# Mechanical Physics problem with linearly increasing force and distance must find time

1. Jul 8, 2011

### jcsopko

Hi, I'm an independent game developer (currently in college) and have been trying to solve a problem and get an equation for use within my physics engine.

1. The problem statement, all variables and given/known data
Here's the problem/situation illustrated through two graphs and a diagram: http://i.imgur.com/qHAsU.jpg

2. Relevant equations
unknown

3. The attempt at a solution
Through the help of others in other forums I think we've solved the problem with force as it increases linearly with the following equation:

t_2 = Log[(B + A d_2 + Sqrt[A d_2 (2 B + A d)])/B]))/Sqrt[A]

It appears to give a reasonable answer although I have no way of double-checking it until it's implemented in the physics engine, however it does not work for decreasing linear force (you typically will get a negative within the squareroot).

If you guys could help me out I'd be forever grateful.

Thanks for your time and consideration!

2. Jul 9, 2011

### FroChro

Re: Mechanical Physics problem with linearly increasing force and distance must find

Hello!
First of all, your 2 graphs are inconsistent. If force is linearly decreasing by time, it is not linearly decreasing by distance, because velocity is not constant.

Let's suppose that 1st graph is the one defining your problem. Your force is like

$$F=4-\frac{3}{x}t .$$

Where x is the unknown time you are searching for. Acceleration is force divided by mass, let's suppose for simplicity that mass is equal to 1. Since we have dependence of acceleration (which is time derivation of velocity) on time, we can directly integrate it to obtain velocity:

$$v=v_0 + 4\cdot t -\frac{3}{2x}t^2$$

Now let's suppose (for simplicity) that initiall velocity ( v(0)=v_0) is zero. We can integrate again to obtain distance traveled by time t:

$$s =2t^2 - \frac{t^3}{2x}$$

Now just put t=x and s = 20 and solve the equation.

3. Jul 9, 2011

### jcsopko

Re: Mechanical Physics problem with linearly increasing force and distance must find

Hi FroChro!

First off, thanks for taking the time to help me out. I read your post and did as instructed and obtained ~3.65 seconds for the time, which sounds reasonable. I have a couple questions though.

1. Looking at your first equation for force I'm unable to figure out how you got it. What equations did you use to arrive at it?

2. You said the two graphs are inconsistent, but I do not understand your logic for that conclusion. You said it's because velocity is not constant, however force linearly decreasing with distance does not imply constant velocity; by my understanding it implies linearly decreasing acceleration which implies the opposite (an increasing velocity at a progressively lower acceleration over time and distance). Keep in mind it is always moving to the right, the force never goes negative because the all force ceases as soon as the distance of 20 meters is reached (and there at 20 meters the force is equal to 1N), the only thing I need is to find the time it takes to reach the 20 meters. I apologize if the confusion was due to my inability to illustrate what was occurring, I hope this short rant clears it up. If not feel free to ask any questions so I can try to understand what you're saying, I haven't taken physics since high school and it was a bit of a joke class. Also I can explain the in-game application of this problem as it may help you better understand.

Again, thanks for the help, I've been working on this for two weeks now and I'm still struggling.

Last edited: Jul 9, 2011
4. Jul 10, 2011

### FroChro

Re: Mechanical Physics problem with linearly increasing force and distance must find

Your force is linearly decreasing with time. Linear function means that if you change independent variable x by adding a number c to it [x->x+c], value of function/dependent variable y will change in following fashion: y-> y + a*c, where a is constant independent of number c. This actually means that if you know a value of linear function in one point x_0, and you know value of constant a, you have complete information about that linear function. (Because y(x)=y(x_0) + a*(x - x_0).) You can simplify last equation and get an expression y(x)=a*x+b, where b is y(x_0)-a*x_0 . Therefore you can see that linear function is determined by values of 2 constants a,b. This is also in agreement with the fact that a graph of linear function is a "straight line" (because everything I wrote so far is true also if you change the word "linear function" for a word "(graph) of straight line"). Straight lines can be altered in two ways: changing their slope or shifting them.

So we need to determine constants a and b. (This time y=a*t+b, we have symbol x reserved for one constant, but yet undetermined value of time.)We know that F(0)=4 and F(x)=1. It's quite straightforward to do so. 1st equation tells us b=4 and second one, after substituing value of b, a= -3/x.
Let's say you have some observable quantity "o" that is linearly dependent on time. For simplicity let's b=0, a=1, so we have o(t)=t. If you are moving with constant velocity, "o" is also linearly dependent on distance passed, o=t=(s/v) . But changing velocity v, you are changing constant "a", therefore altering slope of function o(s), which is no longer linear. Having o=t, this can be also seen quite intuitively, if you are changing speed, you are changing also the dependence of time passed on distance passed. So, to summarize, you can't have linear dependence on both time and distance unless the dependence between time and distance is also linear - unless you have a constant velocity.

Also, most of what I wrote should be familiar to you if you know mathematics on the level let's say that of calculus. In general, if you know a dependence of acceleration on time, you need to know how to integrate it to obtain functions v(t) and s(t); therefore, knowledge of calculus is quite needed here. To solve more general problems of kinematics, you will need some knowledge of differential equations.

I hope this hepled.
FroChro