A skateboarder, starting from rest, rolls down a 14.0-m ramp. When she arrives at the bottom of the ramp her speed is 7.40 m/s.
(a) Determine the magnitude of her acceleration, assumed to be constant.
(b) If the ramp is inclined at 20.5° with respect to the ground, what is the component of her acceleration that is parallel to the ground?
please help! it would be much appreciated!!
To determine the magnitude of the skateboarder's acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity (7.40 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled down the ramp (14.0 m).
(a) We need to rearrange the equation to solve for the acceleration a:
a = (v^2 - u^2)/(2s)
Substituting the given values, we get:
a = (7.40^2 - 0)/(2 * 14.0)
a = (54.76 - 0)/28
a ≈ 1.96 m/s^2
Therefore, the magnitude of her acceleration is approximately 1.96 m/s^2.
(b) To find the component of her acceleration that is parallel to the ground, we can use trigonometry. The component parallel to the ground is given by:
a_parallel = a * sin(θ)
where θ is the angle of incline (20.5°).
Substituting the values, we get:
a_parallel = 1.96 * sin(20.5°)
a_parallel ≈ 0.669 m/s^2
Hence, the component of her acceleration that is parallel to the ground is approximately 0.669 m/s^2.