Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
   distance travel in the reaction time
    d₁ = v x t
    d₁ = 20 x 0.5 = 10 m
    distance travel after you apply brake
    using equation of motion
    v² = u² + 2 a s
    0 = 20² - 2 x 10 x s
     s =  20 m
 total distance traveled by the car
 D = d₁ + d₂ 
 D = 20 + 10 = 30 m
   distance between car and the deer = 38 m - 30 m
                                                               = 8 m
b) now, maximum speed car.
    distance travel in reaction time
     d₁ = s x t 
     d₁ = 0.5 V
distance left between them
    d₂ = 38 - d₁
    d₂ = 38 - 0.5 V
    distance travel after you apply brake
    using equation of motion
     v² = u² + 2 a d₂
     0 = (V)² - 2 x 10 x (38 - 0.5 V)
      V² + 10 V - 760 = 0
now, solving the quadratic equation
   
   
          V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s