A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is

Question

A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 019 (part 2 of 2) 10.0 points If his or her reaction time is 1.21983 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s.

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Hưng Khoa 3 years 2021-07-19T00:17:04+00:00 1 Answers 29 views 0

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  1. thongdat2
    0
    2021-07-19T00:18:58+00:00
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    Answer:

    1.     t_reaction = 1.08 s

    2.     v₀₁ = 16.365 m/s

    Explanation:

    1. This is a kinematics exercise, let’s analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.

    Braking distance

               v² = v₀² + 2 a x

    when he finishes braking the speed is v = 0

                0 = v₀² + 2 a x

                x = -v₀² / 2a

                x = – 17²/2 (-7)

                x = 20.64 m

    the distance for the reaction is

                d = x_reaction + x

                x_reaction = d – x

                x_reaction = 39 – 20.64

                x_reaction = 18.36 m

    as long as it has not reacted the vehicle speed is constant

                v = x_reaction / t_reaction

                t_reaction = x_reaction / v

                t_reaction = 18.36 / 17

                t_reaction = 1.08 s

    2. Let’s find the distance traveled in the reaction time of t1 = 1.21983 s

           as the speed is constant

               v = x / t

               x₁ = v t₁

    the distance traveled during braking is

               v² = v₀² + 2a x₂

               0 = v₀² + 2 a x₂

               x₂ = -v₀² / 2a

             

               v = v₀

    the total distance is

             x_total = x₁ + x₂

             x_total = v₀ t₁ + v₀² / 2a

             39 = v₀ 1.21983 + v₀²/14

             v₀² + 17.08 vo – 546 =0

    we solve the second degree equation

           v₀ = [ -17.08 ±√(17.08²  + 4  546) ]/2

           v₀ = [-17.08 ± 49.81 ]/2

           v₀₁ = 16.365 m/s

           v₀₂ = – 33.445 m/s

    as the acceleration is negative the correct result is v₀₁ = 16.365 m/s

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )