physics!!
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!!
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Answers
a) The magnitude of the skateboarder's acceleration can be found using the equation a = v/t, where a is acceleration, v is velocity, and t is time. The time it takes the skateboarder to travel the 14.0 m is t = d/v = 14.0 m/7.40 m/s = 1.89 s. Therefore, the acceleration is a = (7.40 m/s)/(1.89 s) = 3.91 m/s2. b) The component of acceleration that is parallel to the ground is determined from the equation a_parallel = a*cos(θ), where a is the acceleration, θ is the angle of inclination, and a_parallel is the component of acceleration that is parallel to the ground. The angle of inclination is θ = 20.5°, so the component of acceleration that is parallel to the ground is a_parallel = (3.91 m/s2)*cos(20.5°) = 3.68 m/s2.
