Respuesta :
Answer:
The car travels to a complete stop from the time the deer is first seen and the distance traveled is 10.04 m.
Explanation:
The speed of the car is 20 m/s.
The deer that steps on to the road is 85 m in front of the car.
Seeing the deer the brakes are applied but after a reaction time of 0.5 seconds.
The deceleration of the car is given as 9 m/[tex]s^2[/tex].
The formula for distance is given as
[tex]s = ut + \frac{1}{2} at^2[/tex] Â Â .... Â Â (i)
where s  = distance traveled in meters
      u = initial velocity
      a = acceleration
      t = time
and v = u + at  ....   (ii)
where v  = final velocity
      u = initial velocity
      a = acceleration
      t = time in seconds
Therefore we have from (ii) Â 0 = 20 - 9t
Therefore time it takes for the car to stop, t = [tex]\frac{20}{9}[/tex] = 2.22 seconds.
Therefore from (i) we have
s = 20(t + 0.5) Â - 9[tex]t^2[/tex]
 = 20(2.22 + 0.5) - (9 × [tex]2.22^2[/tex])
  = 54.4  - 44.36
  = 10.04 m
In the above distance formula that is also mentioned in (i) we have accounted for the reaction time which is 0.5 seconds.
The car travels to a complete stop from the time the deer is first seen and the distance traveled is 10.04 m.
