A cyclist travels in a circular path of radius 5 m. The angular position of the cyclist as a functionof time is given by3 3 ) radians( ) (4t t t= − where ‘t’ is in seconds. The angular velocity ofthe cyclist becomes zero after
Question
A cyclist travels in a circular path of radius 5 m. The angular position of the cyclist as a functionof time is given by3 3 ) radians( ) (4t t t= − where ‘t’ is in seconds. The angular velocity ofthe cyclist becomes zero after
Solution
The angular velocity (ω) is the rate of change of the angular position (θ) with respect to time (t). It can be found by taking the derivative of the angular position function with respect to time.
The given function for angular position is θ(t) = 4t^3 - 3t^2.
So, the derivative of this function will give us the angular velocity:
ω(t) = d(θ)/dt = d/dt [4t^3 - 3t^2] = 12t^2 - 6t.
The angular velocity becomes zero when ω(t) = 0.
So, we set the equation 12t^2 - 6t = 0 and solve for 't':
12t^2 - 6t = 0 => 6t(2t - 1) = 0
Setting each factor equal to zero gives the solutions t = 0 and t = 1/2 seconds.
So, the angular velocity of the cyclist becomes zero after 1/2 second.
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