#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

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

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.

#3

a(t)=v'(t)=s''(t) give me a minute to review my notes and i'll let u know for sure

#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.

#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?

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

correct me if i'm wrong but i just learned this like a week and a half ago

#7

a + b = c

#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.

s(t) = -cos t + 1

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

#9

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

#11

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.

#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.

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.

*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

#14

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.

#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?

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.

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

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.

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.

#19

I got it all now, thank you all for your time, you guys were very helpful

#20

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

#21

ahh i haven't gotten to integrals yet..... all i know is that its the opposite of derivatives????

#22

yup, its not as hard as it looks, and ill be joining ur russell peters group now