Answer:
[tex]-0.032 m/s^2[/tex]
Explanation:
To answer the question, we just need to consider the motion along the horizontal direction.
The component of the initial velocity of the ice skater along the x-direction is:
[tex]u_x = u cos \theta =(2.25)(cos 50^{\circ})=1.45 m/s[/tex]
where u = 2.25 m/s is the initial velocity and [tex]50^{\circ}[/tex] is the angle.
The component of the final velocity of the ice skater along the x-direction is
[tex]v_x = u cos \theta =(4.65)(cos 120^{\circ})=-2.33 m/s[/tex]
where u = 4.65 m/s is the final velocity and [tex]120^{\circ}[/tex] is the angle.
The acceleration along the x-direction is given by
[tex]a_x=\frac{v_x-u_x}{t}[/tex]
where
t = 120 s is the time
Substituting,
[tex]a=\frac{-2.33-(1.45)}{120}=-0.032 m/s^2[/tex]
