A 222-kg speedboat is negotiating acircular turn (radius = 30 m) around abuoy. D

Question

A 222-kg speedboat is negotiating acircular turn (radius = 30 m) around abuoy. During the turn, the engine causes a net tangential force ofmagnitude 510 N to be applied to theboat. The initial tangential speed of the boat going into the turnis 5.1 m/s. the tangential acceleration (magnitude) is 2.3m/s2

(b) After the boat is 2.1 s into theturn, find the centripetal acceleration (magnitude).
________________________ m/s2 thank you for your help! the tangential acceleration (magnitude) is 2.3m/s2

(b) After the boat is 2.1 s into theturn, find the centripetal acceleration (magnitude).
________________________ m/s2 thank you for your help!

Explanation / Answer

Given : .            mass (m) = 222 kg   ; radius (r) = 30 m ; Tangential force (F) = 510 N ; Velocity (u) = 5.1 m/s . (a)    Tangential acceleration is : a = F / m = 510 N / 222   = 2.297 = 2.3 m /s2 . (b)    Velocity after 2.1 s   is: .                v = u + at = 5.1 + ( 2.3 * 2.1 ) = 9.93 m /s . Hence centripetal acceleration is : .               aC = m v2 / r  = ( 222* (9.93)2 ) / 30   = 729.676 ˜   730 m/s2 . Hope this helps u!

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