A skateboarder, with an initial speed of 4.2m/s , rolls virtually friction free down a straight incline of length 16m in 3.3 s.
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To find the acceleration of the skateboarder going down the incline, we can use the kinematic equation:
\[x = x_0 + v_0t + \frac{1}{2}at^2\]
Where:
- \(x\) is the final position
- \(x_0\) is the initial position (which can be taken as 0 in this case)
- \(v_0\) is the initial velocity
- \(t\) is the time elapsed
- \(a\) is the acceleration
Rearranging the equation, we can solve for the acceleration:
\[a = \frac{2(x - x_0 - v_0t)}{t^2}\]
Given that the initial velocity (\(v_0\)) is 4.2 m/s, the length of the incline (\(x - x_0\)) is 16 m, and the time (\(t\)) is 3.3 s, we can substitute these values into the equation to find the acceleration:
\[a = \frac{2(16 - 0 - 4.2 \cdot 3.3)}{3.3^2}\]
Now let's calculate the acceleration:
\[a = \frac{2(16 - 13.86)}{10.89}\]
\[a = \frac{2 \cdot 2.14}{10.89}\]
\[a = \frac{4.28}{10.89}\]
\[a \approx 0.393 \, \text{m/s}^2\]
Therefore, the acceleration of the skateboarder going down the incline is approximately 0.393 m/s^2.