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!

Straight-line motion: Projectile in resising medium

  1. Sep 21, 2010 #1

    JJBladester

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    A projectile enters a resisting medium at x=0 with an initial velocity v0=900ft/s and travels 4in before coming to rest. Assuming that the velocity of the projectile is defined by the relation v=v0-kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9in into the resisting medium.

    Answers: (a) -2.43x106ft/s2 (b) 1.366x10-3s

    2. Relevant equations

    xi=0ft
    vi=900ft/s
    xf=4in=.33333ft
    v=vi-kx
    vf=0ft/s (rest)

    a(x)=d/dx[v(x)]

    3. The attempt at a solution

    v=vi-kx
    0=900-k*.33333 ----> k=2700s-1

    a(x)=d/dx[v(x)]=d/dx[vi-kx]=-k=-2700

    The answer of -2.43x106ft/s2 is obviously not -2700.
     
  2. jcsd
  3. Sep 21, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    When have you ever seen acceleration defined as the rate of change of velocity with respect to position?:wink:
     
  4. Sep 21, 2010 #3

    JJBladester

    User Avatar
    Gold Member

    Ummm... Yea...

    So should I be doing the following calculation?

    a=d/dt[v(t)]=d/dt[vi-kx]= ?????

    In other words, "acceleration is the derivative of velocity with respect to time". Still stuck.
     
  5. Sep 21, 2010 #4

    JJBladester

    User Avatar
    Gold Member

    So,

    Let's say the acceleration is constant...

    Then I can use this formula: vf2=vi2+2a(x2-x1)

    a=(vf2-vi2)/[2(x2-x1)]

    a=(0-9002)/[2(.33333-0)]=-1.215x10-6ft/s2

    It seems the answer in the book is double that, -2.430ft/s2. Why?
     
  6. Sep 21, 2010 #5
    why did you divide by 2 for acceleration? you are finding slope. (y1 + y2)/(x1+x2)=m
     
  7. Sep 21, 2010 #6

    JJBladester

    User Avatar
    Gold Member

    Actually, slope is (v2-v1)/(x2-x1).

    Doing that yields (0-900)/(.33333-0)=-2700. Right back where I started. Obviously I'm missing something here.
     
  8. Sep 21, 2010 #7
    lol whoops well its been 2 years.


    d = v1t + .5at^2 would make the most sense

    but that doesn't seem to work either since it would be

    0=.5at^2
     
  9. Sep 22, 2010 #8

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Sure, now use the chain rule:

    [tex]\frac{d}{dt}v(x)=\frac{dx}{dt}\frac{dv}{dx}[/tex]
     
  10. Sep 22, 2010 #9

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    That kinematic equation is only valid for constant acceleration problems. There is no reason to assume that acceleration is constant here (it isn't).
     
  11. Sep 22, 2010 #10

    JJBladester

    User Avatar
    Gold Member

    a = dV/dt

    dV/dt = (dV/dx)(dx/dt) - chain rule, like you said

    therefore, since velocity is just the derivative of position with respect to time,

    a = dV/dx V... or rewritten, a is V'(x)V(x)

    a=V'(x)V(x)=(-k)(V)=(-2700s-1)(900ft/s2)=-2.43x10-6ft/s2 Which is the answer in the book!!!

    Now, the second part of the question asks us to find the time required for the projectile to penetrate 3.9in (0.325ft) into the resisting medium.

    We know that v=vi-kx. So, vf=900-2700(0.325)=22.5ft/s.

    How do I find the time required to get 3.9 inches into the medium? Is it just t=(vf-vi)/(af-ai)?
     
  12. Sep 22, 2010 #11

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, you need to be careful not to blindly apply the kinematic equations you learned in high school; most (if not all) of them are for constant acceleration problems and that is true of the equation [itex]\Delta t= \frac{\Delta x}{\Delta v}[/itex]

    Instead, you will need to use the more general relationship between velocity, position and time. Velocity is defined as the rate of change of position w.r.t. time....what equation represents that?
     
  13. Sep 22, 2010 #12

    JJBladester

    User Avatar
    Gold Member

    [tex]x_{f}-x_{i}=\int_{t_{1}}^{t_{2}}v(t)dt[/tex]

    [tex]0.325ft=\int_{0}^{t}[v_{i}-kx]dt[/tex]

    [tex]0.325ft=[v_{i}-kx]t[/tex]

    [tex]t=\frac{0.325ft}{v_{i}-kx}[/tex]

    [tex]t=\frac{0.325}{900-(2700*0.325)}=0.01\bar{4}[/tex]

    Close, but not exactly the 1.366x10-3s the book has... I am on the cusp of getting this problem... With your help...
     
  14. Sep 22, 2010 #13
    [tex]\int_{0}^{t}[v_{i}-kx]dt\neq[v_{i}-kx]t[/tex]
    Because x is an explicit function of t.

    To solve the question you have to find the position as a function of time and then solve for the time.
    Doing that is a bit trickier than it may seem at first glance. I suggest you start with the general expression for the acceleration, the one you used to find the initial acceleration.
    You posted that [tex]a=\frac{dv}{dx} v[/tex] and we know that [tex]\frac{dv}{dx}=-k[/tex] So we find for the acceleration: [tex]\frac{dv}{dt}=-kv[/tex]

    The solution to that differential equation can be found by integration. Once you have [tex]v(t)[/tex], you should be able to find [tex]x(t)[/tex] and then extract [tex]t(x)[/tex]
     
    Last edited: Sep 22, 2010
  15. Sep 22, 2010 #14
    did you get my message on time? This is simple kinematics, you shouldn't need integrals or derivatives. I solved it on paper without integrals or derivatives and I got your text book answers


    edit i forgot about the second part, i had to use your equation, but somehow i got 1.28x10^-3s
     
    Last edited: Sep 22, 2010
  16. Sep 22, 2010 #15

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member


    No, you do need calculus for this problem. This is not a constant acceleration problem, and so those simple kinematic formulas you learned in high school will not work.
     
  17. Sep 22, 2010 #16

    JJBladester

    User Avatar
    Gold Member

    [tex]\[a=\frac{dv}{dx}v\][/tex]

    [tex]\frac{dv}{dx}=-k[/tex]

    [tex]\frac{dv}{dt}=-kv[/tex]

    [tex]\frac{dv}{v}=-kdt[/tex]

    [tex]\int_{v_{0}}^{v}\frac{dv}{v}=-k\int_{0}^{t}dt[/tex]

    [tex]ln(v)-ln(v_{0})=-kt[/tex]

    [tex]e^{ln(v)}-e^{ln(v_{0})}=e^{-kt}[/tex]

    [tex]\mathbf{v(t)=v_{0}+e^{-kt}}[/tex]

    [tex]\int_{x_{0}}^{x}dx=\int_{0}^{t}v(t)dt[/tex]

    [tex]x-x_{0}=\int_{0}^{t}\left[v_{0}+e^{-kt}\right]dt[/tex]

    [tex]\mathbf{x(t)=v_{0}t-\frac{e^{-kt}+1}{k}}[/tex]

    I now have x(t)..... position as a function of time. Now, getting t(x)... Where to start? I know I need to get t on its own somehow but since it's stuck up in the exponential, not sure how to get it down haha...
     
    Last edited: Sep 22, 2010
  18. Sep 23, 2010 #17
    [tex]ln(v)-ln(v_{0})=-kt[/tex]
    [tex]\ln{\frac{v}{v_0}}=-kt[/tex]
    Your next line, [tex]e^{-kt} = e^{\ln{v}}-e^{\ln{v_0}}[/tex] does not follow. The exponential doesn't work that way.
     
    Last edited: Sep 23, 2010
  19. Sep 23, 2010 #18

    JJBladester

    User Avatar
    Gold Member

    Hummm... Yes it does. I've *always* used e^ to get rid of ln's in my equations. I'm pretty sure I didn't make a math foul there.
     
  20. Sep 23, 2010 #19
    Yes you did. When you took the exponential of both sides of the equation, instead of taking the exponential of the left hand side, you took the exponential of each of the arguments separately, and that is a mistake. Note how you got that something with dimensions, [tex]v-v_0[/tex] is equal to something without dimensions, and that is nonsense! [tex]e^{-kt}[/tex]

    Note also that you can never take the logarithm or exponential of a quantity with dimensions, as that does not make physical sense, and here you have completely separated [tex]\ln{v}[/tex] from [tex]\ln{v_0}[/tex] and lost dimensional sense.
    The reason you can take the integral [tex]\frac{dx}{x}[/tex] when [tex]x[/tex] has dimensions is because it will always reduce, according to the laws of logarithms to a dimensionless quantity inside the argument for the logarithm, [tex]\ln{\frac{x_1}{x_2}}[/tex] which is okay.

    [tex]\ln{v}-\ln{v_0}=-kt[/tex]
    Taking the exponential of both sides, we find:
    [tex]e^{\ln{v}-\ln{v_0}}=e^{-kt}[/tex]
    Applying the rules of the exponential, we find:
    [tex]e^{-kt}=\frac{e^{\ln{v}}}{e^{\ln{v_0}}}[/tex]

    And from here, the final solution for v is easily found.

    I suggest you brush up on your algebra, since this is just a small mistake that can be avoided in the future.
     
  21. Sep 23, 2010 #20

    JJBladester

    User Avatar
    Gold Member

    Last edited by a moderator: Apr 25, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Straight-line motion: Projectile in resising medium
  1. Straight Line Motion (Replies: 1)

  2. Straight line motion (Replies: 4)

Loading...