Velocity and Acceleration homework

AI Thread Summary
The acceleration of point A is given by a = -1.8sin(3t), and the initial conditions are x = 0 and v = 0.6 m/s at t = 0. The integration of the acceleration equation leads to the expression v - v0 = 0.6cos(3t). When substituting v0 = 0.6 m/s and evaluating at t = 0.5s, the calculated velocity is v = 0.6424 m/s. However, the textbook answer is 42.4 mm/s, suggesting that the initial velocity should not be included in the integration.
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Homework Statement


The acceleration of point A is defined by the relation a=-1.8sin(kt), where a and t are expressed in m/s^2 and seconds, respectively, and k=3 rad/s. Knowing that x=0 and v=0.6m/s when t=0, determine the velocity of point A when t=0.5s.


Homework Equations


a=dv/dt
a dt = dv
a=-1.8sin(3t)

The Attempt at a Solution


I began by using the equation a dt=dv. I substituted -1.8sin(3t) in for a. After integrating both sides I got v-v0=0.6cos(3t). I plugged in 0.6 m/s for v0. For t=0.5 I got v=.6424 m/s. The answer in the back of th book was 42.4mm/s. If I left v0 as zero, I would get that answer. But I thought I needed to include that. Am I missing something?
 
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srh said:

Homework Statement


The acceleration of point A is defined by the relation a=-1.8sin(kt), where a and t are expressed in m/s^2 and seconds, respectively, and k=3 rad/s. Knowing that x=0 and v=0.6m/s when t=0, determine the velocity of point A when t=0.5s.


Homework Equations


a=dv/dt
a dt = dv
a=-1.8sin(3t)

The Attempt at a Solution


I began by using the equation a dt=dv. I substituted -1.8sin(3t) in for a. After integrating both sides I got v-v0=0.6cos(3t). I plugged in 0.6 m/s for v0. For t=0.5 I got v=.6424 m/s. The answer in the back of th book was 42.4mm/s. If I left v0 as zero, I would get that answer. But I thought I needed to include that. Am I missing something?

Think about the highlighted text. What function of the form v = f(t) is going to give you v0=0.6m/s?
 
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