Ask Ravi

:lol: :lol: There's a pants half-off sale down the street. :naughty:
 
Originally posted by leone@Oct 18 2006, 12:47 AM
:lol: :lol: There's a pants half-off sale down the street. :naughty:
:lol: :lol: :lol: This entire conversation is archive-worthy but this part had me crying. :lol: :lol:
 
Hey ravi, i'm looking to a buy a new tv, and we've been offered a good deal on a SHARP 37" LCD. Model no. AW5X. we double checked the model number, cos I couldnt' find anything aboot it on the internet, but it's AW5X. Any clues on whether or not it's a decent TV?

We aren't getting this from a store, so i have to ask if Sharp make TVs they don't release commercially...
 
Originally posted by cultclassic+Oct 18 2006, 10:50 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (cultclassic @ Oct 18 2006, 10:50 PM)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-leone@Oct 18 2006, 10:42 PM
:lol: Do tell. :naughty:
It's what inside that counts... :naughty: [/b][/quote]
:lol: :lol: :lol:
 
I keep seeing this type of problem. How would you start?

The position of a particle for the time interval t=0.00 s to t=3.00s is given by:

x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

(a) Find expressions for the velocity as functions of time. (b.) At the time(s) between t=0 and t=3s is the particle at rest? (c.) For each of calculated in part b, is the acceleration of the particle positive or negative? (d) Find the maximum velocity of the particle. (e) Calculate the acceleration of the particle when the velocity is a maximum. (f) Does the particle have uniform ccelerated motion?
 
Originally posted by leone@Oct 19 2006, 11:03 AM
I keep seeing this type of problem. How would you start?

The position of a particle for the time interval t=0.00 s to t=3.00s is given by:

x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

(a) Find expressions for the velocity as functions of time. (b.) At the time(s) between t=0 and t=3s is the particle at rest? (c.) For each of calculated in part b, is the acceleration of the particle positive or negative? (d) Find the maximum velocity of the particle. (e) Calculate the acceleration of the particle when the velocity is a maximum. (f) Does the particle have uniform ccelerated motion?


a) diffrerentiate once WRT t to get the velocity equation v(t)
b.) plot v(t) between t=0 to 3. You will see that velocity increases first then decreases; at t=3, if you plug in the values it's negative; so yes, the velocity does become zero at some point between 0 and 3.
c)set the above velocity equation to zero, v(t) = 0 and solve fot t.
differentiate v(t) once more, to get the acceleration a(t). plug the values for t, see if a(t) is + or - at thet instance...
d)velocity is maximum or minimum when the acceleration is zero; put a(t)=0; solve for t; once you get t, plug it into v(t); that will be the maximum value of v
e)I don't think so; uniform acceleration meand a(t) is a constant... not in this case, because x(t) is a third degree, so a(t) is first degree, not a constant..


i've had 3 beers, so i'll have to revisit these answers in the morning... :P
 
:lol: They're better than what I got! Thanks! :wub: :wub:

What's WRT? :unsure:

Does it matter what values you plug in, as long as they're between 0 and 3?
 
WRT = with respect to

Does it matter what values you plug in, as long as they're between 0 and 3?

you mean for plotting? no. more values the better..

btw, this plotting page is better than the other one, you can see the values as you move the mouse over the graph...
so
b.) velocity zero when t is approx 2.7 something,
c) at that point, when t=2.7, accel is -20 something, yea so it is negative..
d) max velocity is at t=1.16667
e) plug t=1.16667
f ) i don't know what they mean by "uniform": :unsure: what I meant was acceleration is not constant; but it IS uniformly decreasing, because the plot is a straight line sloping down...

btw, btw means "by the way" :P
 
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