February 21, 2017

Homework Help: calc

Posted by jane on Thursday, April 2, 2009 at 9:52pm.

i did this problem and it isn't working out, so i think i'm either making a dumb mistake or misunderstanding what it's asking.

A particle moves along the x axis so that its velocity at any time t greater than or equal to 0 is given by v(t) = 1 - sin(2pi*t).

a. Acceleration a(t) of the particle at any time t?

i said that acceleration was the derivative of velocity, so i took the derivative of the velocity function.

v(t) = 1 - sin(2pi*t)
v'(t) = -cos(2pi*t) x 2pi
a(t) = -2pi*cos(2pi*t)

b. Values of t between 0 and 2 where the particle is at rest?

i said it was at rest when velocity = 0.

0 = 1 - sin(2pi*t)
1 = sin(2pi*t)

sin(pi/2) = 1, so...
2pi*t = pi/2
4pi*t = pi
4t = 1
t = 1/4

c. Position x(t) of the particle at any time t if x(0) = 0?

i didn't understand the part about x(0) = 0. i kind of ignored it and it didn't work out. i figured that the position function was the anti-derivative of the velocity function and did:

anti-derivative of 1 - sin(2pi*t)
1/2pi x anti-derivative of 1 - sin(2pi*t)
1/2pi (t + cos(2pi*t)) = x(t)

but that doesn't equal 0 when t = 0. help?

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions