What Minimum Deceleration Is Required for a Car to Stop on a 35 Meter Shoulder?

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SUMMARY

The minimum deceleration required for a car traveling at 35 mi/h to stop on a 35-meter shoulder is greater than 7 m/s². The initial speed must be converted to kilometers per hour, yielding an average speed of 26.25 km/h, which translates to approximately 7.29166 m/s. The car will cover the stopping distance in less than 5 seconds, necessitating a uniform deceleration to reach a final velocity of 0 m/s. The relevant kinematic equation used is v² = u² + 2ad.

PREREQUISITES
  • Understanding of basic physics concepts, particularly kinematics.
  • Familiarity with unit conversions, specifically miles per hour to kilometers per hour.
  • Knowledge of the kinematic equations, especially v² = u² + 2ad.
  • Ability to perform calculations involving average speed and deceleration.
NEXT STEPS
  • Study the kinematic equations in detail, focusing on their applications in real-world scenarios.
  • Learn about unit conversions and practice converting between different measurement systems.
  • Explore examples of deceleration calculations in various contexts, such as vehicle safety and stopping distances.
  • Investigate the effects of different types of deceleration on vehicle dynamics and passenger safety.
USEFUL FOR

Students beginning their studies in physics, particularly those interested in mechanics and motion, as well as automotive engineers and safety analysts focused on vehicle stopping distances.

Janzen
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Hey just starting physics and i am lost. This is the problem:
A car traveling 35 mi/h is to stop on a 35 meter long shoulder of the road. What minimum deceleration is required?

If you could type out How you would this is would be GREATLY appreciated.

Thanks
 
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okay well 1st of all convert mi/h to km/h because the stopping area is given in metres.

The question essentially asks how rapidly must the car decelerate in order to reach a velocity of 0 over 35 metres.

Well the average speed the car will have is 1/2 its initial velocity if we take decleration to be uniform. Therefore, using 1 mi = 1.5 k, the car will have an average speed of 26.25 km/h.

This is equivalent to 437.5m/minute or 7.29166m/second.

Therefore the car will cover the 35 metres in <5seconds.

Therefore the car must decelerate at >7m/s to stop in time
 
You might also find this formula to be very useful
[tex]v_2^2 = v_1^2 + 2ad[/tex]
 

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