No problem! Good luck with your homework.

AI Thread Summary
The discussion revolves around a particle's motion described by a specific vector equation. Participants confirm that the particle moves with constant speed, as indicated by the constancy of the velocity vector's magnitude. For the acceleration, it's noted that while the dot product of velocity and acceleration is zero, indicating they are perpendicular, this does not necessitate a constant magnitude of acceleration. The conversation emphasizes that the conceptual understanding of acceleration's constancy varies based on the specific conditions of the problem. Overall, the insights clarify the relationships between velocity, acceleration, and their implications in the context of the given motion.
RyanH42
Messages
398
Reaction score
16

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.Theres a chance to be not constant.
Is my answers are true ?
Thanks
 
Physics news on Phys.org
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.
 
  • Like
Likes RyanH42
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##.
 
  • Like
Likes RyanH42
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top