(a) 11.1 m/s
The acceleration of the skier is given by:
[tex]a=\frac{F}{m}[/tex]
where F = 92 N is the net force and m = 68 kg. Substituting,
[tex]a=\frac{92 N}{68 kg}=1.35 m/s^2[/tex]
After 8.2 s, the speed of the skier is
[tex]v=at[/tex]
where
a = 1.35 m/s^2 is the acceleration
t = 8.2 s is the time
Substituting,
[tex]v=(1.35 m/s^2)(8.2 s)=11.1 m/s[/tex]
(b) 10.0 m/s
In this section of the hill, the net force is F = -22 N backwards. So, the acceleration of the skier is
[tex]a=\frac{F}{m}=\frac{-22 N}{68 kg}=-0.32 m/s^2[/tex]
When starting this section, the skier is moving at u = 11.1 m/s. So, the final speed will be:
[tex]v=u+at[/tex]
And substituting t=3.5 s, we find
[tex]v=11.1 m/s+(-0.32 m/s^2)(3.5 s)=10.0 m/s[/tex]
(c) 237.9 m
The distance travelled by the skier in the downhill section is
[tex]d=\frac{1}{2}at^2[/tex]
where a = 1.35 m/s^2 and t = 8.2 s. Substituting,
[tex]d=\frac{1}{2}(1.35 m/s^2)(8.2 s)^2=45.4 m[/tex]
The distance travelled by the skier in the levelled out section is given by
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final speed
u = 11.1 m/s is the initial speed
a = -0.32 m/s^2 is the acceleration
d is the distance
Solving for d,
[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(11.1 m/s)^2}{2(-0.32 m/s^2)}=192.5 m[/tex]
So, the total distance is
d = 45.4 m + 192.5 m = 237.9 m
