# Homework Help: Dynamics Question

1. Jan 7, 2006

### epiphany

My hw problems looks like this:

The acceleration of a particle is directly proportional to the square of the time t. When t=0, the particle is at x=36 ft. Knowing that at t=9 s, x=144 ft. and v=27 ft/s, express x and v in terms of t.

I'm not exactly sure what directly proportional means. Can someone help me out?

Also I have another question I'm stuck on:

The acceleration of a particle is defined by the relation a= 12x-28, where a and x are expressed in m/s2 and meters, respectively. Knowing that v=8 m/s when x=0, determine (a) the maximum value of x, (b) the velocity when the particle has traveled a total distance of 3m.

2. Jan 7, 2006

### Tom Mattson

Staff Emeritus

http://mathworld.wolfram.com/DirectlyProportional.html

Note that $a=\frac{d^2x}{dt^2}$ and solve the resulting differential equation for $x$. You should be able to answer the questions using that solution.

3. Jan 7, 2006

### epiphany

Sorry, I'm kinda slow. I'm still not sure what they mean in the question with directly proportional. Does it mean that a = t squared?

4. Jan 7, 2006

### Benny

$$a = 12x - 28 \Rightarrow \frac{{d^2 x}}{{dt^2 }} - 12x = - 28$$

You have a second order ODE. The ICs and the equation both suggest that you can consider this equation as one in which the independent variable (t) is missing.

$$p\left( x \right) = \frac{{dx}}{{dt}} \Rightarrow \frac{{d^2 x}}{{dt^2 }} = \frac{d}{{dt}}\left( {p\left( x \right)} \right) = \frac{{dp}}{{dx}}\frac{{dx}}{{dt}} = p\frac{{dp}}{{dx}}$$

The DE becomes:

$$p\frac{{dp}}{{dx}} - 12x = - 28$$ which is separable after a little rearrangement.

You should be able to solve it from there.

(if you run into problems just reply to this thread - I haven't done the whole question so I don't know if the problem is easy or not)

As for your other question I would interpret it as meaning $$a = kt^2$$.

Edit: Fixed statement above regarding the type of the DE.

Last edited: Jan 8, 2006
5. Jan 8, 2006

### hotvette

Benny is right. It just means that $a = kt^2$, where k is a constant. The use of the word 'directly' is rather redundant.