# Some kinematics/vector calculus

• Sep 12th 2006, 11:38 PM
scorpion007
Some kinematics/vector calculus
Hi, im having problems solving some questions:

1. The position of a particle is given by $r(t) = (2sin(3t))i + (cos(3t))j$
What is its maximum speed?

PS: what's with the latex system? When will it be back? Should i use standard ASCII for now?

I will post more questions in this thread if need be.

EDIT: sorry, i realized i should post questions in separate threads, so i'll do that instead from now on.
• Sep 13th 2006, 04:19 AM
topsquark
Quote:

Originally Posted by scorpion007
Hi, im having problems solving some questions:

1. The position of a particle is given by $r(t) = (2sin(3t))i + (cos(3t))j$
What is its maximum speed?

PS: what's with the latex system? When will it be back? Should i use standard ASCII for now?

I will post more questions in this thread if need be.

EDIT: sorry, i realized i should post questions in separate threads, so i'll do that instead from now on.

To find the max speed you need to find the time derivative:
v(t) = 6 cos(3t) i - 3 sin(3t) j (This is the velocity, incidentally. You gave the displacement function, a vector.)

What is the speed of this particle?
v(t) = sqrt( v_x ^2 + v_y ^2) = sqrt(36 cos^2(3t) + 9 sin^2(3t))
v(t) = sqrt(27 cos^2(3t) + 9)

So maximum speed will occur when a(t) is 0:
a(t) = (-162 sin(3t) cos(3t)) / (2 [sqrt(36 cos^2(3t) + 9 sin^2(3t))]) = 0

Or when -162 sin(3t) cos(3t) = 0.

This equation has a quite simple solution: t = 0 can be seen by inspection. I leave it to you to show that this is a maximum speed, not a minimum.

v(0) = sqrt(27 + 9) = sqrt(36) = 6 (in whatever units you are using.)

-Dan
• Sep 24th 2006, 09:59 PM
scorpion007
where did you get this line from?

Quote:

So maximum speed will occur when a(t) is 0:
a(t) = (-162 sin(3t) cos(3t)) / (2 [sqrt(36 cos^2(3t) + 9 sin^2(3t))]) = 0
the derivative of v(t) does not yield that does it?
• Sep 24th 2006, 10:54 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
where did you get this line from?

the derivative of v(t) does not yield that does it?

Try:

v^2=27 cos^2(3t)+9,

then

2v.a=-162 cos(3t)sin(3t)

a(t)=-162 cos(3t)sin(3t)/v(t),

which when you substitute the expression topsquark gave for v(t)
will give the expression for a(t).

RonL
• Sep 25th 2006, 06:14 AM
topsquark
Quote:

Originally Posted by scorpion007
where did you get this line from?

the derivative of v(t) does not yield that does it?

Sorry. I had made a mistake when I was writing the post out. I guess I didn't correct all of the mistakes before I posted. :o

-Dan