So you can check yourself the answer is 7.37 sec Gabriela hunting for her favori

Question

So you can check yourself the answer is 7.37 sec

Gabriela hunting for her favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the Gabriela has a constant value of 0.250 rad/s. At the instant Gabriela spots the horse, one-quarter of a turn away, the merry- go-round begins to move (in the direction of Gabriela's running) with a constant angular acceleration of 0.0100 rad's2. What is the shortest time it takes for Gabriela to catch up with the horse?

Explanation / Answer

let they meet after t sec

=> ang disp of kid = ang disp of toy horse + /2

=> boy*t = 1/2 * * t2 + /2

substituting the values boy and

we get a quadratic eqn

solving we get

t = 7.369 sec , 42.63 sec

taking shortest time, t = 7.369 sec = 7.37sec

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