A skateboarder, starting from rest, rolls down a 12.5-m ramp. When she arrives at the bottom of the ramp her speed is 6.45 m/s.
(a) Determine the magnitude of her acceleration, assumed to be constant.
V = sqrt(2 a X) applies when the starting velocity is zero. , therefore
a = V^2/(2 X)= 1.664 m/s^2
Here's how you can derive it (if you can't remember it).
Average velocity = V/2
Time T = X/(V/2) = 2 X/V
Acceleration a = V/T = V^2/(2 X)
To determine the magnitude of the skateboarder's acceleration, we can use the equations of motion. The equation that relates displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) is:
v^2 = u^2 + 2as
In this case, the skateboarder starts from rest (u = 0), and her final velocity is 6.45 m/s (v = 6.45 m/s). The displacement is 12.5 m (s = 12.5 m).
Plugging these values into the equation, we have:
(6.45 m/s)^2 = 0^2 + 2a(12.5 m)
Simplifying, we get:
41.6025 m^2/s^2 = 25a
To find the acceleration, divide both sides of the equation by 25:
a = 41.6025 m^2/s^2 / 25
Calculating this value, we find:
a ≈ 1.6641 m/s^2
Therefore, the magnitude of the skateboarder's acceleration is approximately 1.6641 m/s^2.