Mary and Sally are in a foot race (see the figure). When Mary is 22 m
 from the finish line, she has a speed of 4.0 m/s
 and is 5.0 m
 behind Sally, who has a speed of 5.0 m/s
. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.38 m/s2
 to the finish line. 
What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?
Express your answer to two significant figures and include the appropriate units.

Let's break down the problem step by step:

### Given Information:
- **Mary:**
  - Initial speed (\(v_{M0}\)) = 4.0 m/s
  - Distance to finish line (\(d_M\)) = 22 m
  - Mary is 5.0 m behind Sally.
  
- **Sally:**
  - Initial speed (\(v_{S0}\)) = 5.0 m/s
  - Distance to finish line (\(d_S\)) = 17 m (since Mary is 5 m behind Sally)
  - Deceleration (\(a_S\)) = -0.38 m/s²

### Step 1: Time for Sally to Finish the Race
Sally's motion is uniformly decelerated, so we can use the following kinematic equation to find the time \(t\) it takes for Sally to finish the race:

\[
d_S = v_{S0}t + \frac{1}{2}a_St^2
\]

Substitute the known values:

\[
17 = 5.0t + \frac{1}{2}(-0.38)t^2
\]

This simplifies to:

\[
17 = 5.0t - 0.19t^2
\]

Rearranging the equation:

\[
0.19t^2 - 5.0t + 17 = 0
\]

This is a quadratic equation. We can solve for \(t\) using the quadratic formula:

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

where \(a = 0.19\), \(b = -5.0\), and \(c = 17\).

Let's calculate \(t\).The two solutions for time \(t\) are approximately 22.30 seconds and 4.01 seconds. Since the time for Sally to reach the finish line must be a realistic and smaller value, we choose \(t = 4.01\) seconds.

### Step 2: Find the Constant Acceleration for Mary
Now, Mary must cover her 22 m distance in the same time \(t = 4.01\) seconds. The equation for uniformly accelerated motion is:

\[
d_M = v_{M0}t + \frac{1}{2}a_Mt^2
\]

Substitute the known values:

\[
22 = 4.0 \times 4.01 + \frac{1}{2}a_M(4.01)^2
\]

Simplify and solve for \(a_M\).The constant acceleration Mary needs to finish the race side-by-side with Sally is approximately \(0.74 \, \text{m/s}^2\).

Do you need any further details or have questions about this problem?

Here are 5 related questions for further exploration:
1. What would happen if Sally did not decelerate at all?
2. How would the solution change if Mary started closer to Sally?
3. What if Sally's deceleration rate were higher?
4. How long would the race take if both maintained constant speeds?
5. Can you derive an equation for a similar problem with different initial conditions?

**Tip:** Always double-check the signs in kinematic equations—negative for deceleration, positive for acceleration.

Mary and Sally are in a foot race. When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally decelerates at a constant rate of 0.38 m/s². What constant acceleration does Mary need during the remaining portion of the race to cross the finish line side-by-side with Sally?

Solve this foot race kinematics problem where Mary must accelerate to finish side-by-side with Sally. The solution involves key concepts like uniform acceleration, quadratic equations, and equations of motion. Suitable for Grade 10-12 students.

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