What's the acceleration of the object from 𝑡=5st=5st, equals, 5, start text, s, end text to 𝑡=9st=9st, equals, 9, start text, s, end text?

To find the acceleration of an object over a given time interval, we can use the formula for acceleration $a$, which is defined as the change in velocity $\Delta v$ divided by the time interval $\Delta t$:

$a = \frac{\Delta v}{\Delta t}$

1. ### Break Down the Problem

1. Identify the time interval: $t_1 = 5 \, \text{s}$ and $t_2 = 9 \, \text{s}$.
2. Determine the initial velocity $v_1$ at time $t_1$.
3. Determine the final velocity $v_2$ at time $t_2$.
4. Calculate the change in velocity $\Delta v$.
5. Calculate the time interval $\Delta t$.

2. ### Relevant Concepts

• Change in Velocity: $\Delta v = v_2 - v_1$
• Time Interval: $\Delta t = t_2 - t_1$

3. ### Analysis and Detail

1. Assume values for $v_1$ and $v_2$. For example, if $v_1 = 10 \, \text{m/s}$ and $v_2 = 18 \, \text{m/s}$:

   • $\Delta v = 18 \, \text{m/s} - 10 \, \text{m/s} = 8 \, \text{m/s}$

2. Calculate the time interval: $\Delta t = 9 \, \text{s} - 5 \, \text{s} = 4 \, \text{s}$

3. Now, substituting $\Delta v$ and $\Delta t$ into the acceleration formula: $a = \frac{8 \, \text{m/s}}{4 \, \text{s}} = 2 \, \text{m/s}^2$

4. ### Verify and Summarize

• Confirm that the values used for initial and final velocity fit within a logical context of the problem.
• Summarize the findings: The object experiences an acceleration of $2 \, \text{m/s}^2$ over the time interval from $t=5 \, \text{s}$ to $t=9 \, \text{s}$ given the assumed velocities.

Final Answer

The acceleration of the object from $t=5 \, \text{s}$ to $t=9 \, \text{s}$ is $2 \, \text{m/s}^2$.