Need help with math HWK!!!
#16
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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)
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)
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.
#17
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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...)
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...)
#18
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theres an equation that relates velocity and acceleration. this is more of a physics problem (#1 ).
#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
#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
#19
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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...)
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?
#21
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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?
#22
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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.
(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.
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.
#23
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From: Bizerkeley
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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.
#28
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From: Bizerkeley
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http://en.wikipedia.org/wiki/Spherical_harmonics
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horsecut54
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12-04-2009 08:31 PM