You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s/s. You instantly start running toward the still-open door at
5.5 m/s. What is the maximum time you can wait before starting to run and still catch the bus?

To solve this problem, we need to determine the maximum time you can wait before starting to run and still catch the bus.

### Step 1: Define Variables
- Distance between you and the bus initially: \( d_0 = 9.0 \, \text{m} \)
- Your running speed: \( v = 5.5 \, \text{m/s} \)
- Acceleration of the bus: \( a = 1.0 \, \text{m/s}^2 \)
- Time you wait before starting to run: \( t_{\text{wait}} \)

### Step 2: Set Up Equations
- The bus starts from rest with acceleration, so its position after \( t \) seconds is given by:
  \[
  x_{\text{bus}}(t) = \frac{1}{2} a t^2 = \frac{1}{2} (1.0) t^2 = 0.5 t^2
  \]

- You start running after a delay of \( t_{\text{wait}} \). The time you spend running is \( t - t_{\text{wait}} \), and since you're moving at constant speed, your position is:
  \[
  x_{\text{you}}(t) = v (t - t_{\text{wait}}) = 5.5 (t - t_{\text{wait}})
  \]

Initially, you're 9.0 meters behind the bus, so the relative position between you and the bus must satisfy:
\[
x_{\text{you}}(t) = x_{\text{bus}}(t) + d_0
\]
Substituting the expressions for \( x_{\text{you}}(t) \) and \( x_{\text{bus}}(t) \):
\[
5.5 (t - t_{\text{wait}}) = 0.5 t^2 + 9.0
\]

### Step 3: Solve for \( t_{\text{wait}} \)
We will now solve this equation for the maximum \( t_{\text{wait}} \) that ensures you catch the bus.

First, expand and rearrange the equation:
\[
5.5 t - 5.5 t_{\text{wait}} = 0.5 t^2 + 9.0
\]
\[
0.5 t^2 - 5.5 t + 9.0 = -5.5 t_{\text{wait}}
\]
Now, express \( t_{\text{wait}} \):
\[
t_{\text{wait}} = \frac{0.5 t^2 - 5.5 t + 9.0}{-5.5}
\]
To find the maximum \( t_{\text{wait}} \), we will need to determine the time \( t \) at which the equation holds true by solving for when you and the bus meet.

I'll calculate this now.The times when you and the bus could meet are \( t = 2.0 \, \text{seconds} \) and \( t = 9.0 \, \text{seconds} \). The relevant time for catching the bus is \( t = 9.0 \, \text{seconds} \), since \( t = 2.0 \, \text{seconds} \) represents a scenario where you wouldn't catch up yet.

### Step 4: Maximum Time Before Running
Using the time \( t = 9.0 \, \text{seconds} \), we can now calculate the maximum time you can wait before running, which is \( t_{\text{wait}} \). The equation for \( t_{\text{wait}} \) was:
\[
t_{\text{wait}} = \frac{0.5 t^2 - 5.5 t + 9.0}{-5.5}
\]
Substitute \( t = 9.0 \, \text{seconds} \) into the equation.

I'll compute this now.It seems like I can’t do more advanced data analysis right now. Please try again later. 

In the meantime, if you'd like help understanding the final steps of the problem or want to try manually calculating \( t_{\text{wait}} \), feel free to ask! Here's a reminder of the last equation you would need:

\[
t_{\text{wait}} = \frac{0.5 \times 9^2 - 5.5 \times 9 + 9.0}{-5.5}
\]

Let me know if you'd like further assistance with it or if you have any questions. 

### 5 Related Questions:
1. What happens if your running speed increases?
2. How would the answer change if the bus had a different acceleration?
3. What if the distance to the bus was greater than 9.0 m?
4. How would air resistance affect the time to catch the bus?
5. What if you started running before the bus began accelerating?

### Tip:
Always keep track of units during calculations to ensure accuracy, especially in problems involving multiple physical quantities like speed and acceleration.

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s². You instantly start running toward the still-open door at 5.5 m/s. What is the maximum time you can wait before starting to run and still catch the bus?

Find out how to calculate the maximum time you can wait before running to catch a bus that accelerates at 1.0 m/s², given a running speed of 5.5 m/s. This problem involves kinematic equations and solving a quadratic equation. Ideal for high school students in Grades 10-12.

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