Kinematics: Analyzing circular motion of a particle

[Edward Hillsby]

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

 

[BvU]

Hello @Edward Hillsby, !

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 what is u anyway ? [edit] ah ! $v_0$ !

Tip: Make a plot !

 

[Edward Hillsby]

u/initial velocity =v0, u is popular and very common in UK

SUVAT: v=u+at, r= r0+u0t+1/2 a t^2

 

[BvU]

UK is using inches and still has some catching up to do

So: you know $v = u + at$ and you know $v.\ \$ What can be easier ?

Tip: Make a plot !

 

[Edward Hillsby]

I am not sure how to use trig for the NE direction. It is 45 degree or pi/4, slope= 1

 

[BvU]

Tip: Make a plot ! All will be revealed !

 

[Edward Hillsby]

is it vy sin 45=uy+ayt?, it does not work. What plot?

 

[BvU]

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 $t=0, 1, ...15$ sMay I also advise you to use super- and subscripts ?

 

[Edward Hillsby]

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

 

[BvU]

It is working.

Without the plot ?

1 smart mathematician got similar result

For what ?

pi/4=r_y/r_x...ry=(pi_*r_x)/4...is it right?

I don'know if it's right. But intellegible it sure isn't

 

[Edward Hillsby]

so, can angle in certain condition, here slope =1, be expressed by opposite/adjacent in right angle triangle? It mean angle = tangent??

 

[haruspex]

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 r_x, r_y for coordinates of position.
r_y=(π/4)r_x would mean the position is a bit E of NE from the origin (r_y<r_x), so it is wrong on two counts:
1. For exactly NE you would want the coordinates equal. I.e. (r_y/r_x)=1=tan(π/4).
2. The condition you should be applying is that the velocity is NE.

 

[Edward Hillsby]

thanks