Need help with math HWK!!!

Thread Tools
 
Search this Thread
 
Old 01-13-2009 | 05:33 PM
  #16  
chimchimm5's Avatar
Registered User
iTrader: (9)
 
Joined: Oct 2005
Posts: 3,311
From: San Jose, CA
Car Info: 2011 WRX hatch gray
Originally Posted by chinoyboi
OK, here's one of many problems... I'll post more a little later

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s2. Its maximum cruising speed is 90 mi/h. (Give your answers correct to one decimal place.)
(a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

(for a. I calculated that 90mph is 1.5 feet per hour so I just multiplied times 15 and it was still incorrect)
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

Last edited by chimchimm5; 01-13-2009 at 05:57 PM.
Old 01-13-2009 | 05:51 PM
  #17  
chinoyboi's Avatar
Thread Starter
Registered User
iTrader: (8)
 
Joined: Apr 2008
Posts: 1,493
From: Hercules CA
Car Info: 03 WRX --> 07 STI --> 10 Cayman S
Originally Posted by chimchimm5
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
D90-0 = dist covered when decel from top speed to 0

Total dist = (D0-90) + (D15min) + (D90-0)

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

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

(more in a sec...)
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.
Old 01-13-2009 | 05:53 PM
  #18  
mot's Avatar
mot
Likes The Color Purple
iTrader: (5)
 
Joined: Jan 2004
Posts: 1,895
From: Walnut Creek, CA
Car Info: plug-in prius
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
Old 01-13-2009 | 05:56 PM
  #19  
chinoyboi's Avatar
Thread Starter
Registered User
iTrader: (8)
 
Joined: Apr 2008
Posts: 1,493
From: Hercules CA
Car Info: 03 WRX --> 07 STI --> 10 Cayman S
Originally Posted by chimchimm5
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
D90-0 = dist covered when decel from top speed to 0

Total dist = (D0-90) + (D15min) + (D90-0)

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

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

(more in a sec...)

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?
Old 01-13-2009 | 06:22 PM
  #20  
joltdudeuc's Avatar
Old School
iTrader: (6)
 
Joined: Jan 2001
Posts: 14,983
From: Union City
Car Info: '99 RBP GM6
This pretty much stopped me from becoming a Meteorologist...

I could be on TV by now.
Old 01-13-2009 | 06:26 PM
  #21  
psoper's Avatar
250,000-mile Club President
 
Joined: Nov 2002
Posts: 4,770
From: Bizerkeley
Car Info: MBP 02 WRX wagon
Originally Posted by chinoyboi
How can you calculate the acceleration in terms of distance? Is there cancellation of units involved?

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?
Old 01-13-2009 | 06:34 PM
  #22  
psoper's Avatar
250,000-mile Club President
 
Joined: Nov 2002
Posts: 4,770
From: Bizerkeley
Car Info: MBP 02 WRX wagon
Originally Posted by chinoyboi
Two ***** are thrown upward from the edge of a cliff 432 ft above the ground. The first is thrown with a speed of 48 ft/s and the other is thrown a second later with a speed of 24 ft/s.
(a) Do the ***** ever pass each other?
1
yes
no

(apparently they do pass each other because if ball A goes up and eventually falls, then the second ball will continue to go up and thus pass each other, but calculating when is the tricky part for me)

(b) If they pass each other, give the time when this occurs. If they do not pass each other, enter NONE.
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.
Old 01-13-2009 | 06:41 PM
  #23  
psoper's Avatar
250,000-mile Club President
 
Joined: Nov 2002
Posts: 4,770
From: Bizerkeley
Car Info: MBP 02 WRX wagon
Originally Posted by chinoyboi
Find f. (x>0)
f ''(x) = x-2
x > 0
f(1) = 0
f(5) = 0

(So for this one, do you have to find the anti derivative twice? and then calculate?)
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...

Last edited by psoper; 01-13-2009 at 06:54 PM.
Old 01-13-2009 | 06:43 PM
  #24  
Roo's Avatar
Roo
Forester Specialist
iTrader: (3)
 
Joined: Jan 2003
Posts: 3,985
From: Sonoma County
Car Info: '98/'04 Foresters (S & XT)
ROFL at Pete - now I know why you're so damn good at those Time Speed Distance rallies!! LOL

Old 01-13-2009 | 09:02 PM
  #25  
slugrx's Avatar
Registered User
iTrader: (12)
 
Joined: Apr 2008
Posts: 7,224
From: SF Bay Area - Peninsula
Car Info: 04 psm wrx sedan
my roomate probably could do this hes in some crazy advanced math
Old 01-13-2009 | 10:15 PM
  #26  
bluwrxwgn's Avatar
VIP Member
iTrader: (3)
 
Joined: Nov 2002
Posts: 4,074
From: BAIC Wagon Clique
Car Info: '02 WRB WRX Wagon
I took all these calculus classes and differential equations but I don't remember jack. But I do remember that I hated anti derivatives.
Old 01-13-2009 | 10:52 PM
  #27  
chinoyboi's Avatar
Thread Starter
Registered User
iTrader: (8)
 
Joined: Apr 2008
Posts: 1,493
From: Hercules CA
Car Info: 03 WRX --> 07 STI --> 10 Cayman S
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.
Old 01-14-2009 | 12:40 PM
  #28  
psoper's Avatar
250,000-mile Club President
 
Joined: Nov 2002
Posts: 4,770
From: Bizerkeley
Car Info: MBP 02 WRX wagon
Originally Posted by Roo
ROFL at Pete - now I know why you're so damn good at those Time Speed Distance rallies!! LOL

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
Related Topics
Thread
Thread Starter
Forum
Replies
Last Post
Mach5WRX
Hawaii
11
01-22-2004 11:06 PM
AWDcamera
Bay Area
12
04-28-2003 08:48 PM




All times are GMT -7. The time now is 07:04 AM.