The maximum deceleration of a car

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SUMMARY

The maximum deceleration of a car on a dry road is established at 8.0 m/s². In a scenario where two cars approach each other at 88 km/h (55 mi/h) from a distance of 85 meters, calculations confirm that they will not collide, stopping 10 meters apart. The equations of motion for both cars are derived, with car 1's position represented as x_1(t) and car 2's as x_2(t), incorporating their respective accelerations and initial velocities.

PREREQUISITES
  • Understanding of kinematics, specifically equations of motion.
  • Familiarity with basic physics concepts such as velocity and acceleration.
  • Knowledge of graphing techniques for distance versus time plots.
  • Ability to convert units, specifically between kilometers per hour and meters per second.
NEXT STEPS
  • Explore advanced kinematic equations for varying acceleration scenarios.
  • Learn about collision physics and the principles of momentum conservation.
  • Investigate graphical analysis of motion, focusing on position-time graphs.
  • Study real-world applications of deceleration in automotive safety systems.
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Physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and collision avoidance strategies.

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The maximum deceleration of a car...

Homework Statement


The maximum deceleration of a car on a dry road is about 8.0 m/s^2


The Attempt at a Solution


A)
If two cars are moving head-on toward each other at 88 km/h (55 mi/h), and their drivers apply their brakes when they are 85 m apart, will they collide?
Answer: No


B)
How far apart will they be when they stop?
Answer: 10m


C) HELP Drawing the graph and points

On the same graph, plot distance versus time for both cars.
Assume x = 0 is the midpoint between the cars and t = 0 when the brakes are applied. Label the position versus time plot of the car with the positive velocity as x_1(t).
http://img684.imageshack.us/img684/6803/94527989.png
 
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if x = 0 is the midpoint, car 1 is at x = -85/2 and car 2 is at 85/2 initially.

x(t) = 1/2 a t^2 + v * t + x(i)
so
x_1(t) = 1/2 (-8) * t^2 + 88000/3600 * t - 85/2
x_2(t) = 1/2 8 t^2 - 88000/3600 * t + 85/2
Notice that car 2's velocity is positive (to the right) and its acceleration is negative(to the left). The reverse is true for car 2.
 

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