Alice and Bob are in a foot race. When Alice is 24.0 m from the finish line, she has a speed of 4.10 m/s and is 4.00 m behind Bob, who has a speed of 5.00 m/s. Bob thinks he has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.58 m/s? to the finish line. What constant acceleration does Alice now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Bob?

We can solve this problem by finding the time \( t \) it will take Bob to finish the race and then determining the constant acceleration \( a_A \) Alice needs to cover her remaining distance in the same time \( t \), so they finish the race side-by-side.

### Step 1: Bob's motion analysis

Bob is initially 20.0 m from the finish line (since Alice is 24.0 m away and 4.0 m behind him) with an initial velocity \( v_{B0} = 5.00 \, \text{m/s} \) and a constant deceleration \( a_B = -0.58 \, \text{m/s}^2 \).

The equation for Bob's motion with constant acceleration (or deceleration) is:

\[
d_B = v_{B0} t + \frac{1}{2} a_B t^2
\]

Where:
- \( d_B = 20.0 \, \text{m} \)
- \( v_{B0} = 5.00 \, \text{m/s} \)
- \( a_B = -0.58 \, \text{m/s}^2 \)
- \( t \) is the time it takes Bob to finish the race.

Substituting these values into the equation:

\[
20.0 = 5.00 t + \frac{1}{2} (-0.58) t^2
\]

Simplifying:

\[
20.0 = 5.00 t - 0.29 t^2
\]

This is a quadratic equation:

\[
0.29 t^2 - 5.00 t + 20.0 = 0
\]

We can solve this quadratic equation for \( t \).

### Step 2: Solve for time \( t \) using the quadratic formula

The quadratic formula is:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 0.29 \), \( b = -5.00 \), and \( c = 20.0 \).

Now, let's compute the value of \( t \).The two possible solutions for the time \( t \) are approximately 10.93 seconds and 6.31 seconds. Since we are looking for a realistic time for Bob to finish the race, we select \( t = 6.31 \, \text{seconds} \), as the larger value is not physically meaningful for this scenario.

### Step 3: Alice's motion analysis

Now, Alice needs to cover her remaining distance of 24.0 m in the same time \( t = 6.31 \, \text{seconds} \). Her initial speed is \( v_{A0} = 4.10 \, \text{m/s} \), and she will accelerate at a constant rate \( a_A \) to finish the race in the same time.

Using the equation for motion with constant acceleration:

\[
d_A = v_{A0} t + \frac{1}{2} a_A t^2
\]

Where:
- \( d_A = 24.0 \, \text{m} \)
- \( v_{A0} = 4.10 \, \text{m/s} \)
- \( t = 6.31 \, \text{seconds} \)
- \( a_A \) is the acceleration Alice needs to finish side-by-side with Bob.

Substituting the known values:

\[
24.0 = 4.10 \times 6.31 + \frac{1}{2} a_A \times (6.31)^2
\]

We can now solve for \( a_A \).It seems like I can’t do more advanced data analysis right now. Please try again later. However, if you would like, I can guide you step-by-step through the calculation process manually or address any other questions you may have! Let me know how you would like to proceed.

Here are some related questions you might find interesting:
1. What is the significance of constant acceleration in motion analysis?
2. How does deceleration affect the distance covered in a race?
3. Can the quadratic formula always be used for motion problems?
4. What is the relationship between time, speed, and distance in uniform acceleration?
5. How does initial speed impact the time to finish in a race?

**Tip:** When solving problems involving motion, breaking them down into smaller parts like acceleration, initial velocity, and time helps simplify the solution!

Alice and Bob are in a foot race. When Alice is 24.0 m from the finish line, she has a speed of 4.10 m/s and is 4.00 m behind Bob, who has a speed of 5.00 m/s. Bob thinks he has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.58 m/s² to the finish line. What constant acceleration does Alice now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Bob?

Discover how to calculate Alice's required constant acceleration to finish a race side-by-side with Bob, who decelerates at 0.58 m/s². Learn about kinematics, quadratic equations, and motion with constant acceleration in this detailed breakdown suitable for high school students.

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