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its alot of sums; so deriving is easy
0 +360pi/6 cos(t.pi/6) +180pi/3cos(t.pi/3)
ok i got that part right. now the solving for p(t)=0 part is where im stuck
looks tricky ;)
360/6 = 180/3 so that all the same
cos(t.pi/6) + cos(t.pi/3) = 0 when.....
you agree with this so far?
we gotta get them at 180 from each other than right?
t.60 = t.30 + 180; or, t.30 = t.60 + 180
t(60-30) = 180 t(30) = 180 t = 60 maybe?
6.60 = 360 6.30 = 180; thats a good one right?
a(1) + a(-1) = 0 right
so every multiple of 6 should be good for that one...
t.30 = t.60 + 180 t(30-60) = 180 t = 180/-30 t = -6; same thing
so whenever t = a multiple of 6, we should be good
i kinda lost you at t.60 = t.30 + 180; or, t.30 = t.60 + 180. from cos(t.pi/6) + cos(t.pi/3) = 0 in fact, for that^ i simplified and got 60cos((t x pi/6)+ (t x 2pi/6) = 0
given that t not equal zero perhaps; or some such restriction
there are bound to be multples of 6 that make t.60 = t.30; in that case its of no use
ok; we need angles that are 180 from each other in order for there cosines to be opposites of each other right?
so we start at 60 and 30 and need to at least find the angles that are 180 from each other.... t.60 = t.30 + 180 means that for every multiple of t, we move by 60 and 30 till we find angles that are 180 from each other right?
t.60 - t.30 = 180 t(60 - 30) = 180 t = 180/30 t = 6
now when the angels are 360 from eah other they are the same...so we exclude doing the same thing but with +360 instead of +180 right?
360/30 = 12; so every multiple of 12 wont work.... right?
so odd multiples of 6 work; but even multiples fail; so we restricct it to when t is an odd multiple of 6
i can also think of one more condition, but havent found a way to describe it yet......... it when they angle to the right matches the angle on the left; like a 60 and a 120 would have opposite cosines
lol i think im good. thank you.
another question if you dont mind? lol
i dunno, im pretty much good for one question and then go stupid lol
whats the q
LOL, it's a position, velocity, acceleration question.
Find the position function X(t), of a particle moving along a straight line, with constant acceleration equal to 2, knowing that the position and velocity vanish at t = 2.
is it posted already or you wanna hit me with it
Find the initial position and the time t > 0 at which the particle passes again through the initial position.
i know i should do anti derivative for the first part in order to find the velocity equation.
acceleration is the derivative of velocity is the derivative of position
with a constant acceleration of 2; that means we got linear velocity of 2x+c right?
would i plug in velocity = 0 since it vanishes or no?
and a position of x^2 +bx +cthen
the only way for a linear function to vanish at some point it to make a hole iin the graph right?
which means making the velocity function an extention of a rational function
2x +c ------ at best d-2
but with a d-2 up top
something to make a whole at 2 to make the graph vanish
lol i think you might be over thinking the problem. this was suppose to be an easy math problem my prof gave on the test. haha
i could be over thinking it tho :)
lmao would it be wrong to set the velocity - 0?
velocity = 0*
cant have a constant acceleration whout velcoty can we?
my keyboard hates me lol
haha its okay
there was a similar question to this, and my sis helped me with that.
and in that other problem, my sis taught me to set velocity = 0
then solve for the constant
cant say that i ever came across the problem before, so it might be plausible :)
and then from that, do another anti derivative to find the position fuction
lol alrighty, thanks anyways! enjoy the rest of your day :]
amristre can u help me