A speedboat with a length of 10.2m and a mass of 247 kg is negotiaging a circular turn (radius = 32 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 513 N to be applied to the boat. The intial speed of the boat going into the turn is 5 m/sec.
1)What is the tangential acceleration of the boat, 2 sec into the turn?
2)What is the tangential velocity of the boat, 2 sec into the turn?
3)What is the centripetal acceleration, 2 sec into the turn?
4)What is the total acceleration, 2 sec into the turn?
- NCSLv 71 month agoFavourite answer
1) a = F / m = 513N / 247kg = 2.08 m/s²
2) v = v₀ + at = 5m/s + 2.08m/s*2s = 9.15 m/s
3) a_c = v²/r = (9.15m/s)² / 32m = 2.62 m/s²
4) a_tot = √(a² + a_c²) = 3.34 m/s²
and the direction, measured from the radius, is
Θ = arctan(a / a_c) = 38.4º
Hope this helps!
- oubaasLv 71 month ago
1) tangential acceleration ta of the boat :
ta = tangential force / mass = 513/247 = 2.077 m/sec^2
(angular acceleration aa = ta/r = 2.077 / 32 = 0.065 rad/sec^2)
2) tangential velocity Vt' of the boat, 2 sec into the turn :
Vt' = Vti+ta*2 = 5.0+2.077*2 = 9.15 m/sec
3) centripetal acceleration ca, 2 sec into the turn :
ca = Vt'^2/r = 9.15^2/32 = 2.618 m/sec^2
4) total acceleration a , 2 sec into the turn :
a = √ ta^2+ca^2 = √ 2.077^2+2.618^2 = 3.34 m/sec^2