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a.d.d92
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41.58m?
Finding distance using acceleration and time
- Thread starter a.d.d92
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SUMMARY
The discussion focuses on calculating the total distance between two stop signs for a car that accelerates, coasts, and then decelerates. The car accelerates at 4.2 m/s² for 6.0 seconds, coasts for 2.3 seconds, and decelerates at 3.3 m/s² until it stops. Using the equations of motion, specifically s = ut + 0.5at², the participants derive distances for each segment, ultimately calculating a total distance of 229.77 meters between the stop signs.
PREREQUISITES- Understanding of kinematics and equations of motion (SUVAT equations)
- Familiarity with acceleration and deceleration concepts
- Ability to perform basic algebraic manipulations
- Knowledge of units of measurement, specifically meters and seconds
- Study the SUVAT equations in detail to understand their applications
- Learn how to analyze motion graphs, particularly velocity-time graphs
- Explore real-world applications of kinematics in automotive engineering
- Practice solving similar problems involving acceleration, coasting, and deceleration
Students studying physics, automotive engineers, and anyone interested in understanding motion dynamics and kinematics calculations.