In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose “Pre” was at the 7.85 km

Question

In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose “Pre” was at the 7.85 km7.85 km mark at a time of 25.0 min25.0 min . If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 7.85 km7.85 km mark was the same as his overall average speed up to that time.

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Latifah 4 years 2021-07-25T17:17:46+00:00 1 Answers 61 views 0

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  1. thugiang
    0
    2021-07-25T17:19:42+00:00
    Reply

    Answer:

    a = 0.161 $m/s^2$

    Explanation:

    Given :

    $ d_{total} = 10 km = 10000 m

    $t_{total} $ = 27 min 43.6 s

            = 1663.6 s

    $d_1$ = 7.85 km = 7850 m

    $t_1$ = 25 min = 1500 s

    $t_2$ = 60 s

    Now the initial speed for the distance of 7.85 km is

    $ v_1 = \frac{d_1}{t_1} = \frac{7850}{1500}$  = 5.23 m/s

    The velocity after 60 s after the distance of 7.85 kn is

    $v_2 = v_1 + at_2$

        = 5.23 + a(60)

    The distance traveled for 60 s after the distance of 7.85 km is

    $d_2 = v_1t_2+\frac{1}{2}at_2^2$

    $d_2 = (5.23)(60)+\frac{1}{2}a(60)^2$

        = 313.8 + a(1800)

    The time taken for the last journey where the speed is again uniform is

    $t_3 = t_{total}-t_1-t_2 $

       = 1663.6 – 1500 – 60

       = 103.6 s

    Therefore, the distance traveled for the time $t_3$ is

    $ d_3 = v_2 t_3$

        = (5.23+60a)(103.6)

        = 541.8 + 6216 a

    The total distance traveled,

    $ d_{total}= d_1 + d_2 + d_3$

    Now substituting the values in the above equation for the acceleration a is

    10000 = 7850 + (313.6 + 1800a) + (541.8 + 6216a)

    10000 = 8706.5 + 8016a

    1294.4 = 8016a

    a = 0.161 $m/s^2$

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