the same signs in the force equation? This is quite non - intuitive for me.
     
  5. Apr 2, 2016 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Read the text carefully, the boat
    So the drag force (a vector) is negative with respect to the velocity and so is the acceleration.
    ##\vec F_{drag} = -15.3 v^2\hat v##
    when ##\vec v = v \hat v ## . The equation for the acceleration is ##m \frac{d\vec v}{dt} = \vec F_{drag} = -15.3 v^2 \hat v## You can omit the vector notation writing the equation for the speed v: mdv/dt = -15.3 v2.
     
  6. Apr 2, 2016 #5
    Oh I see. So, just to confirm, v hat is the unit vector that points in the direction of the velocity correct?
     
  7. Apr 2, 2016 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    yes, ##\hat v## is the unit vector in the direction of the velocity .
     
  8. Apr 2, 2016 #7
    THANK YOU!!! I Get it now!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Basic 1-d dynamics with a bit of calculus.
  1. Basic Dynamics (Replies: 3)

  2. Basic Dynamics (Replies: 3)

  3. Dynamics 2-d (Replies: 1)

Loading...