You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 57 m in front of you. Your reaction time before ste

Question

You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 57 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?

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Khang Minh 3 years 2021-08-31T16:44:32+00:00 1 Answers 16 views 0

Answers ( )

  1. thuthuy
    0
    2021-08-31T16:46:29+00:00
    Reply

    Explanation:

    Lets calculate the distance that the car has traveled first.

    a)

    as the car was traveling with the constant speed.

    v = \frac{d}{t}

    d = v×t

    d = (20)(0.5) = 10 m    (1)

    now we will calculate the distance that is required in order to stop the car.

    v^{2} _{f} = v^{2} _{i} + 2ad_{2}

    0 = (20)²+ 2(-10)d_{2}

    d_{2} = 20 m       (2)

    now by using (1) and (2) we can find out the total distance traveled y the car before it stops.

    d_{1} +d_{2} = 30 m

    Distance between the car and the deer when it stops = 57 – 30 = 27 m

    b)

    lets calculate the maximum speed you could have and still not hit the deer.

    we have

    v_{f} = 0

    total distance traveled before stopping = 57 m

    we must calculate the distance for reaction time with constant speed.

    d =v _{max} t = 0.5 v_{max}

    d_{2} = 57 - d_{1}

    d_{2} = 57 - 0.5v_{max}

    v^{2} _{f} =v^{2} _{max} + 2ad_{2}

    0 = v^{2} _{max + 2(-10)(57- 0.5v_{max})

    v^{2} _{max} = 1140 - 10v_{max}

    Apply the quadratic equation on it.

    v_{max} = -b±\sqrt{b^{2}-4ac }/2a

           a = 1

           b = 10

           c = -1140

    by putting these values we got

    v^{} _{max} = 29.13, - 39.13 m/s\\

    we neglect the negative value.

    v^{} _{max} = 29.13 m/s

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