Question

A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 fps. While in the water the ball experiences an acceleration of a=10−0.8v, where a and v are expressed in fps^2 and fps respectively. Knowing the ball takes 3 sec to reach the bottom of the lake, determine the following:
a. the depth of the lake
b. the speed of the ball when it hits the bottom of the lake

Answer 1

@ganeshie8

Answer 2

@satellite73

Answer 3

fps = feet per sec?

Answer 4

yes

Answer 5

Is \(a=10-0.8v\) a function of a variable velocity \(v\)? If so, you can try solving for \(v\) since this is a differential equation.

Answer 6

can one write ball will float and never go down to lake's surface?

Answer 7

yes it is.

Answer 8

can you help me @SithsAndGiggles

Answer 9

Do you know how to solve linear DEs? Find the integrating factor, etc?

I'm not sure what to do beyond that, but you'll have a function of the velocity with respect to time. That along with whatever formulas you've learned should be enough to figure out what you need. I don't know if you treat it like a free fall problem or not, though...

Answer 10

i know how to solve linear DEs. But I don't know how to deal with it.

Answer 11

I mean with this problem

Answer 12

Use SUVAT equations :)

Answer 13

You're given an initial velocity of \(v(0)=16.5\), so the initial acceleration is \(a(16.5)=10-0.8(16.5)=-3.2\).

Let's see what we get for the velocity function:
\[\large\begin{align*}\frac{dv}{dt}+0.8v&=10\\
e^{0.8t}\frac{dv}{dt}+0.8e^{0.8t}v&=10e^{0.8t}\\
\frac{d}{dt}\bigg[e^{0.8t}v\bigg]&=10e^{0.8t}\\
v&=10e^{-0.8t}\left(\int e^{0.8t}~dt\right)\\
v&=10e^{-0.8t}\left(\frac{1}{0.8}e^{0.8t}+C\right)\\
v&=12.5+Ce^{-0.8t}\\
16.5&=12.5+C\\
C&=4
\end{align*}\]
So \(v(t)=12.5+4e^{-0.8t}\)...

The ball takes three second to reach the bottom. Presumably, the ball stops moving at that time (doesn't bounce back up), so (I would think) you have velocity defined by
\[v(t)=\begin{cases}12.5+4e^{-0.8t}&\text{for }0\le t<3\\0&\text{for }t\ge3\end{cases}\]
This makes me think that plugging in \(t=3\) should give the velocity of the ball as it hits the bottom, which would be about 12.86 fps.

Not sure how to approach part (a) just yet...

Answer 14

Actually, I just might... I think it's just a matter of integrating the velocity function over the interval [0,3], which would give the total displacement. Since the ball is traveling straight down (with no obstructions), the depth of the water would be the displacement.
\[d=\int_0^3 v(t)~dt\approx42.05\]

Answer 15

sith's solution looks good.

you could also solve using
a= x'' , v= x' i.e. distance with respect to time
you get the 2nd order diff eq

x'' +0.8 x' = 10

solve the homogeneous equation
x'' + 0.8 x' = 0
which has the characteristic equation
m^2 + 0.8 m =0
m=0 and m= -0.8
thus
\[ x_h= c_1 e^{0t} + c_2 e^{-0.8t} \\ x_h= c_1 + c_2 e^{-0.8t} \]
for the particular solution, we would choose a constant x=C except that the homogeneous sol'n already has a constant. so instead use xp= Ct
putting that into the equation we get
0.8C = 10, C= 10/.8 = 12.5

the solution for distance traveled is
\[ x= x_h+x_p = c_1 + c_2 e^{-0.8t} + 12.5t \]
to find the constants, we use x'= 16.5 at t=0
differentiate x to find
\[ x'= -0.8 c_2 e^{-0.8\cdot 0} +12.5 = 16.5 \]
we find \( c_2= - 5 \)

we set x=0 at t=0 and solve
\[ c_1 - 5 e^{-0.8\cdot 0} + 12.5 \cdot 0 =0 \\ c_1= 5\]
and finally
\[x= 5 - 5 e^{-0.8t} + 12.5t \]
and velocity
\[ x'= v = 4 e^{-0.8t} + 12.5 \]

evaluate at t=3

Answer 16

what is \[x_{h}\] by the way?

Answer 17

the solution of a differential equation has two parts:
1) the particular solution. in this case, the solution that gives 10
2) the homogeneous solution, where we equate the equation to 0 rather than to 10.

See
http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx
for a discussion.

Answer 18

Ok, you mean Xh to be the characteristic solution?

Answer 19

can you just explain briefly where 0.8C = 10 came from?

Answer 20

To find the particular solution, we "guess" it.
if we have
equation = 10
we would guess the particular solution is 10 and all possible derivatives.
But if the homogeneous solution already has a constant as part of its answer (and in your problem it does) we try x= C t (we multiply by the independent variable) as our guess.

see http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

we put our guess into the diff equation
x'' + 0.8 x' = 10

with x= C t, we find x' = C (i.e. d/dt C t = C)
and x'' = d/dt C = 0
we get
0 + 0.8 C = 10
or
0.8 C =10
solve for C to get 12.5 and the particular solution xp= 12.5 t

Answer 21

where does x = Ct equation came? Is that just a guess?

Answer 22

yes, that is the guess. Paul's discussion might help (though I have not read through it, he's generally helpful)

Answer 23

but where does that guess came from? how do you arrive at that guess?

Answer 24

This is the "method of undetermined coefficients" and it only works for when the right hand side is a sum of terms, each term having a finite number of linearly independent derivatives. This restricts the technique to terms such as
\[ c, t^c, e^{ct}, \sin c\ t, \cos c\ t \]
luckily we can use it for constant right-hand side (i.e. the 10)