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Where are we now and where might we be going - Foot spread travel

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A carbon footprint is the total amount of greenhouse gases, primarily carbon dioxide (CO2CO_{2}), released into the atmosphere as a result of the activities of a particular individual, organization, or community.

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Travel is a major contributor to carbon footprints. Different modes of transport have different efficiencies and emission rates per kilometer.

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Speed is defined as the rate at which an object covers distance, calculated using the formula v=dtv = \frac{d}{t}.

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Energy transformation in vehicles often involves converting chemical energy from fuel into kinetic energy (KE=12mv2KE = \frac{1}{2}mv^{2}) and heat energy.

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Sustainable travel involves minimizing the environmental impact by choosing modes of transport with lower emission factors, such as cycling, walking, or using electric vehicles powered by renewable energy.

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The Greenhouse Effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without its atmosphere, accelerated by CO2CO_{2} emissions from combustion engines.

📐Formulae

v=dtv = \frac{d}{t}

d=v×td = v \times t

t=dvt = \frac{d}{v}

Total Emissions=Distance×Emission FactorTotal\ Emissions = Distance \times Emission\ Factor

P=FAP = \frac{F}{A}

💡Examples

Problem 1:

A family travels 450 km450\ km from London to Edinburgh. If their car produces 0.12 kg0.12\ kg of CO2CO_{2} per kilometer, calculate the total carbon footprint for this one-way trip.

Solution:

Total Emissions=450 km×0.12 kg/km=54 kg CO2Total\ Emissions = 450\ km \times 0.12\ kg/km = 54\ kg\ CO_{2}

Explanation:

To find the total carbon footprint, multiply the total distance traveled by the emission rate (factor) of the vehicle.

Problem 2:

A high-speed train travels at a constant speed of 250 km/h250\ km/h. How long will it take to travel a distance of 750 km750\ km?

Solution:

t=dv=750 km250 km/h=3 hourst = \frac{d}{v} = \frac{750\ km}{250\ km/h} = 3\ hours

Explanation:

Using the speed-distance-time relationship, we divide the total distance by the average speed to find the time taken.

Problem 3:

Compare the pressure exerted on the ground by a hiker weighing 600 N600\ N wearing boots with a total surface area of 0.04 m20.04\ m^{2} versus wearing snowshoes with a surface area of 0.2 m20.2\ m^{2}.

Solution:

For boots: Pboots=600 N0.04 m2=15000 PaP_{boots} = \frac{600\ N}{0.04\ m^{2}} = 15000\ Pa For snowshoes: Psnowshoes=600 N0.2 m2=3000 PaP_{snowshoes} = \frac{600\ N}{0.2\ m^{2}} = 3000\ Pa

Explanation:

Pressure is force per unit area. Increasing the surface area (using snowshoes) spreads the 'footprint' force, resulting in lower pressure on the ground.