Question

A skateboarder, starting from rest, rolls down a 13.0-m ramp. When she arrives at the bottom of the ramp her speed is 8.40 m/s.

(a) Determine the magnitude of her acceleration, assumed to be constant.


(b) If the ramp is inclined at 25.5° with respect to the ground, what is the component of her acceleration that is parallel to the ground?

Answer 1

(a) The magnitude of her acceleration, assumed to be constant which will be given as :

using equation of motion 3,   v_f² = v₀² + 2 a x                                                                { eq.1 }

where, v₀ = initial speed of skateboarder = 0 m/s

x = distance travelled = 13 m

v_f = final speed at the bottom of ramp = 8.4 m/s

inserting all these values in above eq.

(8.4 m/s)² = (0 m/s)² + 2 a (13 m)

(70.56 m²/s²) = (26 m) a

a = (70.56 m²/s²) / (26 m)

a = 2.71 m/s²

(b) If the ramp is inclined at 25.5⁰ with respect to the ground, the component of her acceleration that is parallel to the ground which is given as ::

a_x = a cos $\theta$                                                                   { eq.2 }

inserting the values in eq.2,

a_x = (2.71 m/s²) cos (25.5⁰)

a_x = 2.44 m/s²