# Velocity and Acceleration Vectors

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1. Jan 23, 2016

### CGI

1. The problem statement, all variables and given/known data

2. Relevant equations
Rotational Kinematic Equations
Kinematic Equations

3. The attempt at a solution
I honestly have no clue how to get a vector out of this.

Θ = Θ(initial) + ω(initial)*t + .5αt^2

and how maybe that v = wr could play into this, but there is so much I don't know where to start.

Any help would be really appreciated!

2. Jan 23, 2016

### TSny

3. Jan 23, 2016

### CGI

Oh okay, so I should be thinking of something along those lines.

So if my theta = 0 and my R = .8ft, would my position vector just be r = .8êr?
what would my r dot and theta dot be? Sorry, this is still relatively new to me
and I've been watching videos on it as well. I just want to make sure I understand
this.

4. Jan 23, 2016

### haruspex

Assuming the answers are wanted in terms of the x, y coordinates:
Let the diagram position represent time 0. At time t, what will r and theta be? So what will x and y be?

5. Jan 23, 2016

### CGI

Would the x just be rcosΘ and y be rsinΘ?

6. Jan 23, 2016

### haruspex

Yes, but you are trying to find velocities and accelerations, so you need to express them as functions of time.

7. Jan 23, 2016

### CGI

Oh okay. Right. Could that just be rcostheta(t) and rsintheta(t)?

8. Jan 23, 2016

### haruspex

According to the problem statement, both r and theta vary with time. Write each as a function of t.

9. Jan 23, 2016

### CGI

Hmmm...okay. Could I say that r = r_initial + vt and that theta = theta_inital + wt where w = 45 rev/min?

10. Jan 23, 2016

### haruspex

Yes. Now express x and y that way and differentiate as necessary.

11. Jan 23, 2016

### CGI

When you say express x and y in that way, do you mean that I can say

x = (r_initial + vt)cos(Θ_initial + ωt)

And the same for y, only with a "sin?"

12. Jan 24, 2016

### haruspex

Yes. Now differentiate.

13. Jan 24, 2016

### CGI

Okay I was just double checking. So when I take the derivative with respect to t I get,

x = vcos(Θ_inital + ωt) - (r_initial + vt)sin(Θ_initial + ωt)*ω

y = vsin(Θ_inital + ωt) + (r_inital + vt)cos(Θ_initial + ωt)*ω

14. Jan 24, 2016

### haruspex

Yes.

15. Jan 24, 2016

### CGI

Okay great. So would these two x and y be the vector for velocity?

16. Jan 24, 2016

### haruspex

Small correction: in your post #13 I presume you meant $\dot x=$ etc., not x=. Similarly y.
If so, yes they would be the x and y components of velocity.