Question

An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree direction. What is the y-component of her acceleration? PLEASE HELP

Answer 1

The y-component of the acceleration is $0.22 m/s^2$

Explanation:

The y-component of the acceleration is given by

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction $\theta=50.0^{\circ}$

v = 4.65 m/s is the final velocity, in a direction $120^{\circ}$

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

$u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s$

So the y-component of the acceleration is

$a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2$

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