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: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

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