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Applied de

  • Thread starter bobey
  • Start date
  • #1
30
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a depth of mass a00 kg is released from a warship into the sea and allowed to sink. While gravity is pulling it down, a buoyancy force of 1/40 times its weight (=mg) is pulling it up. therefore, water resistance also exerts a force on the weapon that is proportional to its velocity, with a constant of proportionality of 10 kg/s. How long will it take for the weight to reach the velocity of 70 m/s?

i've tried and my de equation is :

f = ma

mg-1/40(mg)-10v = m(dv/dt)

dv/dt = g - )1/40)g - (10/m) v

dv/dt = (39/40)g - (10/m) v

v(t) = (39/40)gv - 5(v^2)/m + c

applying the initial condition (x=0, t=o, v=0) => c=0

thus v(t) = (39/40)gv - 5(v^2)/m


is my de is right??? how i get t expression from this equation???
 

Answers and Replies

  • #2
674
2
When you solved the differential equation you made a couple of mistakes. For example, you can't integrate v in the 4th line, last term with respect to time and get v^2. If you do that you will get x(t) instead since v = dx/dt.

Solving the diff eq will take a little more work.
 
Last edited:

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