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!!
a. V^2 = 2ad,
2ad = V^2,
2a*14 = (7.4)^2,
28a = 54.76,
a = 54.76/28 = 1.96 m/s^2.
b. a(hor) = 1.96*cos20.5 = 1.83 m/s^2.
To solve this problem, we can use the principles of motion along an inclined plane. The key equation we will use is:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
(a) To find the magnitude of the skateboarder's acceleration, we first need to find the initial velocity. Given that she starts from rest, u = 0 m/s.
Now, we can rearrange the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (7.40^2 - 0) / (2 * 14.0)
Simplifying:
a = 54.54 / 28.0
a = 1.95 m/s^2
So, the magnitude of the skateboarder's acceleration is 1.95 m/s^2.
(b) To determine the component of her acceleration parallel to the ground, we need to find the acceleration along the inclined plane. This is given by:
a_parallel = a * sin(theta)
where theta is the angle of inclination. Given that the ramp is inclined at 20.5°, we can substitute the values:
a_parallel = 1.95 * sin(20.5°)
Using a calculator, we find:
a_parallel = 0.67 m/s^2
Therefore, the component of her acceleration parallel to the ground is 0.67 m/s^2.