Having trouble with another problem in my dynamics class, if anyone could be of assistance I'd really appreciate it

 

i don't know but let me know what class you are in and what degree you are pursuing so i know not to do the same.

 

I would answer C, C is always right.

 

Take the integral of acceleration, which is velocity. They give you the acceleration function, so you can integrate that and get velocity. That will give you velocity in terms of a starting constant, so it will look like v(t) = ... + C. They give you initial velocity, v(0) = 1, and they give you v(6) = 14 also, and so using those two points you can find the actual velocity equation. Use that to get C, and then you have the complete velocity equation v(t).

Once you have velocity, you need to integrate that again. That will give you distance in terms of a starting constant. It will look like x (t) = ... + C. But again, they give you two points. At time 0, they tell you x = 0. So x (0) = 0 + C. To get a second point, plug in x = 6 to the velocity equation you found above, and you will get x (6) + C. Do the same thing above to eliminate the constant and get the true distance equation x (t).

So now you have a(t), v(t), and x (t) obtained through calculus. It has been a looong time since dynamics so I can't recall too much on the coeff. of restitution, but see if having found v(t) and x (t) will help you out. I'm not too sure where to go from here, but I'm pretty sure you're going to need those. It asks where to determine where the rocket will land, so you'll probably need to use the x (t) equation you calculated above, you'll have to find out of the coeff. of restitution will affect that at all. When you see a problem that gives you a acceleration, and then they give you velocity at two different times, then they are hinting that you need to perform integration. Try that stuff and see if it gets you anywhere and let us know.

 

Well if the rocket weighs a certain weight at takeoff and is a liquid propellant let's say
and the fuel is half the weight of the rocket(usually alot more) the landing weight is
going to be half the weight of the takeoff weight so is that not even figured in all this
or did they forget that equation? And that is going to be needed to figure out it's
impact.

 

I'm having a hard time because the acceleration is given as a function of velocity. I need to figure out what the velocity is at t=10 seconds, then from there using linear momentum is impact equations figure out how it will bounce off that ramp.

Blue beast, this problem ignores the properties of an actual rocket as far as I can tell. We haven't even covered it in this class.

 

Kinetic energy is a function of mass and velocity, potential energy is a function of mass and the force of gravity; total energy will be a combination of the two. I think the problem is simple enough that they want you to use M = 322 LBS, as stated in the problem, and assume it doesn't change (although you're right, in real life it would). I had dynamics as a sophomore in college and this was a pretty typical problem with typical assumptions.

 

Sorry, I didn't see that it was a function of velocity. This makes the problem pretty hairy, because the derivative of a function is a function of the function itself. That means it's a differential equation. Can you try this:

dv/dt = a
dv/dt = 6 * sqrt(v - k)

Then you have a differential equation, although I'm not sure if that is ODE so it might not be solvable. I dunno man, maybe calculus isn't the way to go with this one.

 

so, dt = 6 * (v-k) ^ .5 dt .. then I can integrate each side, so

t = (integral) 6 * (v-k) ^ .5 dt

Just came in from working on the bronco, I'm going to get a shower .. think I'm on the right path?

 

OK, I pulled out my old DiffEQ notes, and here is what you do:

dv/dt = a

dv/dt = 6 * (v - k)^0.5

dv = [6 * (v - k)^0.5] * dt

dv / [6 * (v-k)^0.5] = dt

(1/6) * (v-k)^(-0.5) * dv = dt

(1/6) * integral [(v-k)^(-0.5) * dv] = integral (dt)

(1/6) * 2 * (v-k)^(0.5) = t + C

(1/3) * sqrt(v-k) = t + C

sqrt(v - k) = 3t + 3C

v - k = (3t + 3C)^2

That gives you:

v(t) = (3t + 3C)^2 + k


You then use the two sample points they gave you to find out what the constant C is for. That will get you v(t) in terms of the constant k. This was solved by the "separate equations" method of solving differential equations.

 

Ok, thanks I think I got the velocity .. now to figure out the impact part