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31 August, 17:56

Cheetahs, the fastest of the great cats, can reach 45 mph in 2.0 sec starting from rest. Assuming that they have constant acceleration throughout that time, find (a) their acceleration (in f t/s2 and m/s2) (b) the distance they travel during that time (in m and ft)

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  1. 31 August, 18:12
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    acceleration is 10.05 m/s² or 32.97 ft/s²

    distance is 40.22 m or 131.95 ft

    Explanation:

    given data

    velocity = 45 mph = 20.1168 m/s

    time = 2 sec

    to find out

    acceleration and distance

    solution

    first we will apply here formula for acceleration that is

    v = u + at ... 1

    here v is velocity and u is initial speed and t is time and a is acceleration

    so put here all value

    20.11 = 0 + a (2)

    a = 10.05 m/s

    so acceleration is 10.05 m/s² or 32.97 ft/s²

    and

    distance is calculated as

    distance = v * t ... 2

    put here value

    distance = 20.11 * 2

    distance = 40.22

    so distance is 40.22 m or 131.95 ft
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