Question

A particle, initially at rest, moves along the x-axis such that its acceleration at time t>0 is given by a(t)=cos(t). At the time t=0, its position is x=3.
A) Find the velocity and position functions for the particle.
B) Find the values of t for which the particle is at rest.

Answer 1

We will start with velocity.

Answer 2

Alright :D

Answer 3

\[
v(t)-v(0) = \int_0^t\cos(t)~dt
\]

Answer 4

They have already given us that \(v(0) = 0\), so that simplifies things for us.

Answer 5

Oh so "initially at rest" means velocity right

Answer 6

Yes

Answer 7

Can you integrate?

Answer 8

Uhm
That would mean that the acceleration would be sin(t) + C

Answer 9

No, there is no constant of integration here, because it isn't an indefinite integral.

Answer 10

We used a definite integral, so any constant of integration was canceled out through subtraction.

Answer 11

I don't think I've learned definite/indefinite yet

Answer 12

Definite integrals are ones where we have limits of integration. In this case our limits were \(0\) to \(t\). Indefinite ones don't have limits of integration.

Answer 13

Actually I think I've only learned indefinite

Answer 14

Oh

Answer 15

Technically, I should have written:\[
v(t)-v(0) = \int_0^t\cos(x)~dx
\]Or something, because your variable of integration can't be a limit of integration as well, but is isn't a bit issue.

Answer 16

Anyway, we should get: \[
v(t)-v(0) = \sin(t)\\
v(t)-0 = \sin(t)\\
v(t) = \sin(t)
\]

Answer 17

That is our velocity. The next thing to find is our position.\[
x(t)-x(0) = \int_0^t \sin(u)~du
\]

Answer 18

So I'm on the first chapter of integration, so I'm not very familiar with the vocab. but why don't we include C into the answer.

Answer 19

Did you learn the fundamental theorem of calculus?

Answer 20

I don't think so

Answer 21

Okay, suppose you have a function \(f(x)\) with a derivative \(f'(x)\).

Answer 22

One of the things the fundamental theorem of calculus says is that: \[
\int_a^bf'(t)~dt = f(b)-f(a)
\]

Answer 23

I also haven't learned the little notation on the ends of the integral sign either :c

Answer 24

So you really haven't learned definite integrals?

Answer 25

Nope, only indefinite

Answer 26

Okay...

Answer 27

Okay, let's just suppose then that: \[
v(t) = \sin(t) + C
\]

Answer 28

Yeah

Answer 29

In that case we need to solve for \(C\) by using \(v(0)=0\).

Answer 30

C=0

Answer 31

So that would make v(t) = sin(t)

Answer 32

and the position function is x(t)= -cos(t)

Answer 33

How did you get that?

Answer 34

The integral of the Velocity function would be Position right

Answer 35

But how did you find \(C\)?

Answer 36

well you said that v(0) = 0
so i plugged it into the V(t) = sint +c
0=sin(0) +C
C=0