Luna and Angela are competitors in a running competition. Both are leading the race and enter at the same time the last 30 meters. Angela en

Question

Luna and Angela are competitors in a running competition. Both are leading the race and enter at the same time the last 30 meters. Angela enters the final segment runnning at 4 m/s and “applies” an acceleration of 2 m/s2 while Luna enters the final segment at 3m/s but “applies ” an acceleration of 3 m/s2. (a) Who finishes the race first? (Do the calculations) (b) Who is running faster at the end of the race?

in progress 0
Jezebel 3 years 2021-08-26T10:01:46+00:00 1 Answers 5 views 0

Answers ( )

  1. thnahdat
    0
    2021-08-26T10:02:49+00:00
    Reply

    Answer:

    (a) Luna

    (b) Luna

    Explanation:

    (b)

    The final speed can be found by using the third equation of motion:

    2as = v_f^2 - v_i^2\\

    where,

    a = acceleration

    s = distance covered

    vf = final speed

    vi = initial speed

    For Angela:

    a = 2 m/s²

    s = 30 m

    vi = 4 m/s

    vf = v_fa

    therefore,

    2(2\ m/s^2)(30\ m) = (v_{fa})^2 - (4\ m/s)^2\\v_{fa}^2 = 120 m^2/s^2 + 16\ m^2/s^2\\v_{fa} = 11.66\ m/s\\

    For Luna:

    a = 3 m/s²

    s = 30 m

    vi = 3 m/s

    vf = v_fl

    therefore,

    2(3\ m/s^2)(30\ m) = (v_{fa})^2 - (3\ m/s)^2\\v_{fa}^2 = 180 m^2/s^2 + 9\ m^2/s^2\\v_{fa} = 13.74\ m/s\\

    Therefore, Luna will be running faster at the end of the race.

    (b)

    To calculate the time of completion we will use the first equation of motion:

    v_f = v_i + at

    FOR ANGELA:

    11.66\ m/s = 4\ m/s + (2\ m/s^2)t_a\\t_a = \frac{7.66\ m/s}{2\ m/s^2}\\\\t_a = 3.83\ s

    FOR LUNA:

    13.74\ m/s = 3\ m/s + (3\ m/s^2)t_l\\t_l = \frac{10.74\ m/s}{3\ m/s^2}\\\\t_l = 3.58\ s

    Hence, LUNA finishes the race first.

Leave an answer

Browse

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