Straight-line motion: Projectile in resising medium

[JJBladester]

Homework Statement

A projectile enters a resisting medium at x=0 with an initial velocity v₀=900ft/s and travels 4in before coming to rest. Assuming that the velocity of the projectile is defined by the relation v=v₀-kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9in into the resisting medium.

Answers: (a) -2.43x10⁶ft/s² (b) 1.366x10^-3s

Homework Equations

x_i=0ft
v_i=900ft/s
x_f=4in=.33333ft
v=v_i-kx
v_f=0ft/s (rest)

a(x)=d/dx[v(x)]

The Attempt at a Solution

v=v_i-kx
0=900-k*.33333 ----> k=2700s^-1

a(x)=d/dx[v(x)]=d/dx[v_i-kx]=-k=-2700

The answer of -2.43x10⁶ft/s² is obviously not -2700.

 

[gabbagabbahey]

a(x)=d/dx[v(x)]

When have you ever seen acceleration defined as the rate of change of velocity with respect to position?

 

[JJBladester]

When have you ever seen acceleration defined as the rate of change of velocity with respect to position?

Ummm... Yea...

So should I be doing the following calculation?

a=d/dt[v(t)]=d/dt[v_i-kx]= ?

In other words, "acceleration is the derivative of velocity with respect to time". Still stuck.

 

[JJBladester]

So,

Let's say the acceleration is constant...

Then I can use this formula: v_f²=v_i²+2a(x₂-x₁)

a=(v_f²-v_i²)/[2(x₂-x₁)]

a=(0-900²)/[2(.33333-0)]=-1.215x10^-6ft/s²

It seems the answer in the book is double that, -2.430ft/s². Why?

 

[Liquidxlax]

why did you divide by 2 for acceleration? you are finding slope. (y1 + y2)/(x1+x2)=m

 

[JJBladester]

why did you divide by 2 for acceleration? you are finding slope. (y1 + y2)/(x1+x2)=m

Actually, slope is (v₂-v₁)/(x₂-x₁₎.

Doing that yields (0-900)/(.33333-0)=-2700. Right back where I started. Obviously I'm missing something here.

 

[Liquidxlax]

Actually, slope is (v₂-v₁)/(x₂-x₁₎.

Doing that yields (0-900)/(.33333-0)=-2700. Right back where I started. Obviously I'm missing something here.

lol whoops well its been 2 years.d = v1t + .5at^2 would make the most sense

but that doesn't seem to work either since it would be

0=.5at^2

 

[gabbagabbahey]

Ummm... Yea...

So should I be doing the following calculation?

a=d/dt[v(t)]=d/dt[v_i-kx]= ?

In other words, "acceleration is the derivative of velocity with respect to time". Still stuck.

Sure, now use the chain rule:

$$\frac{d}{dt}v(x)=\frac{dx}{dt}\frac{dv}{dx}$$

 

[gabbagabbahey]

lol whoops well its been 2 years.


d = v1t + .5at^2 would make the most sense

but that doesn't seem to work either since it would be

0=.5at^2

That kinematic equation is only valid for constant acceleration problems. There is no reason to assume that acceleration is constant here (it isn't).

 

[JJBladester]

Sure, now use the chain rule:

$$\frac{d}{dt}v(x)=\frac{dx}{dt}\frac{dv}{dx}$$

a = dV/dt

dV/dt = (dV/dx)(dx/dt) - chain rule, like you said

therefore, since velocity is just the derivative of position with respect to time,

a = dV/dx V... or rewritten, a is V'(x)V(x)

a=V'(x)V(x)=(-k)(V)=(-2700s^-1)(900ft/s²)=-2.43x10^-6ft/s² Which is the answer in the book!

Now, the second part of the question asks us to find the time required for the projectile to penetrate 3.9in (0.325ft) into the resisting medium.

We know that v=v_i-kx. So, v_f=900-2700(0.325)=22.5ft/s.

How do I find the time required to get 3.9 inches into the medium? Is it just t=(v_f-v_i)/(a_f-a_i)?

 

[gabbagabbahey]

How do I find the time required to get 3.9 inches into the medium? Is it just t=(v_f-v_i)/(a_f-a_i)?

No, you need to be careful not to blindly apply the kinematic equations you learned in high school; most (if not all) of them are for constant acceleration problems and that is true of the equation $\Delta t= \frac{\Delta x}{\Delta v}$

Instead, you will need to use the more general relationship between velocity, position and time. Velocity is defined as the rate of change of position w.r.t. time...what equation represents that?

 

[JJBladester]

Instead, you will need to use the more general relationship between velocity, position and time. Velocity is defined as the rate of change of position w.r.t. time...what equation represents that?

$$x_{f}-x_{i}=\int_{t_{1}}^{t_{2}}v(t)dt$$

$$0.325ft=\int_{0}^{t}[v_{i}-kx]dt$$

$$0.325ft=[v_{i}-kx]t$$

$$t=\frac{0.325ft}{v_{i}-kx}$$

$$t=\frac{0.325}{900-(2700*0.325)}=0.01\bar{4}$$

Close, but not exactly the 1.366x10^-3s the book has... I am on the cusp of getting this problem... With your help...

 

[RoyalCat]

$$\int_{0}^{t}[v_{i}-kx]dt\neq[v_{i}-kx]t$$
Because x is an explicit function of t.

To solve the question you have to find the position as a function of time and then solve for the time.
Doing that is a bit trickier than it may seem at first glance. I suggest you start with the general expression for the acceleration, the one you used to find the initial acceleration.
You posted that $$a=\frac{dv}{dx} v$$ and we know that $$\frac{dv}{dx}=-k$$ So we find for the acceleration: $$\frac{dv}{dt}=-kv$$

The solution to that differential equation can be found by integration. Once you have $$v(t)$$, you should be able to find $$x(t)$$ and then extract $$t(x)$$

 

[Liquidxlax]

did you get my message on time? This is simple kinematics, you shouldn't need integrals or derivatives. I solved it on paper without integrals or derivatives and I got your textbook answersedit i forgot about the second part, i had to use your equation, but somehow i got 1.28x10^-3s

 

[gabbagabbahey]

This is simple kinematics, you shouldn't need integrals or derivatives. I solved it on paper without integrals or derivatives and I got your textbook answers

No, you do need calculus for this problem. This is not a constant acceleration problem, and so those simple kinematic formulas you learned in high school will not work.

 

[JJBladester]

The solution to that differential equation can be found by integration. Once you have $$v(t)$$, you should be able to find $$x(t)$$ and then extract $$t(x)$$

$$\[a=\frac{dv}{dx}v\]$$

$$\frac{dv}{dx}=-k$$

$$\frac{dv}{dt}=-kv$$

$$\frac{dv}{v}=-kdt$$

$$\int_{v_{0}}^{v}\frac{dv}{v}=-k\int_{0}^{t}dt$$

$$ln(v)-ln(v_{0})=-kt$$

$$e^{ln(v)}-e^{ln(v_{0})}=e^{-kt}$$

$$\mathbf{v(t)=v_{0}+e^{-kt}}$$

$$\int_{x_{0}}^{x}dx=\int_{0}^{t}v(t)dt$$

$$x-x_{0}=\int_{0}^{t}\left[v_{0}+e^{-kt}\right]dt$$

$$\mathbf{x(t)=v_{0}t-\frac{e^{-kt}+1}{k}}$$

I now have x(t)... position as a function of time. Now, getting t(x)... Where to start? I know I need to get t on its own somehow but since it's stuck up in the exponential, not sure how to get it down haha...

 

[RoyalCat]

$$ln(v)-ln(v_{0})=-kt$$
$$\ln{\frac{v}{v_0}}=-kt$$
Your next line, $$e^{-kt} = e^{\ln{v}}-e^{\ln{v_0}}$$ does not follow. The exponential doesn't work that way.

 

[JJBladester]

$$ln(v)-ln(v_{0})=-kt$$
$$\ln{\frac{v}{v_0}}=-kt$$
Your next line, $$e^{-kt} = e^{\ln{v}}-e^{\ln{v_0}}$$ does not follow. The exponential doesn't work that way.

Hummm... Yes it does. I've *always* used e^ to get rid of ln's in my equations. I'm pretty sure I didn't make a math foul there.

 

[RoyalCat]

Yes you did. When you took the exponential of both sides of the equation, instead of taking the exponential of the left hand side, you took the exponential of each of the arguments separately, and that is a mistake. Note how you got that something with dimensions, $$v-v_0$$ is equal to something without dimensions, and that is nonsense! $$e^{-kt}$$

Note also that you can never take the logarithm or exponential of a quantity with dimensions, as that does not make physical sense, and here you have completely separated $$\ln{v}$$ from $$\ln{v_0}$$ and lost dimensional sense.
The reason you can take the integral $$\frac{dx}{x}$$ when $$x$$ has dimensions is because it will always reduce, according to the laws of logarithms to a dimensionless quantity inside the argument for the logarithm, $$\ln{\frac{x_1}{x_2}}$$ which is okay.

$$\ln{v}-\ln{v_0}=-kt$$
Taking the exponential of both sides, we find:
$$e^{\ln{v}-\ln{v_0}}=e^{-kt}$$
Applying the rules of the exponential, we find:
$$e^{-kt}=\frac{e^{\ln{v}}}{e^{\ln{v_0}}}$$

And from here, the final solution for v is easily found.

I suggest you brush up on your algebra, since this is just a small mistake that can be avoided in the future.

 

[JJBladester]

I have solved the problem in its entirety. The answers I have match those in the book.

https://docs.google.com/fileview?id=0B1UmO-np5c1tNjRlNTgyOWEtNWU2OC00YWExLTg2ZGItZjFhYzI4ZjYxNGY0&hl=en" is a link to the PDF of my solution.

Thanks all for your persistent guidance.

 

[RoyalCat]

Very well done. :) The only two comments I have is that you wrote $$\ln{v}-\ln{v_0}$$ and kept that separation of the two logarithms all the way to the final solution, while this is not a downright mistake, it could lead to some confusion, and as I relayed to you in the private message, in terms of physical meaning, it is ill-advised to write it like that. Though if you're comfortable with it, and know to keep track of it all, by all means, more power to you. :)

The second thing is also purely aesthetic, you carried the $$x_0$$ throughout the final part of the solution, only to put in $$x_0=0$$ at the end, which just makes for more cluttered writing. Even if $$x_0$$ were not 0, it would still close many doors to error to introduce a new variable, $$\Delta x \equiv x-x_0$$ to simplify the writing.

You're very welcome, and best wishes with your further studies. :)