#1
the lesson is on integration.

So a partile is movong along the x-axis. The acceleration at time t > 0 is a(t) = cos t.
when t = 0 x = 3.

A. find velocity and position functions.

B. find values of t when particle is at rest.

Any help is appreciated
#2
Quote by newaccount
the lesson is on integration.

So a partile is movong along the x-axis. The acceleration at time t > 0 is a(t) = cos t.
when t = 0 x = 3.

A. find velocity and position functions.

B. find values of t when particle is at rest.

Any help is appreciated

Velocity [v(t)] is going to be the integral of acceleration [a(t)].
Position is going to be the integral of velocity [v(t)].

Then you can use t as a y-value, and use the (x,y) coordinates you have to find out what the Cs are.

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Quote by Trowzaa
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#3
a(t)=v'(t)=s''(t) give me a minute to review my notes and i'll let u know for sure
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#4
I think v(t) = sin t + C. Just a little addition if you need that. I can't recall how to solve C or I would.

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#5
Are there supposed to be two different C's in s(t), or is the C the same through the whole problem?

Edit: I got s(t) = -cost + ct + c, but are the c's a different value?
Last edited by newaccount at Oct 30, 2007,
#6
so v(t)=sin t and s(t)=-cos t
correct me if i'm wrong but i just learned this like a week and a half ago
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#8
answer: v(t) = sin t
s(t) = -cos t + 1

Unless I pfeuked up, which is very possible since I haven't done this since last May.

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#9
Quote by riffmasterjosh
so v(t)=sin t and s(t)=-cos t
correct me if i'm wrong but i just learned this like a week and a half ago



you forgot the constants
#10
the + C at the end is just a constant it stays the same
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#11
Quote by newaccount
Are there supposed to be two different C's in s(t), or is the C the same through the whole problem?

Edit: I got s(t) = -cost + ct + c, but are the c's a different value?

Nah bro, solve for C at the v(t) stage before you go on or you get 2 different C's and you don't know enough variables to solve for both of them.

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#12
Hey, I may have pfeuked up actually.. new post when I fix it.

OK I admit it... I am bafflefied. "t" is usually used as an X-value, and you have given another "x" value because it's moving on the x-axis and I have become confused. But my process is right. The numbers are just wrong in my work, so the answer is wrong.

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Last edited by SteveHouse at Oct 30, 2007,
#13
sorry if i'm confusing people what how do u solve for the c and why? i guess i'm getting that confused since i left it out the first time
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#14
Quote by riffmasterjosh
sorry if i'm confusing people what how do u solve for the c and why? i guess i'm getting that confused since i left it out the first time

This problem confuses me less so let's go with it.

when x = 2, y = 0
f(x) = 2x
F(x) = ?
Antidifferentiate and F(x) = x^2 + C
The function notation can be looked at as a Y value.
0 = 2^2 + C
0 = 4 + C
C = -4
F(x) = x^2 - 4
Fin.

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#15
I messed up , i forgot to say the particle is initially at rest. So I got c by using v(0) = 0.
0 = sin0 + c, so c = 0, so does that mean that s(t) = -cos t?
#16
a(t) = cos(t)
v(t) = sin(t) + C (a constant)
x(t) = -cos(t) + Ct + D (another constant)
x(0) = 3 = -cos(0) + C*0 + D
3 = -1 + D
D = 4

x(t) = -cos(t) + Ct + 4
v(t) = x'(t) = sin(t) + Ct

...not enough information is given to find the velocity function, because of the 'C'. Well, this is because ^ I did it with indefinite integrals. Maybe if you did it with definite, and by using '0' and 't' as the upper and lower limits respectively, it would give something better.
#17
Quote by newaccount
I messed up , i forgot to say the particle is initially at rest. So I got c by using v(0) = 0.
0 = sin0 + c, so c = 0, so does that mean that s(t) = -cos t?


Aha. That's what's needed.
#18
Did you mean to say that x = 0 when t = 3? I think some information is missing. Here's what I know.
The particle started at rest, and then accelerated at a rate of cos t.
At time 0, the particle is at position x = 3.
.....
Duh.
At time 0, if it started at rest, the velocity of the particle, will be 0. sin 0 = 0. That means that C = 0 then, so v(t)=sin t.
Then the position = 3. That is, s(t) = -cos t + c = 3. cos 0 = 1, so -1 + c = 3, and therefore c = 4.
QED, v(t) = sin t and s(t) = -cos t + 4.
Also 4 - cos t, if you are like me and don't like starting something with a negative sign.
FIN. On to the next bastard problem.

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#20
Quote by newaccount
I messed up , i forgot to say the particle is initially at rest. So I got c by using v(0) = 0.
0 = sin0 + c, so c = 0, so does that mean that s(t) = -cos t?


Yes, this means v(t) = sin(t),
but s(t), (I was using x(t)), will have a constant on the end:
s(t) = -cos(t) + 4 (which was found based on s(x) = 3
#22
yup, its not as hard as it looks, and ill be joining ur russell peters group now