A skateboarder, with an initial speed of 3.2 m/s , rolls virtually friction free down a straight incline of length 16 m in 3.3 s. At what angle θ is the incline oriented above the horizontal?
s = 3.2t + 1/2 at^2
3.2(3.3) - a/2 (3.3^2) = 16
a = 1.0
so, a = 1/9.8 g
So, sin(theta) = 1/9.8
theta = 5.86 degrees
Well, it seems like this skateboarder is on a roll! Now, let's calculate that angle θ to make sure this isn't just a 'wild ride'.
To start off, we can use the kinematic equation for displacement along an inclined plane:
Δx = v₀t + (1/2)at²
Here, Δx is the displacement, v₀ is the initial velocity, t is the time, and a is the acceleration due to gravity. Since the skateboarder rolls virtually friction-free, the only acceleration acting on them is due to gravity, which we know is roughly -9.8 m/s² (now that's a real up-and-down relationship).
So, plugging in the values we know:
16 m = (3.2 m/s)(3.3 s) + (1/2)(-9.8 m/s²)(3.3 s)²
16 m = 10.56 m + (1/2)(-32.34 m)
16 m = 10.56 m - 16.17 m
Now, to solve for θ, we use a little trigonometry. The angle θ is the angle of the incline above the horizontal, so we can use the sine function:
sin(θ) = opposite/hypotenuse
In this case, the opposite side is the vertical drop of -16.17 m and the hypotenuse is the length of the incline, 16 m.
sin(θ) = -16.17 m / 16 m
θ = sin⁻¹(-16.17/16)
Now, sadly, θ turns out to be greater than 90 degrees, which doesn't really make sense physically. So, it looks like my calculations got a bit 'skewed'. Perhaps this skateboarder's 'airtime' caused some measurement error? Let's try a little harder to solve this puzzle, shall we?
To find the angle θ at which the incline is oriented above the horizontal, we can use the concept of motion along an inclined plane.
The skateboarder's initial speed is 3.2 m/s, the length of the incline is 16 m, and the time it takes to traverse the incline is 3.3 s. We can use these values to find the acceleration and then calculate the angle θ.
Step 1: Calculate the acceleration (a) of the skateboarder using the kinematic equation:
distance = initial velocity * time + (0.5 * acceleration * time^2)
16 m = 3.2 m/s * 3.3 s + (0.5 * acceleration * (3.3 s)^2)
16 m = 10.56 m + (5.44 * acceleration)
16 m - 10.56 m = 5.44 * acceleration
5.44 * acceleration = 5.44 m
acceleration = 1 m/s^2
Step 2: Calculate the angle (θ) using the formula:
acceleration = g * sin(θ)
1 m/s^2 = 9.8 m/s^2 * sin(θ)
sin(θ) = 1 m/s^2 / 9.8 m/s^2
sin(θ) = 0.102
Step 3: Use the inverse sine function to find the angle θ:
θ = sin^(-1)(0.102)
θ ≈ 5.89 degrees
Therefore, the incline is oriented approximately 5.89 degrees above the horizontal.
To find the angle θ at which the incline is oriented above the horizontal, you can use the concept of kinematics and trigonometry.
First, let's identify the given information:
- Initial speed of the skateboarder (u) = 3.2 m/s
- Length of the incline (s) = 16 m
- Time taken to travel down the incline (t) = 3.3 s
Next, use the kinematic equation for motion along an incline:
s = ut + (1/2)at^2
Since the incline is virtually friction-free, the acceleration will be due to gravity and can be represented as a = g * sin(θ), where g is the acceleration due to gravity (9.8 m/s²).
Substituting the given values into the equation, we have:
16 = (3.2) * 3.3 + (1/2) * (9.8) * sin(θ) * (3.3)^2
Simplifying this equation will give you:
16 - (3.2 * 3.3) = (1/2) * (9.8) * (3.3)^2 * sin(θ)
Now, rearrange the equation to solve for sin(θ):
sin(θ) = (16 - (3.2 * 3.3)) / [(1/2) * (9.8) * (3.3)^2]
Evaluate the right side of the equation using a calculator:
sin(θ) = 0.247
Finally, take the inverse sine (sin⁻¹) of both sides to find the angle θ:
θ = sin⁻¹(0.247)
Using a calculator, the angle θ is approximately 14.2 degrees above the horizontal.