Velocity and Acceleration homework

[srh]

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?

 

[OlderDan]

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?

 

[m2003]

Your approach is correct. However, the difference in the answer may be due to rounding errors or different units used. By converting the answer to mm/s, you should get the same value of 42.4 mm/s. It is important to pay attention to units when solving problems in physics.