A body moving in a straight line with constant accelaration

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Homework Help Overview

The problem involves a body moving in a straight line with constant acceleration, specifically focusing on the time taken to cover two successive distances of 1 meter. The goal is to find the acceleration based on the given time intervals of 3 seconds and 5 seconds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for additional information, such as initial or final velocity, to solve the problem. Some suggest setting up equations based on the distance formula for uniformly accelerated motion, while others express confusion about the problem's setup.

Discussion Status

The discussion is ongoing, with some participants providing potential approaches to set up equations, while others are questioning the completeness of the information provided. There is no explicit consensus on how to proceed, but several lines of reasoning are being explored.

Contextual Notes

There is a noted concern regarding the accuracy of the problem statement, with participants questioning whether all necessary details have been included.

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Homework Statement


A body moving in a straight line with constant accelaration takes 3 seconds and 5 seconds to cover two successive distances of 1m.Find the accelaration?


Homework Equations





The Attempt at a Solution



 
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Constant Acceleration. Hint is in the question.
 


i didnt get u
 


Hmm, seems impossible without more information (i.e. initial or final velocity). Did you copy the question down properly?
 


question is right
 


It looks straight forward to me. Take the initial speed to be v and acceleration to be a. Take t= 0 at the beginning of the first second in which the object goes 3 feet.

The distance such an object covers is t seconds is [itex]vt+ (1/2)at^2[/itex]. In one second, it will have gone distance v+ (1/2)a. In two seconds it will have gone distance 2v+ 2a. What distance did it go the second second? Set those equal to 3 and 5 and you have two equations to solve for v and a.
 


HallsofIvy said:
It looks straight forward to me. Take the initial speed to be v and acceleration to be a. Take t= 0 at the beginning of the first second in which the object goes 3 feet.

The distance such an object covers is t seconds is [itex]vt+ (1/2)at^2[/itex]. In one second, it will have gone distance v+ (1/2)a. In two seconds it will have gone distance 2v+ 2a. What distance did it go the second second? Set those equal to 3 and 5 and you have two equations to solve for v and a.

So simple, nice one.
 

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