Question

A skateboarder travels on a horizontal surface with an initial velocity of 4.0 m/s toward the south and a constant acceleration of 1.8 m/s2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.
a)What is her x position at t=0.60s?

b)What is her y position at t=0.60s?

c)What is her x velocity component at t=0.60s?

d)What is her y velocity component at t=0.60s?

Answer 1

Answer:

a) 0.324 m

b) -2.4 m

c) 1.08 m/s

d) -4 m/s

Explanation:

Initial position $(x,y)=(0,0)$

Initial velocity $(u_x,u_y)=(0.0,-4.0)m/s$

Acceleration $(a_x,a_y)=(1.8,0.0)m/s^2$

We need to use the following equations of motion:

$S=ut+\frac{1}{2}at^2$

$v=u+at$

a) $S_x=u_xt+\frac{1}{2}a_xt^2=(0)(0.6)+(0.5)(1.8)(0.6^2)=0.324m$

b) $S_y=u_yt+\frac{1}{2}a_yt^2=(-4.0)(0.6)+(0.5)(0)(0.6^2)=-2.4m$

c) $v_x=u_x+a_xt=0+(1.8)(0.6)=1.08m/s$

d) $v_y=u_y+a_yt=-4.0+(0)(0.6)=-4m/s$