# AP Style Question, Help?

• Oct 14th 2009, 01:46 PM
Jasmina8
AP Style Question, Help?
A particle moves along a line so that at any time t its position is given by x(t)=2pi t + cos 2pi t.

a. Find the velocity at time t.
b. Find the acceleration at time t.
c. What are the values at t, 0<t<3 for which the particle is at rest?
d. What is the maximum velocity?

I would really appreciate your help on how to solve this problem, thanks.
• Oct 14th 2009, 01:50 PM
stapel
Since you've shown no work, I'll guess that you're needing a nudge on how to get started:

Quote:

Originally Posted by Jasmina8
A particle moves along a line so that at any time t its position is given by x(t)=2pi t + cos 2pi t.

a. Find the velocity at time t.

Velocity is the derivative of position with respect to time, so differentiate.

Quote:

Originally Posted by Jasmina8
b. Find the acceleration at time t.

Acceleration is the derivative of velocity with respect to time, so differentiate again.

Quote:

Originally Posted by Jasmina8
c. What are the values at t, 0<t<3 for which the particle is at rest?

To find when the velocity is zero, set the velocity function equal to zero, and solve for the time(s) t.

Quote:

Originally Posted by Jasmina8
d. What is the maximum velocity?

Take whichever techniques they have taught you for finding the maximum of a function, and apply these to the velocity function. (Wink)
• Oct 14th 2009, 02:06 PM
Jasmina8
Am I supposed to find the derivative of the original function to get the velocity, and if so do I do it like this?

x(t)= 2pi t + cos 2pi t

y'= 2pi -sin 2pi t (2pi) ( I guess what I'm thinking is to use chain rule to find the derivative of the first function to get my velocity, so I dont know if I'm right or not?
• Oct 14th 2009, 03:12 PM
stapel
Quote:

Originally Posted by Jasmina8
Am I supposed to find the derivative of the original function to get the velocity...?

You were given a "position" function and asked for a "velocity" function. You were reminded that the "velocity" function is the derivative of the "position" function, and told to differentatet the "position" function to find the "velocity" function.

So, yes, you are supposed to find the derivative of the original "position" function to find the "velocity" function. Sorry for the confusion.

Quote:

Originally Posted by Jasmina8
...and if so do I do it like this?

x(t)= 2pi t + cos 2pi t

y'= 2pi -sin 2pi t (2pi)

Yes. Now simplify. (Wink)
• Oct 19th 2009, 05:21 PM
Jasmina8
AP Question!
thanks.