Question

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 .A. How much distance is between you and the deer when you come to a stop?
B. What is the maximum speed you could have and still not hit the deer?

Answer 1

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

  $x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}$

         V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s