Kinematics Crash Simulation Question

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SUMMARY

The discussion centers on the physics of crash simulations, specifically analyzing the stopping distance of a car when its speed is doubled. The correct equation to determine stopping distance is V² = Vi² + 2ad, which indicates that when the initial velocity is doubled, the stopping distance quadruples. The conclusion drawn is that both the stopping time and stopping distance are affected, leading to the answer being D: both A and C. Participants emphasized the importance of using the correct equations for accurate results in kinematics.

PREREQUISITES
  • Understanding of kinematic equations, specifically V² = Vi² + 2ad
  • Familiarity with concepts of acceleration and deceleration
  • Knowledge of average speed calculations in motion
  • Basic principles of kinetic energy and work-energy theorem
NEXT STEPS
  • Study the derivation and applications of the kinematic equation V² = Vi² + 2ad
  • Learn about the work-energy principle in the context of braking forces
  • Explore the implications of constant acceleration in real-world crash simulations
  • Practice solving problems involving stopping distances and times using various initial velocities
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and professionals involved in crash safety analysis who seek to deepen their understanding of kinematic principles and their applications in real-world scenarios.

tobya93
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1. In a crash simulation, a car traveling at x m/s can stop at a distance d m with a maximum deceleration. If the car is traveling at 2x m/s:

A. Stopping time is doubled
B. Stopping distance is doubled
C. Stopping distance is quadrupled
D. Both A and C


Ok, so I thought this was a relatively easy question. However, I managed to butcher it. My problem is, I do not understand which equation to use. To find if the distance doubled or quadrupled I used the equation: d = ((Vi + V)/2) * t. Since the car will be stopping the V will be zero and you simply get d = (Vi/2) * t. If the velocity is doubled, distance is doubled. However, the book used the equation V^2 = Vi^2 + 2ad. Thereby getting, stopping distance quadrupled.

Also for time they used V = Vi + at.
Answer is D.

Once again, I am lost to know which equation to use. Since both seemed to work..
Thanks for your help!
 
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Hello Toby, and welcome to PF.
Pity you didn't use the template. Any good reason for that ? Because now I don't know what material (equations) you have available to tackle this.
Also, variable names, given/known data are a bit messy now.

Your d = ((Vi + V)/2) * t is not very useful, since you don't know t (or did you assume t is the same for x m/s and 2x m/s ?). It is also very seldom you can use average speed, so better stay away from this equation if possible (by happenstance it is in fact correct here)

1.) Are you familiar with d = x * t + 1/2 * a * t2 for motion under constant acceleration(/deceleration) ?
Usually written as ## S = v_0\, t + {1\over 2}\, at^2##

2.)
for time they used V = Vi + at
gives the impression you don't recognize this. Is that really unfamiliar ?

3.) Do you recognize
V^2 = Vi^2 + 2ad
? In another way: do you know about kinetic energy and the work done by a braking force that acts over a distance d ?
 
I get a notification "Arka420 has just replied ..." time stamp 20140623 15:22:15 +0000 but don't get to see it posted. Some good spirit got crossed at the considerable giveaway therein? But an identical notification must have been mailed to Toby at he same instant, isn't it?

If spiritual activity was indeed called for, leave something less telepathic for mere earthlings, please. Is Toby being helped meaningfully now (or rather, since originally posting, two weeks ago)?
 

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