Two sprinters are competing in a $100 text(myS race.
Sprinter SA$ accelerates uniformly at the start for $50\text{m}$, where they reach their top speed of $15
(frac(\text(m)\texts)S. They then continue at this speed for the rest of the race.
Sprinter $B$ accelerates at $4 \frac(\text(m}H\text(s}^2}$ for $4 \text{s)$ at the start, then continues at
their final speed until they are passed by sprinter SAS. When this happens, they accelerate at $1
\frac(\text(m}H\text{s}^2}S until the end of the race.
Which sprinter finishes the race faster, and by how much time? Give the time rounded to the nearest
millisecond.

We are tasked with determining which sprinter finishes the race faster and by how much time, based on their respective speeds and accelerations. Let's break this down into several steps:

### Sprinter A's Movement:

1. **Acceleration Phase (first 50 meters):**  
   Sprinter A accelerates uniformly for the first 50 meters until they reach a top speed of $15 \, \frac{\text{m}}{\text{s}}$.
   
   Using the kinematic equation:
   \[
   v^2 = u^2 + 2a s
   \]
   where:
   - $v = 15 \, \frac{\text{m}}{\text{s}}$ (final velocity),
   - $u = 0 \, \frac{\text{m}}{\text{s}}$ (initial velocity),
   - $s = 50 \, \text{m}$,
   we solve for the acceleration $a$:
   \[
   (15)^2 = 0 + 2a(50) \implies 225 = 100a \implies a = 2.25 \, \frac{\text{m}}{\text{s}}^2.
   \]

2. **Time to reach top speed (in the first 50 meters):**
   Using the equation $v = u + at$:
   \[
   15 = 0 + 2.25 t \implies t = \frac{15}{2.25} = 6.67 \, \text{s}.
   \]

3. **Time for the remaining 50 meters at top speed:**
   Sprinter A continues at $15 \, \frac{\text{m}}{\text{s}}$ for the remaining 50 meters. The time taken is:
   \[
   t = \frac{50}{15} = 3.33 \, \text{s}.
   \]

4. **Total time for Sprinter A:**
   \[
   t_A = 6.67 \, \text{s} + 3.33 \, \text{s} = 10 \, \text{s}.
   \]

### Sprinter B's Movement:

1. **Acceleration Phase (first 4 seconds):**  
   Sprinter B accelerates at $4 \, \frac{\text{m}}{\text{s}}^2$ for 4 seconds. The velocity at the end of this phase is:
   \[
   v = u + at = 0 + 4 \times 4 = 16 \, \frac{\text{m}}{\text{s}}.
   \]
   
   The distance covered during this phase is:
   \[
   s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4)(4^2) = 32 \, \text{m}.
   \]

2. **Coasting phase before being passed by Sprinter A:**
   Sprinter B maintains their top speed of $16 \, \frac{\text{m}}{\text{s}}$ until they are passed by Sprinter A.  
   To determine when Sprinter B is passed, we need to figure out the distance Sprinter A has covered by the time Sprinter B reaches 32 meters.

   By the time Sprinter B reaches 32 meters in 4 seconds, Sprinter A has traveled:
   \[
   s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} (2.25) (4^2) = 18 \, \text{m}.
   \]
   Thus, Sprinter A has traveled only 18 meters after 4 seconds.

   Now, Sprinter A continues to accelerate for the remaining 32 meters (50 - 18 = 32 meters). We need to calculate how long it takes Sprinter A to cover this remaining distance.

  

Two sprinters are competing in a 100m race. Sprinter A accelerates uniformly at the start for 50m, where they reach their top speed of 15 m/s. They then continue at this speed for the rest of the race. Sprinter B accelerates at 4 m/s² for 4 seconds at the start, then continues at their final speed until they are passed by sprinter A. When this happens, they accelerate at 1 m/s² until the end of the race. Which sprinter finishes the race faster, and by how much time? Give the time rounded to the nearest millisecond.

Explore how to determine which sprinter finishes faster in a 100m race, with one accelerating uniformly and the other adjusting their speed. Learn how to use kinematic equations, including formulas for acceleration, velocity, and time. This problem is ideal for students in Grades 11-12.

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