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

a) 0.324 m

b) -2.4 m

c) 1.08 m/s

d) -4 m/s

Step-by-step 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+(1)/(2)at^2`

`v=u+at`

a)
`S_x=u_xt+(1)/(2)a_xt^2=(0)(0.6)+(0.5)(1.8)(0.6^2)=0.324m`

b)
`S_y=u_yt+(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`