To find the acceleration of the object, we'll use the kinematic equation that relates position, initial velocity, time, and acceleration:

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$

where:

• $x(t)$ is the position at time $t$.
• $x_0$ is the initial position.
• $v_0$ is the initial velocity.
• $a$ is the acceleration.
• $t$ is the time.

Given values:

• Initial position ($x_0$) = 3.09 cm
• Initial velocity ($v_0$) = 14.0 cm/s
• Time ($t$) = 3.55 s
• Final position ($x(t)$) = -5.00 cm

Plugging in these values, we get:

$-5.00 = 3.09 + 14.0 \times 3.55 + \frac{1}{2} a \times (3.55)^2$

Now, we can solve for the acceleration $a$. Let's calculate it step-by-step:

1. Calculate $14.0 \times 3.55$.
2. Subtract this product and the initial position from the final position.
3. Divide the result by $\frac{1}{2} \times (3.55)^2$ to find the acceleration.

Let's compute this:

$-5.00 = 3.09 + 49.7 + \frac{1}{2} a \times 12.6$

Simplifying further:

$-5.00 = 52.79 + 6.275a$

Solving for $a$:

$-5.00 - 52.79 = 6.275a$

$-57.79 = 6.275a$

$a = \frac{-57.79}{6.275}$

Now, let’s calculate the final value for acceleration $a$.The acceleration of the object is approximately $-9.17 \, \text{cm/s}^2$.

This negative value indicates that the object is decelerating in the positive x-direction (or accelerating in the negative x-direction).

Would you like more details or have any questions? Here are some related questions to consider:

1. What would be the velocity of the object at $t = 3.55 \, \text{s}$?
2. How can you determine the displacement of the object over a different time interval?
3. What initial velocity is required for the object to stop exactly at the final position?
4. How would the results change if the initial velocity was in the negative x-direction?
5. How do you determine the time it takes for the object to reach a specific position?

Tip: When dealing with kinematic equations, always keep track of the signs for velocity, acceleration, and displacement, as they indicate direction.