Expert: Paul Klarreich - 12/27/2007

Question
A particle moves along the x-axis in such a way that its acceleration at time t for t is greater than or equal to 0 is given by a(t) = 4cos(2t). At time t = 0, the velocity of the particle is v(0) = 1 and its position is v(0) = 0.
a. Write an equation for the velocity v(t) of the particle.
b. Write an equation for the position x(t) of the particle.
c. For what values of t, 0 greater than or equal to t less than or equal to pi, is the particle at rest?


Answer
Questioner:   Stephano
Category:  Calculus
Private:  No
 
Subject:  AP Calculus
Question:  A particle moves along the x-axis in such a way that its acceleration at time t for t is greater than or equal to 0 is given by a(t) = 4cos(2t). At time t = 0, the velocity of the particle is v(0) = 1 and its position is v(0) = 0.
a. Write an equation for the velocity v(t) of the particle.
b. Write an equation for the position x(t) of the particle.
c. For what values of t, 0 greater than or equal to t less than or equal to pi, is the particle at rest?
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Hi, Stephano,
IF  a(t) = 4cos(2t), then integrate to get:

v(t) = 2 sin(2t) + v0

Since  v(0) = 1,

v(0) = 2 sin(0) + v0

1 = v0

Now

v(t) = 2 sin(2t) + 1

Integrate to get:

x(t) = - cos(2t) + t + x0

If its position is  x(0) = 0 << that's what you meant to write, I think:

x(0) = - cos(0) + 0 + x0

0 = - 1 + x0

x0 = 1

There's your (a) and (b). Now:

c. For what values of t, 0 <= t <= pi, is the particle at rest?

c. For what values of t, 0 <= t <= pi, is v(t) = 0?

v(t) = 2 sin(2t) + 1
Set that = 0 and solve:

0 = 2 sin(2t) + 1

2 sin(2t) = -1

sin(2t) = -1/2

2t = 210 degrees or 330d.

t = 105 or 165 degrees.