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.