Alex and Abdeleselam are running in the 5000m race. When Alex is 22m from the finish line, he has a speed of 4.0 m/s and is 5.0 m behind Abdeleselam, who has a speed of 5.0 m/s. Abdeleselam thinks he has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.40 m/s2 to the finish line. What constant acceleration does Alex need to finish 5m ahead of Abdeleselam?

To solve this problem, we need to calculate the constant acceleration Alex requires to finish 5 meters ahead of Abdeleselam.

### Known variables:

- Distance Alex needs to travel: \(d_{\text{Alex}} = 22 \, \text{m}\)
- Speed of Alex: \(v_{\text{Alex}} = 4.0 \, \text{m/s}\)
- Alex is 5.0 meters behind Abdeleselam: \(d_{\text{Abdel}} = 22 - 5 = 17 \, \text{m}\)
- Speed of Abdeleselam: \(v_{\text{Abdel}} = 5.0 \, \text{m/s}\)
- Abdeleselam's deceleration: \(a_{\text{Abdel}} = -0.40 \, \text{m/s}^2\)
- Alex needs to finish 5 meters ahead, so his total distance should be \(22 + 5 = 27 \, \text{m}\)

#### Step 1: Find the time Abdeleselam takes to finish the race.
We can use the equation for motion with constant acceleration:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
For Abdeleselam:
- Initial velocity \(v_0 = 5.0 \, \text{m/s}\)
- Acceleration \(a = -0.40 \, \text{m/s}^2\)
- Distance \(d = 17 \, \text{m}\)

The equation becomes:
\[
17 = 5.0 t + \frac{1}{2} (-0.40) t^2
\]
Simplifying:
\[
17 = 5.0 t - 0.20 t^2
\]
Rearrange the quadratic equation:
\[
0.20 t^2 - 5.0 t + 17 = 0
\]
Multiply through by 5 to clear the decimals:
\[
t^2 - 25 t + 85 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
t = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(1)(85)}}{2(1)}
\]
\[
t = \frac{25 \pm \sqrt{625 - 340}}{2}
\]
\[
t = \frac{25 \pm \sqrt{285}}{2}
\]
\[
t = \frac{25 \pm 16.88}{2}
\]
Taking the positive root (since time cannot be negative):
\[
t = \frac{25 + 16.88}{2} = \frac{41.88}{2} \approx 20.94 \, \text{seconds}
\]

#### Step 2: Find the acceleration Alex needs.
Alex must cover 22 + 5 = 27 meters in the same amount of time, \(t = 20.94\) seconds. Using the same equation for motion with constant acceleration:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Substitute \(d = 27\), \(v_0 = 4.0 \, \text{m/s}\), and \(t = 20.94 \, \text{seconds}\):
\[
27 = 4.0 \times 20.94 + \frac{1}{2} a \times (20.94)^2
\]
Simplifying:
\[
27 = 83.76 + \frac{1}{2} a \times 438.56
\]
\[
27 - 83.76 = 219.28 a
\]
\[
-56.76 = 219.28 a
\]
\[
a = \frac{-56.76}{219.28} \approx -0.259 \, \text{m/s}^2
\]

### Final Answer:
Alex needs a constant **deceleration** of approximately \(-0.259 \, \text{m/s}^2\) to finish 5 meters ahead of Abdeleselam.

Do you need any more details or further explanation?

Here are 5 related questions:
1. What is the significance of negative acceleration in this context?
2. How would the solution change if Abdeleselam's deceleration was not constant?
3. How can we calculate the maximum speed Alex could have and still finish 5 meters ahead?
4. What if Alex needed to finish 10 meters ahead of Abdeleselam? How would that affect the acceleration?
5. How does the quadratic formula help in solving kinematic problems like this?

**Tip:** Always double-check whether the given acceleration is positive or negative; it changes the interpretation!

Alex and Abdeleselam are running in the 5000m race. When Alex is 22m from the finish line, he has a speed of 4.0 m/s and is 5.0 m behind Abdeleselam, who has a speed of 5.0 m/s. Abdeleselam thinks he has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.40 m/s^2 to the finish line. What constant acceleration does Alex need to finish 5m ahead of Abdeleselam?

Learn how to calculate the constant acceleration Alex needs to finish 5 meters ahead of Abdeleselam in a 5000m race. This problem covers key kinematic equations, including distance, velocity, and acceleration, with the quadratic formula solution. Suitable for Grades 9-12.

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