chimchimm5

iTrader: (9)

I don't have time to do (and shouldn't be doing) the math for you, but to pass college level (engineering level) calculus, you must understand everything you're doing to the point of being able to derive all the equations you use.

In this case, the integral (antiderivative) acts as the summation. So you can think of the summation in 3 major parts for part (a):

D0-90 = dist covered when accel from 0 to top speed (90)
D15min = dist covered at top speed for 15min

Total dist = (D0-90) + (D15min)

I don't have "integral" symbol so I'll use: |

D15min is easy = velocity * time = 90 m/h * 15 min (you convert the units)

But Acceleration is the derivative of Velocity, Velocity is the derivative of Distance.

So D0-90 = |0,90 D(dv), where D(dv) = dv * dt

crap I don't have time to finish... g/l

chinoyboi

iTrader: (8)

Thanks I appreciate your help! I'm not really asking for you to solve it for me, more of, where should I be going. I'm not exactly an "engineering" major in particular, I'm more of Information Systems, which don't really require a whole lot of engineering application-type math. It's been nearly 3 quarters since I last took math and I'm very rusty on it.

mot

iTrader: (5)

theres an equation that relates velocity and acceleration. this is more of a physics problem (#1 &#2).

#3: ya, integrate twice

#4: f should = 4x^(3/2) + 4x^(5/2) + 2 (c = 2)

#5: f = -sin(theta) - cos(theta) +c1(theta) +c2 (c1 = 2 and c2 = 3)

i gotta run to class. good luck

chinoyboi

iTrader: (8)

Wait, I can calculate most of the things, but the most confusing part I'm having trouble is calculating the acceleration, which is 4ft/s^2. And most of the problems have to account the acceleration as well.

I looked up that 1 foot is 0.000189393939, which 4 ft is roughly 0.00076 miles, which is .00076 miles/s^2. How can you calculate the acceleration in terms of distance? Is there cancellation of units involved?

joltdudeuc

iTrader: (6)

This pretty much stopped me from becoming a Meteorologist...

I could be on TV by now.

psoper

Acceleration is given, you need to calculate the distance based on that;

a=dv/dt, so velocity is your first intergral, you can then determine at what time it reaches the cruising speed, after which velocity is constant- but remembering that v=dx/dt so distance (x) during the acceleration interval comes out of the second integral of the acceleration over that duration.

The "ft per seconds squared" becomes "ft per second" for velocity and goes away on the second integral to give you just feet (distance unit).

Does that help?

psoper

In this instance both are accelerating at -32 ft/sec^2 it is just the initial conditions that have changed, for ball one the initial position is 432 ft, t=0 and the initial velocity is 48 ft/sec, while the second ball starts from the same position a second later, so t=1 and initial velocity is 24 ft/sec-
I wouldn't bet on them passing each other, because the first ball might still be going upward or at least still be enough higher in the air when the second ball reaches its apogee that it won't catch it on the way down-

you'll need to do the math, but its the same set of equations that applied during the train's acceleration interval, just 2 different cases with different initial conditions.

psoper

Yes, each anti-derivative (or integral) will generate an unknown constant, so you will wind up with three vairiables and three equations;
f"(x) is given, f'(x) will give you a second order equation with an unknown "c" while f(x) will be a third order equationwith another unknown constant "b".

But you also were given know two data points so it should all fall together with that info.

and mot gave you the answers for the other two- you should be able to do the work to show how he got there...

Roo

iTrader: (3)

ROFL at Pete - now I know why you're so damn good at those Time Speed Distance rallies!! LOL

slugrx

iTrader: (12)

my roomate probably could do this hes in some crazy advanced math

bluwrxwgn

iTrader: (3)

I took all these calculus classes and differential equations but I don't remember jack. But I do remember that I hated anti derivatives.

chinoyboi

iTrader: (8)

Thanks for all your help guys. I appreciate it, there are still some rough areas, but I'll go over this with my TA or professor.

psoper

TSD and this stuff is all pretty simple- in comparison, here's an introduction to the stuff I have to dig through for work these days;

http://en.wikipedia.org/wiki/Spherical_harmonics