# Homework Help: Kinematics motion - Given position, find velocity and acceleration

1. Dec 17, 2013

### negation

Kinematics motion -- Given position, find velocity and acceleration

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

A particle's position as a function of time is given by xf = xi . sin wt, where xi and w are constants.

a) Find expressions for velocity and acceleration.
b) What are the maximum values of velocity and acceleration.

2. Relevant equations
xf = xi. sin(wt)

3. The attempt at a solution

a) xf = xi. sin(wt)

velocity v = xf' = d/dt xi sin(wt) + xi d/dt sin(w.t) . d/dt w.t

xf' = xi cos(wt).w = xi.w.cos(wt)

acceleration = v' or xf " = [d/dtxi.w + xi d/dt w]cos(w.t) + xi.w d/dt (cos(w.t) . d/dt (w.t)

xf " = x dw/dt cos (wt) - xw^2 sin(w.t)

Can someone verify the above steps for me before I proceed to part (b)?

Last edited: Dec 17, 2013
2. Dec 17, 2013

### Staff: Mentor

Looks OK, but use the fact that ω is constant to simplify that expression. (Your notation is difficult to read. Consider using Latex in the future.)

3. Dec 17, 2013

### negation

Is it possible to simplify further?

On a tangent, how do I utilize Latex? Matlab?

Last edited: Dec 17, 2013
4. Dec 17, 2013

### adjacent

On the advanced editor of physicsforums,you can use latex by pressing the top right button.

5. Dec 17, 2013

### negation

I just realized. But the nitty gritty details would have to wait.

6. Dec 17, 2013

### Staff: Mentor

Yes.

7. Dec 17, 2013

### negation

8. Dec 17, 2013

### Staff: Mentor

dω/dt = ?

9. Dec 17, 2013

### negation

Why do I have to express the equation as dω/dt?

Last edited: Dec 17, 2013
10. Dec 17, 2013

### Staff: Mentor

Huh? Your expression for the acceleration contains dω/dt.

11. Dec 17, 2013

### negation

Last edited: Dec 17, 2013
12. Dec 17, 2013

### Staff: Mentor

That is incorrect. Your first post was correct, just not completed.

13. Dec 17, 2013

### negation

I believe I'm stuck. Even if ω were a constant, hasn't the equation been differentiated to it's second derivative? I could factor out some variables in the second derivative but I don't think that's what you were going at.

14. Dec 17, 2013

### Staff: Mentor

What do you mean "even if"? You are told that ω is a constant.

Answer this. What's the derivative of a constant?

15. Dec 17, 2013

### Staff: Mentor

Another hint: The point I am making is so trivial you will slap yourself once you realize it.

16. Dec 17, 2013

### Staff: Mentor

17. Dec 17, 2013

### negation

it's zero

y" = [d/dt(xw) + (x)d/dt(w)]cos(wt) + (xw) d/dt cos(wt) d/dt(wt)
=
dw/dt cos(wt) - (xw)sin(wt)w
=
dw/dt cos(wt) - xw^2 sin(wt)

Sent from iPhone

Last edited: Dec 17, 2013
18. Dec 18, 2013

### negation

dw/dt cos (wt) - xw sin(wt)w
dw/dt =0
So,
y" = -xw^2 sin(wt)
Is the workings above correctly reduced?

Last edited: Dec 18, 2013
19. Dec 18, 2013

### Vphysics2013

I would suggest you that while calculating derivative of an function if you see any constant in that equation ,you can substitute 0 directly which makes your equation simple.
In your equations
substitute 0 in place of dw/dt of your acceleration equation .

Last edited by a moderator: Dec 18, 2013
20. Dec 18, 2013

### Staff: Mentor

Of course!

Keep going.

Exactly.

21. Dec 18, 2013

### negation

Part (b)

In find maximum velocity:

y' = xw cos (wt)
xw cos (wt) = 0
Cos(wt) = 0
Cos(pi/2) = 0 A cos(3pi/2) = 0
wt = pi/2
t = pi/2w

wt = 3pi/2
t = 3pi/2w

y"(pi/2w)= -xw^2 sin(wt)
= -xw^2 sin(pi/2)
= xw^2 < 0 local max

y"(3pi/2w) = -xw^2
= xw^2 > 0 local min

Max velocity occurs at t = pi/2w

In finding maximum acceleration:

To determine maximum acceleration, we first determine the rate of change of jerk.

Jerk is defined as the 3rd derivative of displacement with respect to time, t.

y'' = -xw^2 sin(wt)

y"' = [d/dt(-xw^2) + (-x)d/dt(w^2)] sin(wt) + (-xw^2) d/dt sin(wt) d/dt(wt)

= [0 + 0 ]sin(wt) - (xw^2) cos(wt)w

= -(xw^3)cos(wt)

y" = 0 = -xw^2 sin(wt)

-xw^2 sin(wt)= 0

sin(wt) = 0

sin(0) = 0 and sin(pi)= 0

wt = 0 and wt = pi

t = 0 and t = pi/w

y"'(0) = -xw^3 cos(w.0)
= -xw^3(1)
= -xw^3

y"'(pi/w) = -xw^3 cos(w.pi/w)
= -xw^3 cos(pi)
= -xw^3(-1)
= xw^3

y"'(0) = -xw^3 < 0
local max

therefore, maximum acceleration occurs at t = 0

Last edited: Dec 18, 2013
22. Dec 18, 2013

### Staff: Mentor

Why are you setting that to zero?

No.

OK.

OK.

Why are you setting the acceleration = zero? You calculated the jerk, so use it!

No.

Try once more.

23. Dec 18, 2013

### negation

What was I thinking. I was rushing to type these.

Will retry

24. Dec 18, 2013

### negation

But in finding maximum velocity, shouldn't I find the derivative of velocity, which gives me acceleration, then set it to zero to determine the roots(max velocity implies y"(root)<0 )?
or are you implying that because I have derived the function for jerk, I should work on jerk instead?

25. Dec 18, 2013

### negation

In finding maximum velocity:

y''= 0 = -xw^2 sin (wt)
sin(wt) = 0 if sin(0) or sin(pi)

wt = 0
t = 0
and
wt = pi
t = pi/w

roots are t = 0 and t = pi/w

y'(0) = xw cos(wt) = xw cos(0) = xw
y''(pi/w) = xw cos(w. pi/w) = xw cos(pi) = -xw

max velocity occurs at t = 0

In finding max acceleration:

y''' = -xw^3 cos(wt) = 0
cos(wt) = 0 if cos(pi/2) and cos(3pi/2)

wt = pi/2
t = pi/2w
and
wt = 3pi/2
t = 3pi/2w

y''(pi/2w) = -xw^2 sin (w . pi/2w)
= -xw^2 sin (pi/2)
= -xw^2

y''(3pi/2w) = -xw^2 sin(w. 3pi/2w)
= -xw^2 sin(3pi/2)
xw^2

max acceleration occurs at t = 3pi/2w

Last edited: Dec 18, 2013
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