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?
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.
To determine the magnitude of the skateboarder's acceleration, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity = 6.06 m/s
u = initial velocity = 0 m/s (starting from rest)
a = acceleration (to be determined)
s = displacement = 13.9 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (6.06^2 - 0^2) / (2 * 13.9)
a = 36.7236 / 27.8
a ≈ 1.32 m/s^2
Therefore, the magnitude of her acceleration is approximately 1.32 m/s^2.
To determine the component of her acceleration parallel to the ground, we can use trigonometry. We know that:
Acceleration parallel to the ground = Acceleration * sin(θ)
Where:
Acceleration = 1.32 m/s^2 (from part a)
θ = 33.2° (angle of the ramp with respect to the ground)
Substituting the values, we have:
Acceleration parallel to the ground ≈ 1.32 * sin(33.2°)
Acceleration parallel to the ground ≈ 0.706 m/s^2
Therefore, the component of her acceleration that is parallel to the ground is approximately 0.706 m/s^2.