Velocity/displacement 11th grade physics

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Homework Help Overview

The discussion revolves around the effects of halving the initial velocity of a skidding truck on its total displacement, as compared to its original velocity. The subject area is kinematics in physics, specifically focusing on displacement and velocity relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between initial velocity and displacement, with one suggesting that the displacement would be less than half due to reduced velocity and time to stop. Others propose using variables in the displacement function to analyze the situation mathematically.

Discussion Status

The discussion includes various attempts to understand the problem, with some participants offering mathematical expressions and others questioning the use of trial and error methods. There is no explicit consensus, but guidance on using mathematical reasoning is present.

Contextual Notes

Assumptions discussed include that the truck starts from rest and that acceleration is constant. There is also a note on the distinction between displacement and distance, highlighting the importance of direction in displacement calculations.

lax1113
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Homework Statement


A truck that is skidding out of control has its initial velocity halved. How will this effect the total displacement of the skidding truck, vs the amount of displacement if the truck were at its original velocity


Homework Equations


displacement= Vi(t)+1/2AT^2 Vi=initial velocity... A=acceleration...T=time


The Attempt at a Solution


I believe that the result would be around less than half the displacement. My reasoning is that the total time to stop would be less, because there is less of a velocity, and also that since the velocity is lower, that the truck would be covering less distance throughout the whole time... maybe 1/3 or around that as a ballpark guess.
 
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Try putting some variables in the displacement function, for example Vi = a m/s, A = a/4 m/s^2. Now put Vi = 2a. Compare the expressions then.
 
kbaumen said:
Try putting some variables in the displacement function, for example Vi = a m/s, A = a/4 m/s^2. Now put Vi = 2a. Compare the expressions then.
dear frnd... kbaumen never encourage your colleagues to put values and solve questions...

trial and error methods are very uncivilized and baseless methods..

they may solve a problem for you but they shall cease to help you in the long run...

use maths to help you always...

v=u+at
a=v/t
v/2=u+a't
a'=v/2t

s1=1/2 v/t *t^2
s2=1/2 v/2t *t^2

s1/s2=2vt/vt
s1/s2=2
s1=2s2

assumptions=the truck starts from rest ..the acceleration is constant.. and the displacements are measured in the same time interval.
 
physixguru said:
dear frnd... kbaumen never encourage your colleagues to put values and solve questions...

trial and error methods are very uncivilized and baseless methods..

they may solve a problem for you but they shall cease to help you in the long run...

use maths to help you always...

Thanks for the advice.
 
The formula for displacement that you used is the same formula i use for solving for distance? why is that? this is to lax113
 
displacement is distance, but it takes into account direction, whereas distance doesn't...so its the same formula for both
 

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