velocity and acceleration for motion along a curve

[dusty_girl1144]

question 1
The acceleration of a particle moving in the x−y plane is −gj. The particle is initially at O
with velocity given by V cos(a)i + V sin(a)j for some positive real number a.

a. Find r(t), the position vector at time t.
b. Prove that the particle follows a path with cartesian equation
y = x tan a − gx^2/2V^2*sec^2 a.



question 2

Particles A and B move in the x−y plane with constant velocities.
˙r A(t) = i + 2j and ˙r B(t) = 2i + 3j ( theres meant to be a dot above "r")
Also rA(2) = 3i + 4j and rB(3) = i + 3j
Prove that the particles collide, by finding:

a. the time of collision
b. the position vector of the point of collision

[Tea.bag]

a=-gj
v=-gtj+c
when t=0 (initially) v=Vcos(a)i + Vsin(a)j
c=Vcos(a)i + Vsin(a)j
therefore, v=Vcos(a)i + (Vsin(a)-gt)j

r(t)=Vtcos(a)i + (Vtsin(a)-gt^2/2)j + c
when t=0 r(0)=0i+0j
r(t)=Vtcos(a)i + (Vtsin(a)-gt^2/2)j

[dusty_girl1144]

a=-gj
v=-gtj+c
when t=0 (initially) v=Vcos(a)i + Vsin(a)j
c=Vcos(a)i + Vsin(a)j
therefore, v=Vcos(a)i + (Vsin(a)-gt)j

r(t)=Vtcos(a)i + (Vtsin(a)-gt^2/2)j + c
when t=0 r(0)=0i+0j
r(t)=Vtcos(a)i + (Vtsin(a)-gt^2/2)j

wow as simple as that?
 
what happen to the changing of the sin and cos tho?

whos up for challenge number 2

[Tea.bag]

lol
ii)
we know that r(t)= Vtcos(a)i + (Vtsin(a)-gt^2/2)j
to convert it into cartesian equation (what the equation look like)
 
we know that:
x(t)=Vtcos(a)
y(t)=Vtsin(a)-gt^2/2

t=x/(vcos(a))
sub this into y(t)
we get,

y(x)=V*(x/(vcos(a)))*sin(a)-g*(x/(vcos(a)))^2/2
after cancellations and stuff,

y(x)=x*sin(a)/cos(a)-g/2*x^2/(V^2*cos(a))
y(x)=x*tan(a)-gx^2/2v^2*sec^2(a)

[Mao]

wow as simple as that?
 
what happen to the changing of the sin and cos tho?

a is not a variable in this sense, it is not changing.

also, for the sake of eyes and brains, please learn LaTeX.
http://vcenotes.com/forum/index.php/topic,3137.0.html

[fredrick]

question 2

let position vector of a= position vector of b, (same position, theres a collision)

rA(t)=ti+2tj+c
3i+4j=2i+4j+c
c=i

r(t)A=(t+1)i+2tj---   1

rB(t)=2ti+3tj+c
i+3j=6i+9j+c
c=-5i-6j

rB(t)=(2t-5)i+(3t-6)j---   2

rA(t)=rB(t)
equate i with i and the j component with the j component.
t+1=2t-5
2t=3t-6

ie t=6

Now for position vector sub into either position vector of A or B
Hence: rA(6)=7i+12j

[cara.mel]

In first line of working there is a typo, t^2 should be 2t (sorry for being picky! )

Mao - People don't have to do what you want them to do...
I see no point in most of it anyway
I find it much easier to do r(t) = ai + bj etc myself, and I don't see how the readability is reduced.
If you want to do whacky sums and integrals then fine use some typesetting language
Most of the time it's not necessary on these forums.

[Mao]

"y = x tan a − gx^2/2V^2*sec^2 a"

if it is simple addition of terms, i have no problem with those
when you move on to indices and fractions... it can still be read, just with great pain.

people don't have to use LaTeX, but they should keep in mind that help is freely given by others, if you make it a pain for them, they can choose to ignore the question completely.

the paradigm exists for those who are unfamiliar with the typesetting that it is not essential, and it is their choice, I put forward the suggestion that it will look nicer and be easier to comprehend.

[Pandemonium]

i thought they would've read it with far greater detail to attention, as opposed to ignoring it altogether.
oh well.

[fredrick]

i like reading em in Latex too, easier to understand. But it would have taken me ages to type it in Latex so slow at it.

EDIT:thnk caramel fixed it