application for differential equation!need help ASAP!!?

Question

1. Atomic waste is placed in sealed canisters and dumped in the ocean. It has been determined that the seal will not break and leak the waste when the canister hits the bottom of the ocean as long as the velocity of the canister is less than 12 m/s when it hits the bottom.

Using Newton’s second law, the velocity satisfies the equation m dv/dt=W-B-kv, where ,
W is the weight of the canister,
B is the buoyancy force, and
the drag is given by kv .

a. Solve the first-order linear equation for v(t) and then integrate to find the position y(t) .

b. If W=2254Newtons, B=2090 Newtons, and k=0.637kg/s, determine the time at which the velocity is 12 m/s.

c. Determine the depth H of the ocean so that the seal will not break when the canister hits the bottom.

Answer 1

The differential equation you have is a non-homogeneous linear equation, written as

dv/dt + kv = (W-B).

The solution for that equation is

v(t)=C*e^-kt + (W-B)/k, where C is a constant to be determined from initial conditions. The derivation is too difficult to enter here because of limitation on mathematical notation. I have made an image of the derivation which goes through all the steps. Please review that here:

http://img247.imageshack.us/img247/6452/linearnonhomogdiffeqpt9.png

If we assume that the descent of the canister starts at v(0) = 0, then

V(0) = C+(W-B)/k = 0,

and C = -(W-B)/k and the velocity equation is

V(t) = -[(W-B)/k]*e^-kt +(W-B)/k,

Integrating this, remember that integral of e^ax is (1/a)e^ax, so the integral of the velocity is

[(1/k)(W-B)/k]*e^-kt + [(W-B)/k]*t + C

y(t) = [(W-B)/k^2]*e^-kt + [(W-B)/k]*t + C

If we call the origin for y = 0, then y(0) = 0 and

C = -(W-B)/k^2

and

y(t) = [(W-B)/k^2]*e^-kt + [(W-B)/k]*t - (W-B)/k^2

simplifying

y(t) = [(W-B)/k]* [(1/k)e^-kt + t - (1/k)]

You should now be able to plug in the values for B, W, and k in the velocity equation, and find t when the velocity is the given value. The general solution for this type of equation is found as follows:

v1 = ae^-kt + b; move (v1) to the right and (ae^-kt) to the left to get

ae^-kt = b - v1

e^-kt = (b-v1)/a

and take the log of both sides so -kt = ln[(b-v1)/a].

Finally, once you find the time for the canister to reach the breaking velocity, put that into the equation for y(t) to find the depth of the ocean to reach that time.

Answer 2

Ok well I'm not quite sure where to start to be honest!

Facts that I know:

dv/dt is the acceleration.

This then leads us to the use of the following equation, part of the suvat formulae.

s = distance, we want to find out depth of the ocean, this is needed.

u = initial velocity, = 0, could or could not be used.

v = final velocity, needs to be less than 12, so this is also needed.

a = W-B-kv, this is known so also should be used.

t = time, and will depend on the acceleration, also needs to be used for part b.


It is quite clear that you will need to use a number of the suvat equations to solve this.

First a.

To find the velocity at time t, when given the acceleration, you should integrate.

dv/dt = W - B - kv

However, I don't know how to do that, as you must intergrate that equation with respect to t, but there are no t's in there to intergrate. All I know is that you have to!

If you can discover this then you should be able to find the answer to b fairly easily. You would just make one side of the equation 12, and submit the other values in to find the value of t.Though i still find it difficult to understand how you would differentiate it. I think that the kv is the important part, because v is the only thing that will change.

At first acceleration will be W - B, or 2254 - 2090 = 164.

Oh crap, sorry just realised that there is a little m before the acceleration, lol, I was beginning to think that it was weird especially as F=MA is Newton's second law.
Argh, ok well m is simply weight divided by gravity, which is 2254/9.8, or 10 to be easier to it would weigh 225.4 kilogrammes.

So ok, hope that everything still applies, I think that it does. You would actually have to divide the force by the mass to find the acceleration.
So in this case that is, 164/225.4 = 0.7275

That is the initial acceleration in m/s/s.

Problem is that as soon as you accelerate, v changes, so the whole equation will change, where as before it was a little easier because kv = 0.

Ok, that is as far as I can get at the moment.

You may find the following equation useful:

v = u +at

s = ut + 1/2 a(t)^2

s = vt - 1/2 a(t)^2

v^2 + u^2 = 2SA

f = ma

These are constant acceleration formulae, so you will need to differentiate to find changing acceleration formulae.

Hope this at least gives you a starting point, and sorry I wasn't of more help!

All the best,

Chris Damant

Answer 3

-to make certain the form of flights fairly flights that ought to bypass great quick so it needs differential equations -make certain the shortest lenght of certain distance like even as air deliver needs to bypass a particular planet then what's the shortest direction (not ideal it determined thinking lot of circumstances) i recognize that's all :(