1. ## Newtonian Mechanics

A diver of mass 60 kg, on entering water with an initial velocity of 15.35 ms-1, experiences a force of resistance, due to the water opposing motion, of magnitude 180v Newtons, where v is the velocity of the diver in ms-1. In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720 N. Take the origin of motion to be at the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10 ms-1.

Calculate the maximum depth, in metres, the diver will submerge. Give your answer to 3 decimal places.

The answer is 4.41. I cant get it no matter how i try. Hopefully someone could enlighten me

From what i know Fnet= 720+180v-mg and that Fnet=ma

Therefore, i got a=2+3v after simplifying. From there i'm stuck because from what i noticed, acceleration is not constant, so the uniform acceleration/deceleration formula could not be applied.

2. ## Re: Newtonian Mechanics

$v < 0 \implies F_r = -180v > 0$

$ma = -180v + 720 - mg$

$\dfrac{dv}{dt} = 2 - 3v$

try again ...

3. ## Re: Newtonian Mechanics

One approach would be to solve the initial value problem:

$x''(t)+3x'(t)=2$ where $x(0)=0$ and $x'(0)=-15.35$

We see the characteristic equation has roots:

$r\in\{0,-3\}$

And so the homogeneous solution is:

$x_h(t)=c_1+c_2e^{-3t}$

Now, the particular solution will take the form:

$x_p(t)=At$

Using the method of undetermined coefficients, we compute:

$x_p'(t)=A$

$x_p''(t)=0$

And substituting into the ODE, we obtain:

$0+3A=2\implies A=\frac{2}{3}$

And so the particular solution is:

$x_p(t)=\frac{2}{3}t$

Hence, by the principle of superposition, we obtain the general solution:

$x(t)=x_h(t)+x_p(t)=c_1+c_2e^{-3t}+\frac{2}{3}t$

And hence:

$x'(t)=-3c_2e^{-3t}+\frac{2}{3}$

So, we now use the given initial conditions to determine the parameters $c_i$, thereby finding the solution to the IVP...can you proceed?

4. ## Re: Newtonian Mechanics

I'll try to continue from there thanks for the help

5. ## Re: Newtonian Mechanics

On a side note, the answer given is wrong?

6. ## Re: Newtonian Mechanics

Originally Posted by noobpronoobpro
On a side note, the answer given is wrong?
$y=-4.41 \, m$ is correct.

7. ## Re: Newtonian Mechanics

Originally Posted by noobpronoobpro
On a side note, the answer given is wrong?
Let's continue where I left off:

$x(t)=c_1+c_2e^{-3t}+\frac{2}{3}t$

$x'(t)=-3c_2e^{-3t}+\frac{2}{3}$

Using the initial velocity, we find:

$x'(0)=-3c_2e^{-3(0)}+\frac{2}{3}=-3c_2+\frac{2}{3}=-\frac{307}{20}\implies c_2=\frac{961}{180}$

Using the initial position, we find:

$x(0)=c_1+c_2e^{-3(0)}+\frac{2}{3}(0)=c_1+c_2=0\implies c_1=-c_2=-\frac{961}{180}$

And so the solution to the given IVP is:

$x(t)=-\frac{961}{180}+\frac{961}{180}e^{-3t}+\frac{2}{3}t=\frac{1}{180}\left(961e^{-3t}+120t-961\right)$

Now, to find the maximum depth, we need to find the time when the diver's velocity is zero:

$x'(t)=-\frac{961}{60}e^{-3t}+\frac{2}{3}=0\implies t=\frac{1}{3}\ln\left(\frac{961}{40}\right)$

And so the position at this time is:

$x(t)= \frac{1}{180}\left(961\cdot\frac{40}{961}+ 120\cdot\frac{1}{3}\ln\left(\frac{961}{40}\right)-961\right)= \frac{1}{180}\left(40+ 40\ln\left(\frac{961}{40}\right)- 961\right)\approx-4.41020112114303$

8. ## Re: Newtonian Mechanics

woah thx for the explicit explanation. i was able to get the solution to the given IVP but i think i made some mistake while finding the time which result in me not getting the answer really thankful for ur help!