A diver is trying a new more and is attempting to increase his angular acceleration in order to increase the angular displacement and thus produce more rotations from a standing start on the board the diver accelerates at 950 degree/s/s.what is the angular displacement over 0.15s?
x = Xi + Vi t + (1/2) a t^2
x = .5 (950 deg/s^2)(.15s)^2
= 10.7 degrees
Could u explain Damon how do u got .5 for xi
Hint the tutor gave for this problem is to use more than one equation to get this Damon thanks for answering could u explain these please in detail
To find the angular displacement, we can use the equation for angular acceleration:
angular acceleration (α) = change in angular velocity (Δω) / time (t)
Since we know the angular acceleration (950 degrees/s/s) and the time (0.15 s), we can rearrange the equation to solve for the change in angular velocity:
Δω = α * t
Δω = 950 degrees/s/s * 0.15 s = 142.5 degrees/s
Now, we can use the equation for angular displacement:
angular displacement (θ) = initial angular velocity (ω_i) * time (t) + (0.5 * α * t^2)
Since the diver is starting from a standing position, the initial angular velocity is zero (ω_i = 0):
θ = 0 + (0.5 * 950 degrees/s/s * (0.15 s)^2)
= 0 + (0.5 * 950 degrees/s/s * 0.0225 s^2)
= 0 + 0.5 * 21.375 degrees
= 10.6875 degrees
Therefore, the angular displacement over 0.15 seconds is approximately 10.69 degrees.