In 1991 at smith college, in massachusetts, ferdie adoboe ran 1.00 × 102 m backward in 13.6 s. suppose it takes adoboe 2.00 s to achieve a velocity equal to her average velocity during the run. find her average acceleration during the first 2.00 s.

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aristocles aristocles · Answer 1 · 2018-10-03T03:39:02+02:00

as it is given that it covers a total distance 1 * 10^2 m

total time taken by it = 13.6 s

now the average speed is given as ratio of total distance and total time

[tex]v = \frac{d}{t}[/tex]

[tex]v = \frac{1* 10^2 }{13.6}[/tex]

[tex]v = 7.35 m/s[/tex]

so the average speed will be 7.35 m/s

now if it starts from rest and achieve the final speed as 7.35 m/s

now we can use kinematics

[tex]v_f = v_i + at[/tex]

[tex]7.35 = 0 + a* 2[/tex]

[tex]a = 3.68 m/s^2[/tex]

so its acceleration will be 3.68 m/s^2