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Find the instantaneous acceleration at  t=7s

To find the instantaneous acceleration at \( t = 7 \) seconds from the velocity-time graph, we need to determine the slope of the velocity curve at that point.

### Step 1: Identify the relevant segment of the graph
From the graph, the velocity is constant from \( t = 4 \) to \( t = 8 \) seconds. The velocity remains at \( v_x = 25 \, \text{m/s} \) during this interval.

### Step 2: Calculate the acceleration
Since the velocity is constant between \( t = 4 \) and \( t = 8 \) seconds, the slope of the velocity-time graph is zero. The acceleration \( a \) is the derivative of velocity with respect to time, which represents the slope of the graph.

\[
a(t = 7 \, \text{s}) = \frac{\Delta v_x}{\Delta t} = 0 \, \text{m/s}^2
\]

Thus, the instantaneous acceleration at \( t = 7 \) seconds is **0 m/s²**.

Would you like more details on this solution, or do you have any questions?

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Here are five related questions to expand on this topic:
1. How would you find the instantaneous acceleration at \( t = 10 \) seconds?
2. What is the total displacement of the object from \( t = 0 \) to \( t = 14 \) seconds?
3. How does the acceleration change between \( t = 8 \) and \( t = 12 \) seconds?
4. What is the physical interpretation of the area under the velocity-time graph?
5. How can you find the average velocity of the object over the entire time interval?

**Tip:** Always remember that the slope of a velocity-time graph gives the acceleration.

Learn how to find the instantaneous acceleration at t=7 seconds from a velocity-time graph. Follow a step-by-step explanation with calculations showing that the acceleration at this point is 0 m/s². Suitable for high school students studying calculus and physics.

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