A skateboarder, with an initial speed of 2.4m/s , rolls virtually friction free down a straight incline of length 19m in 3.5m/s .At what angle theta is the incline oriented above the horizontal?
To find the angle theta at which the incline is oriented above the horizontal, we can use the concept of trigonometry and the kinematic equations. Here are the steps to solve the problem:
Step 1: Understand the problem
A skateboarder is rolling down a straight incline with an initial speed of 2.4 m/s and covering a distance of 19 m. We need to find the angle theta at which the incline is oriented above the horizontal.
Step 2: Identify the relevant equations
We can use the kinematic equation for motion along an inclined plane to solve this problem. This equation relates the acceleration (a), initial velocity (u), final velocity (v), distance (s), and angle of incline (theta).
v^2 = u^2 + 2as
In this case, we are given:
u = 2.4 m/s (initial speed)
v = 3.5 m/s (final speed)
s = 19 m (distance covered)
Step 3: Solve the equation for angle theta
Given the values for u, v, and s, we can rearrange the equation to solve for theta:
v^2 = u^2 + 2as
(3.5 m/s)^2 = (2.4 m/s)^2 + 2a(19 m)
Simplifying this equation will give us the value of the acceleration (a).
Step 4: Find the angle theta
Using the value of acceleration (a), we can calculate the angle theta using the following equation:
a = g * sin(theta)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
By substituting the value of a and g into the equation, we can solve for theta.
Step 5: Calculate theta
Let's solve the equation to find the value of theta:
(3.5 m/s)^2 = (2.4 m/s)^2 + 2a(19 m)
9.8 m^2/s^2 = 5.76 m^2/s^2 + 38a
Simplifying further:
9.8 m^2/s^2 - 5.76 m^2/s^2 = 38a
3.04 m^2/s^2 = 38a
Dividing both sides by 38:
a = 0.08 m^2/s^2
Now, we can substitute the value of a into the equation a = g * sin(theta):
0.08 m^2/s^2 = 9.8 m/s^2 * sin(theta)
Dividing both sides by 9.8:
sin(theta) = 0.08 m^2/s^2 / 9.8 m/s^2
sin(theta) = 0.00816
Now, to find theta, we need to take the inverse sine (arcsine) of 0.00816:
theta = arcsin(0.00816)
Using a calculator, we find that:
theta = 0.469 degrees (approximately)
Therefore, the incline is oriented approximately 0.469 degrees above the horizontal.