Problem 12 - An Introduction to Physical Science

Question

Modern oil tankers weigh more than a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distance of \(5.0 \mathrm{km}\) (about \(3.0 \mathrm{mi}\) ) and a time of 20 min to come to a stop from a top speed of \(30 \mathrm{km} / \mathrm{h}\). (a) What is the magnitude of such a ship's average acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) in coming to a stop? (b) What is the magnitude of the ship's average velocity in \(\mathrm{m} / \mathrm{s}\) ? Comment on the potential of a tanker running aground.

Step-by-Step Solution

Answer

(a) 0.00694 m/s²; (b) 4.17 m/s. Huge stopping distances lead to grounding risk.

Step 1: Convert Units for Initial Velocity

We are given the initial speed of the tanker in kilometers per hour, which is 30 km/h. To find the acceleration, we need this unit in meters per second. Use the conversion factor:\[1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s}\]Thus, the initial velocity \(v_i\) is:\[v_i = 30 \times \frac{5}{18} = \frac{150}{18} \approx 8.33 \text{ m/s}\]

Step 2: Calculate Average Acceleration

The formula for average acceleration \(a\) is:\[a = \frac{v_f - v_i}{\Delta t}\]where \(v_f\) is the final velocity (0 m/s, since the ship stops), \(v_i\) is the initial velocity (8.33 m/s from Step 1), and \(\Delta t\) is the time in seconds. Convert time from minutes to seconds:\[20 \text{ min} = 20 \times 60 = 1200 \text{ s}\]Now, compute \(a\):\[a = \frac{0 - 8.33}{1200} = \frac{-8.33}{1200} \approx -0.00694 \text{ m/s}^2\]The magnitude of the acceleration is \(0.00694 \text{ m/s}^2\).

Step 3: Calculate Average Velocity

Average velocity \(v_{avg}\) can be found using the equation:\[v_{avg} = \frac{d}{\Delta t}\]where \(d\) is the distance (5000 m, converted from 5.0 km) and \(\Delta t\) is the time in seconds (1200 s):\[v_{avg} = \frac{5000}{1200} \approx 4.17 \text{ m/s}\]

Step 4: Comment on Potential of Running Aground

Due to the large stopping distance and time required, a tanker has a significant risk of running aground if not slowed down well before reaching shallow waters.

Key Concepts

Average Acceleration

Average acceleration is a fundamental concept in kinematics, dealing with how quickly an object's velocity changes. It's calculated using the equation:\[ a = \frac{v_f - v_i}{\Delta t} \]where:

• \(v_f\) is the final velocity
• \(v_i\) is the initial velocity
• \(\Delta t\) is the time period over which the change occurs

In the context of our exercise, the tanker comes to a stop, making the final velocity zero. Its initial velocity was converted from kilometers per hour to meters per second.
With this information, the average acceleration was calculated as approximately \(-0.00694\ \text{m/s}^2\).
Remember, the negative sign indicates deceleration, which is a decrease in speed over time.
The magnitude, which focuses only on the size of the acceleration without regard to direction, is \(0.00694\ \text{m/s}^2\).
This typically small number reflects the extended time over which the tanker slows down.

Unit Conversion

Unit conversion is crucial in kinematics to work within compatible units, especially when using formulas for calculations.
In our initial scenario, the tanker’s speed was given in kilometers per hour (km/h). However, the acceleration needs to be in meters per second squared (m/s²). To do this, we used the conversion factor: \[ 1 \ \text{km/h} = \frac{5}{18} \ \text{m/s} \]Here are some steps generally useful for conversion:

• Multiply the value in kilometers per hour by \(\frac{1000}{3600}\) to convert it to meters per second.
• Remember that to convert minutes to seconds, multiply the number of minutes by 60.

Such conversions ensure that the calculations are accurate and consistent with standard units used in physics, allowing for proper assessment and computation of quantities like speed and acceleration.

Velocity

Velocity measures how fast something is moving in a specific direction. It is a vector quantity, which means it has both magnitude and direction, differentiating it from speed, which only indicates magnitude. In the discussed exercise, we were interested in the tanker's average velocity, which gives us a helpful understanding of its motion over a period of time.
To find the average velocity, we used the equation:\[ v_{avg} = \frac{d}{\Delta t} \]where:

• \(d\) is the total distance traveled
• \(\Delta t\) is the time interval

In our calculation, with a total distance of 5000 meters (converted from kilometers) and a time span of 1200 seconds, we found the tanker's average velocity to be approximately \(4.17\ \text{m/s}\).
This average velocity gives a snapshot of the tanker's overall pace of motion towards stopping, excluding details about changes at each moment which are captured by acceleration.

Stopping Distance

Stopping distance is the total distance a vehicle travels before it comes to a complete stop.
This concept is key in assessing factors like safety and efficiency, particularly in large ships like oil tankers, where the momentum and mass require careful consideration of stopping procedures.
The stopping distance in our example was 5 kilometers (5000 meters).
The large distance highlights the considerations necessary in navigation and the potential risks of approaching hazards, like shallow waters, without adequate foresight.
With this in mind, maritime operations need to plan stopping maneuvers well in advance to prevent accidents like running aground.
Effective navigation requires calculating stopping distances based on speed, weight, sea conditions, and the capacity to decelerate, accounting for the massive momentum inherent in large ships.