A skateboarder, with an initial speed of 2.6m/s , rolls virtually friction free down a straight incline of length 18m in 3.3 s. What angle is the incline plane?
53
To determine the angle of the incline plane, we can use the equation for the displacement along an inclined plane:
\(d = v_0t + \frac{1}{2}at^2\)
where \(d\) is the displacement, \(v_0\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration.
In this case, the skateboarder starts from rest (since there is no mention of an initial velocity), so \(v_0 = 0\). The displacement \(d\) is given as 18m, and the time \(t\) is given as 3.3s.
Substituting these values into the equation, we have:
\(18 = 0 + \frac{1}{2}a(3.3^2)\)
Simplifying, we get:
\(18 = \frac{1}{2}a(10.89)\)
Now, to find the acceleration \(a\), we need to rearrange the equation:
\(a = \frac{2 \cdot 18}{10.89}\)
Evaluating this expression, we find:
\(a \approx 3.303~\text{m/s}^2\)
The acceleration represents the force acting down the incline due to gravity. To find the angle of the incline plane, we can use the formula:
\(a = g \cdot \sin(\theta)\)
where \(g\) is the acceleration due to gravity (approximately \(9.8~\text{m/s}^2\)), and \(\theta\) is the angle of the incline plane.
Rearranging the equation to find \(\theta\), we have:
\(\theta = \arcsin(\frac{a}{g})\)
Substituting the values, we get:
\(\theta = \arcsin(\frac{3.303}{9.8})\)
Evaluating this expression, we find:
\(\theta \approx 20.64^\circ\)
Therefore, the angle of the incline plane is approximately \(20.64^\circ\).