Question

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.

Answer 1

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

$v = \frac{d}{t}$

$v = \frac{1* 10^2 }{13.6}$

$v = 7.35 m/s$

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

$v_f = v_i + at$

$7.35 = 0 + a* 2$

$a = 3.68 m/s^2$

so its acceleration will be 3.68 m/s^2