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Homework Help Overview

The discussion revolves around the motion of a particle described by a vector equation, with specific tasks related to velocity and acceleration. The subject area includes concepts from kinematics and vector calculus.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of the velocity being constant and the relationship between velocity and acceleration. There are attempts to derive expressions for velocity and acceleration from the given position vector.

Discussion Status

Some participants have provided guidance on the conceptual aspects of the problem, particularly regarding the nature of constant speed and its implications for acceleration. There is an ongoing exploration of the relationships between the variables involved, with no explicit consensus reached.

Contextual Notes

Participants are navigating the requirements of the problem, including the need to verify conditions for constant speed and the nature of acceleration in this context. There are indications of confusion regarding the interpretation of the position vector and its derivatives.

RyanH42
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Homework Statement


A particle moves so that its equation of motion in vector form given is given by ##\vec{R}=((sint^-1)/2+t/2√(1-t^2))\vec{i}+1/2t^2\vec{j}## , ##0≤t<1##
a)Show that particle moves with a constant speed.
b)Compute ##\vec{v}## and ##\vec{a}##,and verify that ##\vec{v}##.##\vec{a}##=0 (dot product).(As it should be when speed is constant
c)Since the magnitude of speed is constant ,must the magnitude of acceleration also be constant ?

Homework Equations


##\vec{v}=d\vec{R}/dt##
##\vec{a}=d\vec{v}/dt##

The Attempt at a Solution


For a), I have to do ##\vec{v}=d\vec{R}/dt## and then I have to do this ##\|\vec{v}\|##.If ##\|\vec{v}\|## this do not contain t it means ##\|\vec{v}\|## is constant
I understad b)
For c)##\vec{v}##.##\vec{a}##=0 this means ##\|\vec{v}\|.\|\vec{a}\|.cosθ=0## we know that ##\|\vec{v}\|## is not zero so there's two option 1) ##\|\vec{a}\|## will be zero or ##cosθ## will be ##0##.If ##cosθ## is zero then there's no need to be constant magnitude of ##\vec{a}##.I think answer is no.there's a chance to be not constant.
Is my answers are true ?
Thanks
 
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I think 'c' is a conceptual question. You don't need any form of equation to answer the question.
 
Your reply did not help me
 
Your reasoning for c) is good.
I can't comment on a) and b) because I'm not sure what the given equation for R is saying, but you appear happy with your answers to those.
 
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RyanH42 said:
Your reply did not help me

Sorry about that. Here, to clear things up.

a.) Yes, this is correct. The velocity of the particle is constant if the first derivative of your position vector is constant, or if its second derivative is zero (both with respect to time).
c.) This is conceptual if you consider the question for general cases, that is, for all position vectors with a constant velocity. However, if the question is only for this problem in particular, you can calculate ##\theta##.
 
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ecastro said:
Sorry about that. Here, to clear things up.

a.) Yes, this is correct. The velocity of the particle is constant if the first derivative of your position vector is constant, or if its second derivative is zero (both with respect to time).
c.) This is conceptual if you consider the question for general cases, that is, for all position vectors with a constant velocity. However, if the question is only for this problem in particular, you can calculate ##\theta##.
Ok,I get the idea.I can calculate the ##cosθ## Actually question b is asked for that I guessThanks for help.
 

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