Car merging to main road: can it really be so simple?

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SUMMARY

The discussion focuses on calculating the minimum distance a car must be behind another vehicle on a main road to avoid being caught up during merging. The merging car accelerates at 3 m/s² from rest, while the other car travels at a constant speed of 50 km/h (13.89 m/s). Using the formula s = (v^2 - v0^2) / 2a, the initial calculation yielded a distance of 4.38 meters, which was later corrected to 31.7 meters after addressing a potential error in the calculations. This highlights the importance of accurate computations in physics problems involving motion.

PREREQUISITES
  • Understanding of basic kinematics, including acceleration and velocity.
  • Familiarity with the equations of motion, specifically v^2 = v0^2 + 2as and s = (v^2 - v0^2) / 2a.
  • Ability to convert units, particularly speed from km/h to m/s.
  • Basic problem-solving skills in physics to analyze motion scenarios.
NEXT STEPS
  • Study the principles of kinematics in greater detail, focusing on acceleration and distance calculations.
  • Practice converting various units of speed and distance to ensure accuracy in calculations.
  • Explore real-world applications of motion equations in traffic scenarios and vehicle dynamics.
  • Learn about the implications of merging behavior in traffic safety and vehicle spacing regulations.
USEFUL FOR

This discussion is beneficial for physics students, driving instructors, traffic safety analysts, and anyone interested in understanding vehicle dynamics during merging scenarios.

mmoadi
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Homework Statement



Upon merging from side to main road and starting from rest with a constant acceleration 3 m/s². At the exact time of our merging, another car is driving on the main road in the same direction with speed 50 km/h. At least how far behind us should the car be, if we don’t want it to catch up with us?

Homework Equations



v^2 = v0^2 + 2as
s = (v^2 - v0^2) / 2a

The Attempt at a Solution



First I converted 50 km/h into 13.89 m/s.
For initial velocity I took 0, because the car starts from rest. For velocity I took the velocity of the second car, which is 50km/h (13.89m/s).

After plugging the numbers into the formula s = (v^2 - v0^2) / 2a, a got the result 4.38 m.
Somehow the distance seems too small?! Were my calculations correct?
 
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I think you made a typo on the caclulator
s = 13.8^2 / 2*3 = 31.7m
 
Thank you.
 

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