A skateboarder, starting from rest, rolls down a 13.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.06 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 33.2 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?

Question

Henry · Accepted Answer

a. a = (Vf^2 - Vo^2) / 2d, a = ((6.06)^2 - 0) / 27.8 = 1.32m/s^2. b. X = hor. = 1.32cos33.2 = 1.11m/s^2 = component parallel to ground.