What is the Best Way to Approach a Basic Dynamics Problem?

[amondellio]

Homework Statement

I've attached a picture of the problem

Homework Equations

I'm assuming I need to use ads=vdv

The Attempt at a Solution

I've attached a picture of my attempt at solving the problem, but I really have no idea what I'm doing. I'm really having a hard time in this class because my professor only ever goes over theory and proofs. It doesn't help that it's been 5 years since I took calculus.

 

[SteamKing]

The image showing the problem statement is barely legible. Can you take another picture and repost please?

 

[BiGyElLoWhAt]

Where did you get that equation?

If you know calculus, why don't you try to derive a vector position function P(t)-><x,y> using your acceleration and initial conditions. You can take the ground to be y = 0 (or wherever really, but watch your heights) and the left side of the cliff to be x = 0. You have a constant acceleration, g, so it should be relatively easy to figure out.

 

[BiGyElLoWhAt]

@steamking.

Dude standing at edge. First cliff is 200' high, second cliff is 180' high. If he jumps with an initial velocity 8ft/s ihat how far can the cliff be if he wants to not fall to his death.

 

[BiGyElLoWhAt]

Also to finish my thought (oops haha) :

With that position function, you can solve for the x value where y = a certain height (i.e. the height of the second cliff)

 

[amondellio]

Here's a better picture of the problem, sorry

 

[amondellio]

Where did you get that equation?

If you know calculus, why don't you try to derive a vector position function P(t)-><x,y> using your acceleration and initial conditions. You can take the ground to be y = 0 (or wherever really, but watch your heights) and the left side of the cliff to be x = 0. You have a constant acceleration, g, so it should be relatively easy to figure out.

I haven't taken calculus in 5 years, so I don't really think I would say I know calculus haha how would I derive the vector position function you're talking about?

 

[gneill]

The usual kinematic formulas should suffice for this. No calculus required.

Hint: His initial velocity is horizontal. What does that tell you about his initial vertical velocity? How long does it take him to fall 20 ft (200 - 180)?

 

[amondellio]

I think I've solved it with the kinematics equations, but the problem is part of a practice test and my professor wants me to show my work using calculus for all problems. I'm so confused.

 

[MadAtom]

The gentleman (judging by the hat) must move by a distance d at the exact same moment that he reaches high h2.

So, you have to find the time t required for him to move by a distance h1-h2 horizontally.

Then just plug this time on the equation for horizontal motion of the gentleman. (the motion is decomposed in two different motions along y and x axis)

 

[amondellio]

does this look right?

 

[MadAtom]

I think the only calculus you could apply is in the derivation of the motion equations (as
BiGyElLoWhAt said). For that you need the "calculus definition" of the concepts of velocity and accelaration, which, since you professor focus too much on theory, I believe you've covered on class (if not check the introductory part of any calculus-based physics textbook).

 

[MadAtom]

does this look right?

the motion is decomposed into two different motions: on along X and the other along Y. pay attention to what type of motion each one is.

 

[BiGyElLoWhAt]

does this look right?

Not so much.

Whats the velocity vector at t =0 ? (the instant he jumps)

Whats the acceleration vector at t = 0?

What's the position vector at t= 0?

How do these three quantities relate to each other? What can I do with P(t) to get to V(t)? what can I do with V(t) to get to A(t)? How can I reverse these processes?

 

[amondellio]

I don't know what you mean

 

[amondellio]

I just don't get what to do without time or the final x position

 

[BiGyElLoWhAt]

Ok, basic calculus lesson.
Velocity is the rate of change of position with respect to time meaning $v(t) = \frac{d}{dt}P(t)$
Acceleration is the rate of change of velocity with respect to time meaning $a(t)=\frac{d}{dt}v(t) = \frac{d^2}{dt^2}P(t)$

You can reverse these processes by calculating the antiderivative (basically the derivative operation but in reverse, it differs a little for more complex functions)

If you have an acceleration, in this case g, you can write that as a vector. Which way does gravity pull on objects in terms of unit vectors? (or whatevery notation you prefer) <x,y,z> $x\hat{i} + y\hat{j} + z\hat{k}$

When you calculate the antiderivative, you end up with a constant of integration (when you take the derivative of a constant, you get 0, so you have to add a constant into your function) and you need to use boundery conditions to solve for that constant. So when you do this problem, you start with the acceleration vector, and you move to the velocity vector by way of antiderivatives. You take some kind of condition that you know the value of the velocity for and use that to solve for your constant. So in this case you know the initial velocity, that's how fast sir hat wearer jumps. The cool thing about this problem is that you know the initial velocity at t=0, or the beginning of your motion you're trying to model. So $v(t) = \int a(t)dt \ \ \text{&}\ \ v(0) = v_0 \text{(initial velocity)}$ See if you can calculate this. Post back what you have.
If you need a refresher, I personally like Paul's notes: http://tutorial.math.lamar.edu/Classes/CalcI/IndefiniteIntegrals.aspx

Also, this might be a better page, but ultimately I don't really know what you know/don't know, so you might be the best person to gather resources.
http://tutorial.math.lamar.edu/Classes/CalcIII/Velocity_Acceleration.aspx

 

[dean barry]

Do you need calculus ?
The core of the problem is the the time it takes to fall vertically 20 feet (solvable with Newtons rules and a value for g), then use the time to solve the width.

 

[amondellio]

thanks for the help @BiGyElLoWhAt. I don't know why I was having such a difficult time with this problem.

 

[amondellio]

@BiGyElLoWhAt I was hoping you might take a look at this problem and see if it looks right. According to the answer key my professor sent out today, the velocity is correct but the acceleration is not. My professor has a=80.04m/s^2

 

[BiGyElLoWhAt]

Your velocity was correct? Or just close? It looks like you're missing out on the application of the chain rule for derivatives of composite functions.

when you say $V_{r} = \dot{r}$ there's nothing wrong with that, but you're expression for $\dot{r}$ isn't sitting well with me. What you have for $\dot{r} \ \text{&} \ \ddot{r}$ is not $\frac{dr}{dt} \ \text{&} \ \frac{d^2r}{dt^2}$, you have $\frac{dr}{d\theta} \ \text{&} \ \frac{d^2r}{d\theta^2}$ which are both useful, but it's not the final product. What you have to do for this is go through and use the chain rule, which it seems as though you've done it a couple times, but not everywhere it needs to be done. r is a function of theta and theta is a function of time, so $\frac{dr}{dt} = \frac{dr}{d\theta}\frac{d\theta}{dt}$

Also for your a sub theta I'm really not sure where that 2 is coming from. I have a very similar expression, but there's no 2 and you're missing a squared (which happens to work out in this case, but I'm being nit picky).

I would double check your a_r expression, I got something a little different than what you have. Keep in mind you have a product rule and one function is composite in your v_r expression.

 

[BiGyElLoWhAt]

and sorry for getting back so late =[