A skateboarder, starting from rest, rolls down a 13.4-m ramp. When she arrives at the bottom of the ramp her speed is 6.14 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 32.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?
a. V^2 = Vo^2 + 2a.d
a = (V^2-Vo^2)/2d.
a = (6.14-0)/26.8 = 0.229 m/s^2
b. a = 0.229*cos 32.6 = 0.193 m/s^2,
c.
To solve this problem, we can use the equations of motion for constant acceleration.
(a) To find the magnitude of her acceleration, we can use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (6.14 m/s)
- u is the initial velocity (0 m/s, since she starts from rest)
- a is the acceleration (unknown)
- s is the displacement (13.4 m, since she travels down the ramp)
Plugging in the values, the equation becomes:
(6.14 m/s)^2 = (0 m/s)^2 + 2a * 13.4 m
Simplifying and solving for a:
a = (6.14 m/s)^2 / (2 * 13.4 m)
a = 37.6996 m^2/s^2 / 26.8 m
a = 1.407 m/s^2
So the magnitude of her acceleration is 1.407 m/s^2.
(b) Since the ramp is inclined at an angle of 32.6 degrees with respect to the ground, we need to find the component of the acceleration parallel to the ground.
We can use the equation:
a_parallel = a * sin(θ)
Where:
- a_parallel is the component of acceleration parallel to the ground (unknown)
- a is the magnitude of acceleration (1.407 m/s^2)
- θ is the angle of the ramp (32.6 degrees)
Plugging in the values, the equation becomes:
a_parallel = 1.407 m/s^2 * sin(32.6 degrees)
a_parallel = 1.407 m/s^2 * 0.529
a_parallel ≈ 0.744 m/s^2 (rounded to three decimal places)
So the component of her acceleration that is parallel to the ground is approximately 0.744 m/s^2.