Kinematics: Analyzing circular motion of a particle

  • #1
Edward Hillsby
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Homework Statement:
A particle moves with constant acceleration and its velocity at time t :

v= (5-0.1t)i +(3+0.2t)j

1. find a, u

2. find the time t when particle is travelling in direction NE and find its speed.

3. find the distance and bearing of the particle from the starting point at that time.
Relevant Equations:
SUVAT, TRIG, Bearing
Help please
 

Answers and Replies

  • #2
BvU
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Hello @Edward Hillsby, :welcome: !

Apparently you know about SUVAT. Which ones in particular can be used here ?

Help please
Sure! However, there are some requirements to get help at PF. Read here


:smile: what is u anyway ? [edit] ah ! ##v_0## !

Tip: Make a plot !
 
  • #3
Edward Hillsby
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u/initial velocity =v0, u is popular and very common in UK

SUVAT: v=u+at, r= r0+u0t+1/2 a t^2
 
  • #4
BvU
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UK is using inches and still has some catching up to do :-p

So: you know ##v = u + at## and you know ##v.\ \ ## What can be easier ?
Tip: Make a plot !
 
  • #5
Edward Hillsby
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I am not sure how to use trig for the NE direction. It is 45 degree or pi/4, slope= 1
 
  • #6
BvU
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Tip: Make a plot ! All will be revealed !
 
  • #7
Edward Hillsby
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is it vy sin 45=uy+ayt?, it does not work. What plot?
 
  • #8
BvU
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is it vy sin 45=uy+ayt?, it does not work
No. ##v_y = u_y + a_y t\ \ ## and ##\ v_x = u_x + a_x t ##

I am beginnning to fear you overlook the vector character of ##\vec r, \ \vec v,\ \vec a ## ?

What plot?
v= (5-0.1t)i +(3+0.2t)j ##\ \ ## You know what the ##\ \hat\imath\ ## and ## \ \hat\jmath\ ## stand for ?
Plot ##v_y(t)## as a function of ##v_x(t)## for ##t=0, 1, ...15## s


May I also advise you to use super- and subscripts ?
1611583179145.png
 
  • #9
Edward Hillsby
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It is working. Thanks, guys from yahoo and overflow struggle to answer. I did not overlook vectors, but was to obsessed with trigs. 1 smart mathematician got similar result, using trigs: tangent, arcus tangent and got assumption pi/4=r_y/r_x...ry=(pi_*r_x)/4...is it right? I do not know what is different between ry and r_y. Thanks
 
  • #11
Edward Hillsby
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so, can angle in certain condition, here slope =1, be expressed by opposite/adjacent in right angle triangle? It mean angle = tangent??
 
  • #12
haruspex
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so, can angle in certain condition, here slope =1, be expressed by opposite/adjacent in right angle triangle? It mean angle = tangent??
The slope is the tangent of the angle. An angle π/4 gives a slope of 1. tan(π/4)=1.
pi/4=r_y/r_x...ry=(pi_*r_x)/4...is it right?
I assume you are using rx, ry for coordinates of position.
ry=(π/4)rx would mean the position is a bit E of NE from the origin (ry<rx), so it is wrong on two counts:
1. For exactly NE you would want the coordinates equal. I.e. (ry/rx)=1=tan(π/4).
2. The condition you should be applying is that the velocity is NE.
 
Last edited:
  • #13
Edward Hillsby
7
0
thanks
 

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