We are given the acceleration function:

$a(t) = 8\sin(t)$

Step 1: Find the Velocity Function

Since acceleration is the derivative of velocity, we integrate $a(t)$ to find $v(t)$:

$v(t) = \int 8\sin(t) \, dt$

$v(t) = -8\cos(t) + C$

Using the initial condition $v(0) = -11$, we solve for $C$:

$-8\cos(0) + C = -11$

$-8(1) + C = -11$

$C = -3$

Thus, the velocity function is:

$v(t) = -8\cos(t) - 3$

Step 2: Find the Displacement Function

Displacement is the integral of velocity:

$s(t) = \int v(t) \, dt = \int (-8\cos(t) - 3) \, dt$

$s(t) = -8\sin(t) - 3t + C$

The displacement over $[0,3]$ is:

$s(3) - s(0)$

Since we are looking for displacement, we only need to compute the difference:

$s(3) = -8\sin(3) - 3(3) + C$ $s(0) = -8\sin(0) - 3(0) + C$

[ s(3) - s(0) = [-8\sin(3) - 9 + C] - [-8\sin(0) + C] ]

$= -8\sin(3) - 9 + 8\sin(0)$

$= -8\sin(3) - 9$

Approximating:

$\sin(3) \approx 0.1411$

$-8(0.1411) - 9 = -1.1288 - 9 = -10.1288$

Final Answer:

$\mathbf{-10.1288 \text{ m}}$

Let me know if you need more clarification!

Related Questions:

1. What is the velocity of the object at $t = 3$?
2. How would the displacement change if the initial velocity were different?
3. What is the total distance traveled instead of displacement?
4. How does the velocity function behave over time?
5. What would happen if acceleration were $8\cos(t)$ instead?

Tip:

When solving motion problems, always remember that displacement can be negative, but distance traveled is always positive!