Oh, yeah, whoops. It is: v(t)= -9.8t - 4. Okay, now I make the problem:
ds/dt = -9.8t - 4
Am I on the right track now?
I am suppose to integrate next, correct?
Hmm.. so to find v, I would integrate dv/dt, which would give me: v = 9.8t + C? Then, to solve for c, I do 4 = c and that gives me the equation for time which is v(t) = 9.8t + 4?
Is this correct so far, or what do I need to reconsider?
Hmm... it seems it would be, but how would I set-up the formula to determine where the object is at time t? I can't do that with just dv/dt = 9.8, can I?
The at time = 0, velocity is 4, so assuming no air resistance, does that mean it takes 125 seconds to reach the bottom, or does gravity alter...
Homework Statement
I've been given a problem that an object falls from a distance of 800 feet with an initial velocity of 4 f/s. Gravity acts on the object, but air resistance can be disregarded. Find the location at time t and find when the object hits the ground.
The Attempt at a...
e^\int^-^1^/^x = e^-^l^n^x = -x. So this means I would multiply the whole problem by -x which gives me what I start out with.
Or am I doing something wrong?
Okay, so if I divide by X, I get y' - \frac{y}{x} = 0
So the integrating factor would be -x? That can't be right because that just gives me what I started out with.
Oh.. you're right.
It's not separable though, is it? When I try and separate it I cannot get the Ys on one side and the Xs on one side. Unless \frac{dy}{y} = \frac{dx}{x} is the equation correctly separated.
Oh, I understand! Just multiply by e^-x, so the problem looks like this:
-ye^-^x + xe^-^xy' = 0
Therefore:
xe^-^xy = c
Correct?
I'm having trouble with the next problem also. The next problem is similar, but with a x^3 added to it. I can't think of any other way to do it other...
Homework Statement
-y + xy' = 0 and y(2)=5
The Attempt at a Solution
This first part trips me up. I am supposed to find the perfect derivative, which is (xy') = 0 ? Is this legal, or does the -y not allow for that?
If that is correct, then I know that I integrate that, which...
So, here's the problem:
\frac{dy}{dx} = \frac{8y}{5x}
To start off, I separate the integrals, which gives me:
\frac{dy}{8y} = \frac{dy}{5x}
After that, I integrate both sides, which gives me:
\frac{ln8y}{8} = \frac{ln5x}{5} + c
Now, the question says that it runs through (4, 1), so that...