We are given the acceleration function:

$a(t) = 9\sin(t)$

with the initial velocity:

$v(0) = -10 \text{ m/s}$

We need to find the displacement of the object over the interval $[0,3]$.

Step 1: Find the Velocity Function

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

$v(t) = \int 9\sin(t) \, dt$

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

Using the initial condition $v(0) = -10$:

$-9\cos(0) + C = -10$

$-9(1) + C = -10$

$C = -1$

Thus, the velocity function is:

$v(t) = -9\cos(t) - 1$

Step 2: Find the Displacement

Displacement is the integral of velocity over the given interval:

$s = \int_0^3 v(t) \, dt = \int_0^3 (-9\cos(t) - 1) \, dt$

$s = \int_0^3 -9\cos(t) \, dt - \int_0^3 1 \, dt$

$s = (-9\sin(t)) \Big|_0^3 - (t) \Big|_0^3$

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

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

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

Approximating $\sin(3) \approx 0.1411$:

$s \approx -9(0.1411) - 3$

$s \approx -1.27 - 3$

$s \approx -4.27 \text{ m}$

Final Answer:

The displacement over the interval $[0,3]$ is approximately -4.27 meters, meaning the object moves 4.27 meters in the negative direction.

Would you like a breakdown of any step or further clarifications?

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Related Questions:

1. How do we find the total distance traveled instead of displacement?
2. What would the displacement be if the initial velocity were different?
3. How does the choice of the time interval affect displacement?
4. Can we determine when the object momentarily stops?
5. What would happen if acceleration were a different function, such as $a(t) = 9\cos(t)$?

Tip:

Displacement considers direction, while total distance traveled is always positive and requires checking when velocity changes sign!