To find the velocity of the object from $t = 0$ to $t = 3$ seconds, we need to integrate the acceleration function $a(t) = 9 \sin(t)$ with respect to time. The velocity function $v(t)$ is the integral of acceleration:

$v(t) = \int a(t) \, dt$

Given $a(t) = 9 \sin(t)$, the integral of $9 \sin(t)$ is:

$v(t) = \int 9 \sin(t) \, dt = -9 \cos(t) + C$

Where $C$ is the constant of integration. To find the value of $C$, we use the initial condition. The initial velocity is given as $v(0) = 0$ (assuming it’s at rest at $t = 0$):

$v(0) = -9 \cos(0) + C = 0$

Since $\cos(0) = 1$, we have:

$-9(1) + C = 0$

$C = 9$

So, the velocity function becomes:

$v(t) = -9 \cos(t) + 9$

Now, to find the velocity at $t = 0$ and $t = 3$, we substitute these values into the equation:

1. For $t = 0$: $v(0) = -9 \cos(0) + 9 = -9(1) + 9 = 0 \, \text{m/s}$

2. For $t = 3$: $v(3) = -9 \cos(3) + 9$

You can compute $\cos(3)$ using a calculator to get a precise value for $v(3)$.

Would you like to proceed with evaluating the velocity at $t = 3$ or need further explanation?

Here are some related questions:

1. How does the velocity of the object change between $t = 0$ and $t = 3$?
2. What happens to the object's motion when the velocity function has a negative value?
3. How would the velocity function change if the acceleration was given by $a(t) = 9 \cos(t)$?
4. What would happen to the velocity if the initial velocity were not zero?
5. How can you calculate the total distance traveled by the object in this time interval?

Tip: The velocity function is derived from the acceleration function, and the constant of integration depends on the initial velocity. Be sure to always apply the initial conditions when solving for the constant.