Does Slower Deceleration Mean Longer Stopping Distances in Physics?

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Discussion Overview

The discussion revolves around the relationship between deceleration and stopping distances in physics, particularly whether a slower rate of deceleration results in longer stopping distances compared to acceleration. The context includes both theoretical exploration and practical application, with participants engaging in a problem related to kinematics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions if a deceleration rate that is two times slower than the acceleration rate requires twice the distance to stop as compared to the distance needed to accelerate.
  • Several participants suggest using kinematic equations to derive the relationship between acceleration, deceleration, and distance, emphasizing the need to understand uniformly accelerated motion.
  • There is a discussion about the importance of average speed in calculating distance, with one participant noting that average speed should be considered as half the final velocity when deriving distance equations.
  • Another participant proposes a derived relationship, suggesting that acceleration and stopping distance are inversely related, indicating that if acceleration is twice that of deceleration, the stopping distance would be halved.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to solving the problem, with no clear consensus on the initial question regarding the relationship between deceleration and stopping distance. Some participants provide hints and guidance, while others express uncertainty about how to start the problem.

Contextual Notes

Participants rely on kinematic equations and concepts of average speed, but there are unresolved assumptions regarding the specific conditions of the problem, such as the values of acceleration and deceleration. The discussion does not resolve the initial question definitively.

Who May Find This Useful

This discussion may be useful for students learning about kinematics, particularly those interested in the relationships between acceleration, deceleration, and distance in physics problems.

willworkforfood
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Is it true that if the rate of decceleration is two times slower than the rate of acceleration for the same object, that it will need two times the distance to come to a stop as it did to accelerate?
 
Last edited:
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The answer to this question depends on if this is a homework question... :biggrin:
 
It is but it is for extra credit :smile:
 
It's not a very dificult question. Find the distance traveled under constant acceleration for a given period of time. Use what you know about how much the velocity change to solve for the time taken in terms of the maximum velocity and acceleration. Plug this into the equation you had for the distance travelled. Now you have the distance traveled in terms of the acceleration and maximum velocity. By seeing how the acceleration varies with distance your problem is solved.
 
Figure it out for yourself! How does the distance required to accelerate from a speed of 0 to V (or decelerate from V to 0) depend on the acceleration? (Work out or look up the kinematic formulas for uniformly accelerated motion.)
 
Doc Al said:
Figure it out for yourself! How does the distance required to accelerate from a speed of 0 to V (or decelerate from V to 0) depend on the acceleration?

I have no idea where to start or how the distance is related to the acceleration.
 
Are you familiar with the kinematics of uniformly accelerated motion? (If not, then you are not ready to solve this problem.)
 
I know that the equation for acceleration is final velocity - initial velocity divided by time, but I really don't see how that helps answer my original question :(
 
willworkforfood said:
I know that the equation for acceleration is final velocity - initial velocity divided by time
That's a start: You have a relation connecting acceleration and velocity (and time). Now figure out how to get the distance. (Hint: Distance equals average speed X time.)
 
  • #10
The problem doesn't give me average speed or time though.
 
  • #11
willworkforfood said:
The problem doesn't give me average speed or time though.
You don't need actual values. You are trying to find the relationship between acceleration and distance. So far you have:
(1) a = v/t (the initial speed is zero)

I suggest you combine that with my "hint" about distance and average speed. (If something goes from 0 to V, what is its average speed?) Write my statement mathematically. Then you'll have two equations that you can combine to relate distance with acceleration.
 
  • #12
So equation 2 would be x = v*t and you substitute v = at and get

x = a*t^2

Is that right?

Wow I feel stupid thanks for the help on that. :smile:
 
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  • #13
willworkforfood said:
So equation 2 would be x = v*t
No... you need average speed, which is v/2, not v.
and you substitute v = at and get

x = a*t^2
When you combine the equations, don't eliminate v; instead, eliminate t. You want to end up with an equation relating v, a, and x.
 
  • #14
So instead equation 2 would be

x=(vt)/2

Solving for t would be

t = 2x/v

Substituting that into equation 2 would be:

a= v * (v/2x) or a = v^2 / 2x

Meaning when acceleration is 2 times decceleration then you need half the distance to accelerate? Thanks I think I get it now, I am not worthy :P
 
  • #15
Exactly. The derived relationship a = v^2 / 2x tells you that acceleration and stopping distance are inversely related (for a given maximum speed).

Note that this same equation can help you understand why, for a given deceleration, that if you are moving twice as fast it takes 4 times the distance to stop.
 

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