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 6.1 m/s. How long does it take for you to reach the open door and jump in?

To solve this problem, we need to set up equations of motion for both you and the bus and determine when you will reach the door.

### Step 1: Define the variables and equations
1. **For you (running towards the bus):**
   - Initial position: \( x_{\text{you}}(0) = 0 \) m (starting point)
   - Velocity: \( v_{\text{you}} = 6.1 \) m/s (constant speed)

   Your position as a function of time \( t \) is given by:
   \[
   x_{\text{you}}(t) = 6.1t
   \]

2. **For the bus:**
   - Initial position: \( x_{\text{bus}}(0) = 9 \) m (starting point)
   - Initial velocity: \( v_{\text{bus}} = 0 \) m/s (starting from rest)
   - Acceleration: \( a_{\text{bus}} = 1.0 \) m/s\(^2\)

   The bus’s position as a function of time \( t \) is given by:
   \[
   x_{\text{bus}}(t) = 9 + \frac{1}{2} \cdot 1.0 \cdot t^2 = 9 + 0.5t^2
   \]

### Step 2: Set up the equation to find the time when you reach the bus
You reach the bus when your position equals the bus's position:

\[
6.1t = 9 + 0.5t^2
\]

### Step 3: Rearrange the equation into a standard quadratic form
Rearrange the terms to form a quadratic equation:

\[
0.5t^2 - 6.1t + 9 = 0
\]

Multiply everything by 2 to clear the fraction:

\[
t^2 - 12.2t + 18 = 0
\]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -12.2 \), and \( c = 18 \):

\[
t = \frac{-(-12.2) \pm \sqrt{(-12.2)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}
\]

\[
t = \frac{12.2 \pm \sqrt{148.84 - 72}}{2}
\]

\[
t = \frac{12.2 \pm \sqrt{76.84}}{2}
\]

\[
t = \frac{12.2 \pm 8.76}{2}
\]

Now, calculate the two possible times:

\[
t_1 = \frac{12.2 + 8.76}{2} = 10.48 \, \text{s}
\]

\[
t_2 = \frac{12.2 - 8.76}{2} = 1.72 \, \text{s}
\]

Since the only reasonable solution is the positive time when you actually reach the bus:

\[
\boxed{t = 1.72 \, \text{s}}
\]

### Conclusion
It takes you approximately **1.72 seconds** to reach the open door of the bus and jump in.

Would you like any further details on the solution, or do you have any questions?

### Related Questions:
1. What would happen if the bus had a different acceleration?
2. How would your running speed affect the time to reach the bus?
3. What if the bus started at a higher initial velocity?
4. How does the initial distance between you and the bus impact the time required?
5. What happens if you start running a second or two after the bus starts?

### Tip:
When working with equations of motion, always identify your initial conditions and variables clearly to set up accurate equations for the situation.

Discover how to solve a kinematics problem where a person runs towards a bus that accelerates away. This example uses the quadratic formula and equations of motion to determine the time needed to catch up. Ideal for students in Grades 9-12.

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