A skateboarder, starting from rest, rolls down a 13.9- m ramp. When she arrives

Question

A skateboarder, starting from rest, rolls down a 13.9- m ramp. When she arrives at the bottom of the ramp her speed is 6.92 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 35 degrees with respect to the ground, what is the component of her acceleration that is parallel to the ground? ( I have been using the formula a= v^2/2S, but I am not getting the number, so I must be plugging in the wrong numbers from the equation. Then I can't use the answer to answer the second cosine/sin equation)

Explanation / Answer

a. V^2 = 2aS

a = 6.92^2/2*13.9 = 1/72 m/s^2

b. 2aS = v^2 cos ^ = 6.92*6.92 * = 32.13

a = 1.15 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.