Consider a skateboarder who starts from rest at the top of a 14.6 m-long ramp that is inclined at an angle of 13.0° to the horizontal. Assuming that the skateboarder's acceleration is g sin 13.0°, find his speed when he reaches the bottom of the ramp in 3.13 s.
To find the speed of the skateboarder at the bottom of the ramp, we can use the equations of motion. The equation we will use is:
v^2 = u^2 + 2as
Where:
- v is the final velocity (speed) of the skateboarder at the bottom of the ramp
- u is the initial velocity (which is zero since the skateboarder starts from rest)
- a is the acceleration of the skateboarder (given as g sin 13.0°)
- s is the distance traveled down the ramp (given as 14.6 m)
First, we need to calculate the acceleration using the given angle:
a = g sin θ
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- θ is the angle of the ramp (given as 13.0°)
Plugging in the values:
a = 9.8 m/s^2 * sin 13.0°
We can now calculate the acceleration:
a = 9.8 m/s^2 * 0.224951054
a ≈ 2.2 m/s^2
Now let's substitute the values into the equation of motion:
v^2 = 0^2 + 2 * 2.2 m/s^2 * 14.6 m
Simplifying:
v^2 = 2 * 2.2 m/s^2 * 14.6 m
v^2 = 60.88 m^2/s^2
To find v, we take the square root of both sides:
v = √(60.88 m^2/s^2)
v ≈ 7.8 m/s
Therefore, the skateboarder's speed when he reaches the bottom of the ramp after 3.13 s is approximately 7.8 m/s.