Question

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 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?

Answer 1

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

Where:

$V_{f}$: is the final speed = 8.89 m/s

$V_{0}$: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

Where:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!