Constant speed and deaccelaration

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Homework Help Overview

The problem involves a car traveling at a constant speed of 18 m/s that suddenly brakes. The original poster seeks to determine the distance traveled before the brakes are applied, considering a reaction time of 0.52 seconds, and the distance traveled during deceleration with a given deceleration of 5 m/s².

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify the equations needed to calculate the distances involved, questioning whether speed divided by time could yield distance. Some participants suggest using the formula for distance as speed times time. Others explore the implications of constant acceleration and deceleration, discussing how to apply the relevant equations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the equations and concepts involved. There is a recognition of the need to clarify the definitions and assumptions related to speed and deceleration. Multiple interpretations of the problem are being explored, particularly regarding the calculations for distance before braking and during deceleration.

Contextual Notes

There is some confusion regarding the terminology used, particularly with the terms "break" and "brake." Additionally, participants are navigating the constraints of the problem, including the reaction time and the effects of deceleration on distance traveled.

access
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Homework Statement


R is traveling in a car at a constant speed of 18ms when she suddenly breaks.

if reaction time is .52s for applying foot to break how far is traveled before break is applied?

and if the break cause a deceleration of 5m/ssquared, how is traveled before stopping?
 
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ok sorry i not sure on the equations needed i have read the textbook. can you give me the equations so i can work them out? i thought mayb speed / time would give distance?
 
is it constant speed times time? if i do this tho my equation is 18 X .52 yes? 9.36 metres?
 
then the distance of deeceration is constant speed / 5mssquared?
 
sorry i don't mean 5mssquared i ment 5 ms per s
 
The constant speed of the car is 18 m/s and not 18 ms.

In case of constant speed the distance traveled is speed times time. Yes, the car traveled 9.36 meters before braking.
Constant acceleration means that the change of velocity is proportional to the time. (v2-v1)=at. In case of deceleration, v2<v1 and a is negative. You can get t from the change of velocity and a.
In case of accelerating motion, the displacement is
x= v0*t +a/2 *t^2.
Plug in the stopping time for t, and the appropriate value for a.

ehild
 
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?
 
access said:
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?

It is deceleration, so you need to plug in a=-5.

ehild
 
  • #10
props to you.
 
  • #11
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
 
  • #12
SteamKing said:
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
Not the car breaks, but the lady who drives it. :smile:

ehild
 

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