(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
[tex]v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s [/tex]
(b) The initial speed of the skater is
[tex]v_i = 4 m/s[/tex]
while the final speed is
[tex]v_f = 5.3 m/s[/tex]
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
[tex]a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2 [/tex]
(c) The initial speed of the skater is
[tex]v_i = 13.0 m/s[/tex]
while the final speed is
[tex]v_f=0[/tex]
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
[tex]2aS=v_f^2 -v_i^2[/tex]
from which we find
[tex]a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2 [/tex]
where the negative sign means it is a deceleration.
