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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 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 4.38) .\) When the tanker is 500 \(\mathrm{m}\) 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 \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be sate? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.

Step by step solution

01

Understanding the Problem

An oil tanker is moving towards a reef at speed of 1.5 m/s. When it is 500 m away, it starts accelerating backwards due to a force exerted by the engines. We need to determine if the tanker will hit the reef, and if so, at what speed.

02

Calculate the Deceleration

The deceleration (backwards acceleration) can be calculated using Newton's second law: \[ a = \frac{F}{m} = \frac{8.0 \times 10^4 \, \mathrm{N}}{3.6 \times 10^7 \, \mathrm{kg}} = 0.0022 \, \mathrm{m/s}^2 \] This is the rate at which the tanker will slow down.

03

Determine Stopping Distance

Using the formula for stopping distance, \[ v^2 = u^2 + 2as \]we find the stopping distance where final velocity \(v = 0\), initial velocity \(u = 1.5\, \mathrm{m/s}\), and acceleration \(a = -0.0022\, \mathrm{m/s}^2\). Rearranging gives:\[ 0 = (1.5)^2 + 2(-0.0022)s \]\[ s = \frac{(1.5)^2}{2 \times 0.0022} = 511.36 \, \mathrm{m} \]The stopping distance is 511.36 meters.

04

Compare Stopping Distance to Reef Distance

The stopping distance (511.36 m) is greater than the distance to the reef (500 m). This means the tanker will hit the reef.

05

Calculate Impact Speed

To find the speed upon impact, use the same kinematic equation solving for \(v\) when \(s = 500 \, \mathrm{m}\):\[ v^2 = (1.5)^2 + 2(-0.0022)(500) \]\[ v^2 = 2.25 - 2.2 \]\[ v^2 = 0.05 \]\[ v = \sqrt{0.05} = 0.2236 \, \mathrm{m/s} \]The impact speed is 0.2236 m/s, higher than 0.2 m/s, meaning the oil will not be safe.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Deceleration Calculation

In the scenario provided, we need to calculate the deceleration — which is actually negative acceleration — to determine how quickly the tanker can slow down. This uses Newton's Second Law of Motion.

Newton's Second Law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, described by the equation: \[ F = m imes a \]Here, to find the acceleration (or deceleration, since we're slowing down), we rearrange the formula to:\[ a = \frac{F}{m} \]

• Force, \( F = 8.0 \times 10^{4} \, \mathrm{N} \)
• Mass, \( m = 3.6 \times 10^{7} \, \mathrm{kg} \)
• Deceleration, \( a = \frac{8.0 \times 10^4}{3.6 \times 10^7} \approx 0.0022 \, \mathrm{m/s^2} \)

This deceleration value indicates how much the tanker's velocity reduces over time as it moves against the force of the engines.

Stopping Distance

Stopping distance is crucial in this exercise as it helps determine whether the tanker will stop before hitting the reef.

The stopping distance relies on the kinematic equation:\[ v^2 = u^2 + 2as \]* \( v \) is the final velocity, intended to be 0 for a complete stop.* \( u \) is the initial velocity, \( 1.5 \, \mathrm{m/s} \).* \( a \) is the negative acceleration (or deceleration) \( -0.0022 \, \mathrm{m/s^2} \).* \( s \) is the stopping distance. Rearranging the formula to solve for \( s \), we have:\[ s = \frac{u^2}{-2a} = \frac{(1.5)^2}{2 \times 0.0022} = 511.36 \, \mathrm{m} \]
This means the tanker will stop after traveling 511.36 meters backward. Unfortunately, the reef is only 500 meters away. Therefore, the tanker will collide with the reef.

Kinematic Equations

Kinematic equations are essential tools in physics, used for describing a moving object's motion, often requiring only a few known variables. In this problem, it helps determine how fast the ship is going to move when it reaches the reef.

One of the primary kinematic equations used here is:\[ v^2 = u^2 + 2as \] After determining that the tanker will indeed hit the reef, the next step involves determining the speed at impact by solving for \( v \) with:

• Initial velocity, \( u = 1.5 \, \mathrm{m/s} \)
• Acceleration (deceleration), \( a = -0.0022 \, \mathrm{m/s^2} \)
• Distance to reef, \( s = 500 \, \mathrm{m} \)

Plugging in these values:\[ v^2 = (1.5)^2 + 2(-0.0022)(500) \]\[ v^2 = 2.25 - 2.2 \]\[ v^2 = 0.05 \] \[ v = \sqrt{0.05} = 0.2236 \, \mathrm{m/s} \]
The result indicates the tanker impacts the reef at a speed of 0.2236 m/s, above the safety limit of 0.2 m/s, implying potential damage to the oil cargo.