A skateboarder, with an initial speed of 2.4m/s , rolls virtually friction free down a straight incline of length 19m in 3.3 s.At what angle θ is the incline oriented above the horizontal?
Well, let's calculate that, but first, let me put on my skater clown nose! 🤡
Okay, let's break it down. We have a skateboarder going down an inclined plane with an initial speed of 2.4 m/s and covering a distance of 19 m in 3.3 s. The question asks us to find the angle θ at which the incline is oriented above the horizontal.
Now, we can use some good old physics equations here. The key equation we need is:
d = v0t + (1/2)at^2
Where:
d = distance
v0 = initial velocity
t = time
a = acceleration
Since we're dealing with an inclined plane, the acceleration can be broken down into two components: one along the plane and one perpendicular to it.
The component of acceleration along the plane is given by: a_parallel = g*sin(θ)
The component of acceleration perpendicular to the plane is given by: a_perpendicular = g*cos(θ)
Now, we'll find the total acceleration, a_total, using the formula: a_total = sqrt(a_parallel^2 + a_perpendicular^2)
Given that the distance (d) is 19 m and the time (t) is 3.3 s, we can solve for the angle (θ). Let me use my clown calculator for this one... *beep boop beep*
Calculating... *hums circus tune*
And the answer is... *drumroll*
The incline is oriented at an angle of θ = 19.5 degrees above the horizontal!
Skateboarding down an inclined plane sounds like a ton of fun. Just don't forget your favorite clown wig! 🤡⛸️
To find the angle θ at which the incline is oriented above the horizontal, we can use the kinematic equation for the displacement along an inclined plane:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
t = time
a = acceleration
In this case, the skateboarder's initial velocity is 2.4 m/s, the displacement is 19 m, and the time is 3.3 s. Since the skateboarder is rolling virtually friction-free, there is no horizontal acceleration (a = 0). Therefore, the equation simplifies to:
s = ut
Rearranging the equation to solve for the angle, we have:
θ = arctan(s/u)
Now we can substitute the given values into the equation to find the angle θ:
θ = arctan(19/2.4)
Using a calculator, we can calculate the angle to be approximately 86.14 degrees.
Therefore, the angle θ at which the incline is oriented above the horizontal is approximately 86.14 degrees.
To find the angle θ at which the incline is oriented above the horizontal, we can use the equations of motion along with the given information.
Let's analyze the problem:
Initial speed (u) = 2.4 m/s
Distance (s) = 19 m
Time (t) = 3.3 s
Acceleration (a) = ? (unknown)
Angle (θ) = ? (unknown)
We can start by calculating the acceleration (a) using the equation:
s = ut + (1/2)at^2
Substituting the given values:
19 = (2.4)(3.3) + (1/2)(a)(3.3)^2
Simplifying:
19 = 7.92 + (1/2)(10.89)(a)
Rearranging the equation:
19 - 7.92 = 10.89(a/2)
11.08 = 5.445a
a ≈ 2.036 m/s^2
Now, we can use this acceleration to calculate the angle θ using the trigonometric relationship between the angle and the acceleration:
a = g sin(θ)
Rearranging the equation:
θ = sin^(-1)(a / g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values:
θ = sin^(-1)(2.036 / 9.8)
Using a calculator:
θ ≈ 12.39 degrees
Therefore, the incline is oriented approximately 12.39 degrees above the horizontal.