# A hovercraft starts from rest and undergoes a pure rotation with a constant angular acceleration. If it takes?

A hovercraft starts from rest and undergoes a pure rotation with a constant angular acceleration. If it takes 1.1 sec to rotate 45 degrees after starting from rest, how long in seconds would it take to instead rotate 78 degrees after starting from rest under the same constant angular acceleration? Express the numerical portion of your answer rounded to one decimal place.

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• 6 years ago

Basic rotational kinematic equation for constant alpha:

theta-theta0=omega0*t+1/2*alpha*t^2

Substitute known values, total rotation is 45 degrees, omega0=0 (from rest), time is 1.1 seconds.

45=0*t+1/2*alpha*(1.1)^2

Therefore acceleration is 85.811 deg/sec/sec.

Now using the same equation, plug in scenario two. Total rotation is 78 degrees, omega0=0, alpha is 85.511 deg/sec/sec. Solve for t.

78=0*t+1/2*85.511*t^2

Quadratic formula, or graphing should yield t=1.4 seconds.