Question

A skateboarder rolls from rest down an inclined ramp that is 15.0m long and inclined above the horizontal at an angle of theta = 20.0 degree. When she reaches the bottom of the ramp 3.00s later her speed is 10.0m/s. Show that the average acceleration of the skateboarder is g sin theta where g = 9.81m/s2

Answer 1

The initial velocity of the skateboard, u=0 m/s

The length of the ramp, s=15.0 m

The angle of inclination, θ=20.0°

The time duration, t=3.00 s

The final velocity of the skateboard, v=10.0 m/s

The acceleration due to gravity, g=9.81 m/s²

Let us calculate gsinθ

$\begin{gathered} a_1=g\sin \theta \\ =9.81\times\sin 20.0^{\circ} \\ =3.3m/s^2 \end{gathered}$

From the equation of the motion,

$v=u+a_2t$

Where a₂ is the acceleration of the skateboard.

On substituting the known values,

$\begin{gathered} 10.0=0+a_2\times3.00 \\ a_2=\frac{10.0}{3.00} \\ =3.3m/s^2 \end{gathered}$

On comparing we find that,

$a_1=a_2$

Thus, the acceleration of the skateboard is g sinθ.