Understanding Linear Motion: Solving for Travel and Remaining Distance

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Homework Help Overview

The discussion revolves around a problem related to linear motion, specifically focusing on travel distance, remaining distance, and the calculations involved in determining acceleration and time required for a car to stop after applying brakes.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the definitions of travel distance and remaining distance, questioning which should be used in calculations for acceleration and time. There are attempts to clarify the relationship between different segments of the motion, including the distances covered before and after braking.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the application of equations in different segments of the motion, but no consensus has been reached on the specific distances to be used in calculations.

Contextual Notes

There is some confusion regarding the timing of braking and the distances involved, particularly whether to consider the distance from the initial position to the point of braking or from the braking point to the final position. The problem constraints include a specified reaction time and the need to differentiate between uniform motion and accelerated motion.

manal950
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http://store2.up-00.com/May12/Ueu86896.jpg

someone here told me that travel distance is = vs = 30 m

now remaining is 60 - 30 = 30 m

now my questions is ...

to solve this Q we use which distance is travel distance or remaining distance
 
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How far is he from the man when he hits the brakes?
 
Is travel distance - total distance = 60 - 30 = 30 m
 
Right, 30m.So that should answer your part b, and it changes your s from 60 to 30 in your calculation of acceleration.
 
Ok

now for time required and acceleration we use not this distance we use travel distance which s =vt right ?
 
That equation is only for uniform velocity, it can't be used when the object is accelerating.

And the question asks for the time required to stop the car. So your answer has two parts, the time while the driver is reacting (given to be 5s) and the time it takes him to stop after hitting the brakes (which we said covers a distance of 30 meters. What other things do you know about this time period?).
 
see may you don't understand me

I mean

acceleration
v^2 = u^2 + 2as
the answer will be - 0.66

now for this we use s travel distance which s =vt not 60 - 30 = 30 m

But in part B we use the distance is 60 - 30 because question wants distance cover after applying break

correct ??
 
What values are you using for your variables?
I'm not sure what you're plugging in.
 
let me give give step by step

X ...... 0 ..... Z

see now form x to 0 is distance travel when he apply brakes that we got
s = vt = 30 m
now from 0 to z that is remaining distance 60 - 30 = 30 m

so we have now 3 distance total which is 60 and from starting to where he stop (X to 0 ) is 30

and from where he stop to a man = 30 (X to Z )

now the Q ask find acceleration (retardation )

v^2 = u^2 + 2as
the answer will be - 0.66

so my question is here we must use distance form X to 0 or from 0 to Z

so I want you tell me any of those distances used to solve time required to stop the car and uniform retardation .

I hope that clear
 
  • #10
Well in which stretch (X to O or O to X) are the brakes being applied?
 
  • #11
X to O
 
  • #12
Which letter is the man and which is the initial position of the car?
 
  • #13
initial position of the car? is X
man is Z

I told you before
 
  • #14
So if the man is driving at the man, and after he reaches point O he starts to brake, doesn't that mean the accelerating is done from O to Z?

The number is the same, but you need to understand what is going on in each of the two sections of this problem to understand it.
 
  • #15
Ok I want know some think after O is the car move Or at O the car stop fully ?
 
Last edited:
  • #16
The man doesn't hit the brakes until he gets to O, so how can he be stopped at O?
 

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