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Homework Help: How do i solve these ?

  1. Feb 25, 2006 #1
    how do i solve these ????

    consider a river flowing with a speed v along positive X axis . a man starts swimming from a point on the bank of the river ( which is considered to be as origin ) . l is the width of the river and the velocity of man is always directed towards the point (0,l) . speed with which man can swim in still water is u . find the equation of the path that the man follows ?

    i took a general point where man's direction of velocity makes an angle q with the vertical and i assume this point to be (x,y)
    now i m just able to get 2 equation which are
    1. x = (v-u)integeral(cosp)dt
    2. y = (v-u)integeral(sinp)dt
    now i have 4 variables (t,p,x,y) and have just 3 equations (which also dont seem to be solvable) ..... what do i do ?
     
  2. jcsd
  3. Feb 25, 2006 #2

    quasar987

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    I think this is just a matter of solving the equation of motion (F=ma) in the x and y component and combining them vectorially by

    [tex]\vec{r}(t) = x(t)\hat{x} + y(t)\hat{y}[/tex]

    which gives a paramatrization of the path by the time.
     
  4. Feb 26, 2006 #3
    1. x = (v-u)integeral(cosp)dt
    2. y = (v-u)integeral(sinp)dt
    here i have 2 variables p and t ...... so there are 2 parameters . moreover i want to find the relation between x and y independent of any other parameters (sorry i did not mention that in my initial post ) . is it possible to get a relation between x and y independent of any other parameters?
     
  5. Feb 26, 2006 #4

    HallsofIvy

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    NOT multiplied by v-u! v and u are speeds and the corresponding velocities are not in the same direction: v is in the x direction.

    The man's velocity vector is (u cos(p)+ v)i+ (u sin(p))j . Of course, since he is always aiming at the point (0,1), tan p= x/(1- y).
    [tex]sin(p)= \frac{x}{\sqrt{x^2+ (1-y)^2}}[/tex]
    [tex]cos(p)= \frac{1-y}{\sqrt{x^2+ (1-y)^2}}[/tex]

    You actually have two differential equations:
    [tex]\frac{dx}{dt}= \frac{ux}{\sqrt{x^2+ (1-y)^2}}+ v[/tex]
    [tex]\frac{dy}{dt}= \frac{uy}{\sqrt{x^2+ (1-y)^2}}[/tex]

    Since t does not appear explicitely in those,
    [tex]\frac{dx}{dy}= \frac{x}{y}+ \frac{v\sqrt{x^2+(1-y)^2}}{uy}[/tex]
    is an equation for x as a function of y.
     
  6. Feb 26, 2006 #5
    thanx a lot sir i was missing the fact that tan p= x/(1- y) .....
     
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