# Finding location of an object as a function of time

1. Sep 15, 2011

### NCampbell

1. The problem statement, all variables and given/known data
a) Starting From Newtons second law, find the location of the rock as a function of time in horizontal and vertical Cartesian coordinates
b) If r(t) is the distance the rock is from its starting point, what is the maximum value of θ for which r will continually increase as the rock flies through the air?

2. Relevant equations
F=ma

3. The attempt at a solution
a) So as the question asks, I start with newtons second law, that being, f=ma so Fnet=dp/dt (where p=mv), Knowing this am I just to integrate m*dv/dt in vector form?
So I would get something like r(t) = v(t)-1/2a(t)^2 i + v(t)-1/2a(t)^2 j?

b) haven't got that far

2. Sep 15, 2011

### cepheid

Staff Emeritus
If a is not constant, then x(t) (or y(t)) is NOT equal to v(t) - 1/2a(t)^2, you actually have to integrate to figure out what the position function is.

BTW it's confusing because you use parentheses around t to indicate function notation in some places, and seemingly to indicate multiplication in other places. So, I'm not entirely sure what exactly you've written.

3. Sep 15, 2011

### NCampbell

I am sorry I am not grasping the concept of integrating vectors could you provide some direction?

4. Sep 15, 2011

### cepheid

Staff Emeritus
Basically, the components evolve with time independently so that:

r(t) = x(t)i + y(t)j

and,

x(t) = vx(t)dt
y(t) = vy(t)dt

Similarly:

vx(t) = ax(t)dt
vy(t) = ay(t)dt

From Newton's second law:

ax(t) = Fx(t) / m
ay(t) = Fy(t) / m

In all of the above, boldface quantities are vectors, subscripted quantities are vector components, and parentheses indicate function notation.

In case it wasn't clear from the above, the velocity, acceleration, and force vectors are of course given by:

v(t) = vx(t)i + vy(t)j = dr/dt
a(t) = ax(t)i + ay(t)j = dv/dt

F(t) = Fx(t)i + Fy(t)j

So, the summary is that to integrate or differentiate vectors, you can just integrate or differentiate component-wise.

I'm assuming you've been given F(t), otherwise I don't know how you'd solve the problem.

5. Sep 15, 2011

### NCampbell

Thank You Cepheid! that cleared things up for me I was lost after my lecture on vector calculus today amongst the different notations and how each of the components are affected differently, and no I wasn't given F(t) just that a rock was through off a cliff at angle (theta) from the horizontal

6. Sep 15, 2011

### cepheid

Staff Emeritus
Oh okay. So F(t) = 0i - mgj

(gravity: constant with time, and only vertical).

The angle gives you initial conditions on the velocity (i.e. you know both components of v at t = 0). Two initial conditions (initial velocity v(0) and initial position r(0)) are necessary in order to solve the problem, because otherwise when you integrate, you get a whole family of functions as solutions (represented by the arbitrary constants "C" that you get when you integrate each time). The initial conditions help you to find what the value of "C" is in this situation when you integrate.

7. Sep 17, 2011

### Vector Boson

I have this exact same problem (we are probably in the same class and I am stuck on part b). I would greatly appreciate any insight into solving this. Here is the full question:

A rock is thrown from a cliff at an angle θ to the horizontal. Ignore air resistance.
a) Starting From Newton's second law, find the location of the rock as a function of time in horizontal and vertical Cartesian coordinates.
b) If r(t) is the distance the rock is from its starting point, what is the maximum value of θ for which r will continually increase as the rock flies through the air? (Suggestion: write out an expression for r2, using the results of part a), and consider how it changes with time).

My expressions for part a) are:
x(t) = v0cosθt +x0
y(t) = v0sinθt - 4.9t2 + y0
$\vec{r}$(t) = $\hat{x}$x(t) + $\hat{y}$y(t) = $\hat{x}$(v0cosθt +x0) + $\hat{y}$(v0sinθt - 4.9t2 + y0)

So for part b) I have:
r2 = x2 + y2 = (v0cosθt +x0)2 + (v0sinθt - 4.9t2 + y0)2

I am just unsure what I am supposed to do with an expression for r2 and how that relates to finding θ.

Last edited: Sep 17, 2011
8. Sep 17, 2011

### ohms law

This isn't actually an answer (sorry), but...
r represents distance from the starting point. It's a function of x2 and y2. Just thinking about this problem, if you simply hold the rock and let it fall, than θ would be 0 and so would r. If you threw it exactly horizontal, than θ would be 90 degrees, and I'd empirically expect that r2 would reach it's maxima.

Of course, proving that is the question here. My thinking off the cuff here is to use fairly standard algebra to resolve the problem. Manipulate the equation so that you have v0cosθt by itself on the right (or left, whatever...) side, you know? I may be completely off here, but that's my thinking after having just read this now.

9. Sep 17, 2011

### Vector Boson

Actually if you dropped the rock straight off of the cliff then θ = -90 degrees (relative to the horizon as it states in the question) and r is a positive number $\leq$ to the height of the cliff that is always increasing with time as the rock falls. The answer definitely is not throwing it exactly horizontal because that means θ = 0 and we are looking for when θ is at the max angle above the horizontal where the distance to the rock is always increasing.

To get a better sense of the question, imagine throwing the rock at about 85 degrees above the horizon. The rock will go nearly straight up at first (i.e. r is increasing with time) but as soon as it reaches its peak height, it will begin to fall again nearly straight down and r will start decreasing with time. We are looking for the max angle where r will always be increasing over time.

That being said, I still don't know how to solve for θ but I do know that it will be somewhere in the range of ~45 degrees or so.

Last edited: Sep 17, 2011
10. Sep 18, 2011

### ohms law

Right, because... well, because. That makes intuitive sense to me, as well. As a matter of fact, I'm fairly sure that 45 degrees will end up being the answer.

You could try brute forcing the problem for the moment, just to make sure we're on the correct path. Stick the formula into a spreadsheet. I'm fairly convinced that this comes down to standard algebra (in that you simple need to manipulate the formula and solve for theta).

11. Sep 18, 2011

### NCampbell

If we know r (and using polar coordinates) that x=rcos$\theta$ and y=rsin$\theta$ then isn't $\theta$=arctan(y/x)? I am having a difficult time with this question so I am not entirely sure on what to do either.

12. Sep 19, 2011

i have a question, if i got vector r= (something)x-hat+(something)y-hat, how would i counvert it into polar form? i was thinking that if we find the first derivative of the polar form of r, the distance will be contineously increasing if the first derivative is greater than 0?

13. Sep 19, 2011

### NCampbell

To convert to polar form we use r=sqrt(x^2+y^2) and theta= arctan(y/x) where x and y are your x and y components

14. Sep 19, 2011

then how to start part b?

15. Sep 19, 2011

### NCampbell

I am not sure? I don't see the connection between r^2 and how to find theta that way, are you also in this class?

16. Sep 19, 2011

### cepheid

Staff Emeritus
The distance r = |r| =(x² + y²)½ is the distance of the object from the origin. If r is to be always increasing with time, then this suggests that the derivative of r with respect to time must always be positive (for all t), does it not? I think that is the key idea you need in order to get part (b). The problem may have instructed you to work with r² instead because it makes the math easier. I think that if r is always increasing then so is r².

The theta you are referring to here is the theta-coordinate, in a polar coordinate system. In general, this depends on time (since x and y depend on time). However, the problem want you to find the theta at launch, i.e. the initial value of theta at t = 0. This is just one particular value of theta at one particular time, and can be regarded as a constant in the equations. Vector Boson had the right idea.

17. Sep 19, 2011