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21 September, 21:09

You're driving down the highway late one night at 17 m/s when a deer steps onto the road 45 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 12 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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Answers (2)
  1. 21 September, 23:34
    0
    Answer:A) 33m

    B) 32.86m/s

    Step-by-step explanation:

    Using the equation of motion

    V^2=U^2-2aS

    Where V=final velocity of the car=0

    U=initial velocity=17m/s

    a=deceleration=-12m/s^2

    S = distance covered by car before stopping

    Substituting into the equation

    0=17^2-2*12*S

    0=289-24S

    24S=289

    S=289/24

    S=12m.

    The distance between the car and the deer=45-12=33m

    B) Using same equation

    Distance, S is now 45m, which is the distance between the car and the deer. while deceleration remains same

    V^2=U^2-2aS

    0=U^2-2*12*45

    U^2=1080

    U=√1080

    U=32.86m/s
  2. 22 September, 00:26
    0
    A. So therefore distance between deer and car after stopping = 36.5 - 12.04 = 24.46m

    B. u = 29.6m/s = initial velocity

    Step-by-step explanation:

    Given the reaction time as 0.5s

    So the distance traveled before applying breaks = u*rt

    Where u = initial velocity = 17m/s

    rt = reaction time = 0.5s

    So distance traveled = 0.5 * 17 = 8.5m

    So distance between deer and car before applying breaks = 45 - 8.5 = 36.5m

    So therefore using kinematics formula to solve for distance covers after applying breaks

    V² = u² + 2as

    Where v = final velocity = 0

    u = initial velocity = 17m/s

    a = deceleration = - 12m/s²

    s = distance covered

    Solving for s = u²/2a

    s = 17² / (2*12) = 12.04m

    So therefore distance between deer and car after stopping = 36.5 - 12.04 = 24.46m

    B. Using the same kinematic formula to derive maximum initial velocity attainable to cover distance of 36.5m before stopping

    V² = u² + 2as

    V = 0

    a = deceleration = - 12m / s

    s = distance covered = 36.5m

    u = (2as) ^½

    u = (2 * 12 * 36.5) ^½

    u = 29.6m/s
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