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Newton's Law of Motion for a Straight Line Motion 
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#1
Oct1608, 10:32 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, 05:29 AM

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Don't you think that a distance of 10^{4} meter and a time of 10^{4} seconds is a bit unrealistic? :)
In fact I don't think even dividing 10^4 by 10^7 will give you something of order 10^3, so you better check your acceleration. Probably if you fix that error and follow it through your calculation, you will find the correct answer (i.e. if your acceleration is off by a factor 10^6 your distance will be off by the same factor, leading to an answer of order 5.02 10^2 instead). Since your answer will be so close to the critical distance (e.g. it will barely make it or not) be especially careful about rounding: wrong rounding may produce 499.8 meter while it should actually be 500.4 for example, leading to the wrong conclusion. 


#3
Oct1708, 10:32 AM

P: 11

Okay, so I calculated the ship's engine's acceleration again:
a = f/m = 8.0 x 10^{4}N / 3.6 x 10^{7} kg = 2.22222 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 + (2.22222 x 10^{3})(t) t = 675s. 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})(675)^{2} x = 505.74375 = 506 m. Oh, thank you, guys, for correcting my miscalculation. And if I find the speed to determine if the hull can withstand its impact or not? v_{x} = v_{0x} + a_{x}t v_{x} = 0 + 2.22222 x 10^{3}(675) v = 1.5 m/s I don't know how to get 0.17 m/s ... 


#4
Oct1708, 11:01 AM

P: 149

Newton's Law of Motion for a Straight Line Motion
You found the right distance and determined that it will hit the reef. Now you want the speed after it's travelled a certain distance(500m)... you want [tex]{v_x}^2= v_{0x}^2 + 2a_x(xx_0)[/tex] Make sure you pick the right distances and forumlae! 


#5
Oct1708, 01:19 PM

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