Answer:
The total distance covered is [tex]D = 27.76 \ m[/tex]
Explanation:
From the question we are told that
  The initial speed  of the car is  [tex]u = 19.3 \ m/s[/tex]
  The time taken before breaks where applied is  [tex]t = 0.236 \ s[/tex]
   The deceleration is  [tex]a = 8.00 \ m/s^2[/tex]
The first distance covered before break was applied is mathematically evaluated as
   [tex]d = u * t[/tex]
substituting values
   [tex]d = 19.3 * 0.236[/tex]
   [tex]d = 4.484 \ m[/tex]
The distance covered after break has been applied is mathematically evaluated using the equation f motion as follows
   [tex]v^2 = u^2 - 2 as[/tex]
The negative sign is because the car is decelerating  and the final velocity is zero
    [tex]u^2 = 2as[/tex]
=> Â Â [tex]s = \frac{u^2}{2a}[/tex]
substituting values
    [tex]s = \frac{19.3^2}{2(8)}[/tex]
    [tex]s = 23.28 \ m[/tex]
The total distance covered is Â
  [tex]D = d+s[/tex]
substituting values
   [tex]D = 4.484+ 23.28[/tex]
   [tex]D = 27.76 \ m[/tex]
   Â
