A skateboarder travels on a horizontal surface with an initial velocity of 4.2 m/s toward the south and a constant acceleration of 2.6 m/s^2 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. Required:
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?

Explanation:

Let the skateboarder's movement in x -direction be taken into consideration .

a )

initial velocity in x direction u = 0

acceleration a = 2.6 m /s²

time t = .6 s

displacement in x direction

s = ut + 1/2 a t²

= 0 + .5 x 2.6 x .6²

= .468 m

= 46.80 cm

c )

velocity after .6 s

v = u + at

= 0 + 2.6 x .6

= 1.56 m /s

Let the skateboarder's movement in y -direction be taken into consideration .

b ) initial velocity in y direction u = - 4.2 m /s ( velocity is towards south or - y direction )

acceleration a = 0

time t = .6 s

displacement in y direction

s = ut + 1/2 a t²

= - 4.2 x .6  + 0

= - 2.52  m

2.52 m towards south .

d )

velocity after .6 s

v = u + at

= - 4.2  + 0

= - 4.2  m /s

or 4.2 m /s towards south .

the device is called “ rocket engine”