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Newton's Law of Motion for a Straight Line Motion 
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#1
Oct1608, 10:55 PM

P: 11

1. The problem statement, all variables and given/known data
An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 m/s. When the tanker is 500 m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6 x 10^{7} kg, and the engines produce a net horizontal force of 8.0 x 10^{4}N on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of 0.2 m/s or less. You can ignore the retarding force of the water on the tanker's hull. 2. Relevant equations F = ma v_{x} = v_{0x} + a_{x}t x = x_{0} + v_{0x}t + 1/2 a_{x}t^{2} v_{x}^{2}= v_{0x}^{2} + 2a_{x}(xx_{0}) x  x_{0} = (v_{0x} + v_{x} / 2)t 3. The attempt at a solution I don't exactly know what to do first, so I first found the acceleration of the ship's engines. a = f/m = 8.0 x 10^{4}N / 3.6 x 10^{7} kg = 2.22 x 10^{3} m/s^{2} Then I tried to find the time it takes for the ship to hit the reef: v_{x} = v_{0x} + a_{x}t 1.5 = 0 + (2.22 x 10^{3})(t) t = 6.757 x 10^{4} s. And plugged it into the distance traveled: x = x_{0} + v_{0x}t + 1/2 a_{x}t^{2} x = 0 + 0 + 1/2 (2.22 x 10^{3})(6.757 x 10^{4})^{2} x = 5.02 x 10^{4} m. The book's answer said that it's 506 m so the ship will hit the reef, and the speed at which the ship hits the reef is 0.17 m/s, so the oil should be safe. But I don't know how to get to the correct answers. :( 


#2
Oct1708, 05:38 AM

P: 31




#3
Oct1708, 01:26 PM

P: 64

a = f/m = 8.0 x 10^{4}N / 3.6 x 10^{7} kg = 2.22 x 10^{3} m/s^{2} 


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