# Two dimensional motion

1. Jul 15, 2010

I have a question in first year college physics about two dimensional motion.
It isn't that I don't know the answer it's just that I am not sure that I started correctly and want to make sure I have the right idea so I understand the questions of this nature
so I didn't understand exactly what they meant by equation for the velocity as a function of time do they want equation for the magnitude or the vector

1. The problem statement, all variables and given/known data
http://img338.imageshack.us/img338/3289/rocketa.png [Broken]

2. Relevant equations
I am not sure what they are asking for in part a so I don't know the relevant equations

3. The attempt at a solution
V = Vx (i^) + Vy (j^)
Vx= v(x0) +a(x)t =1+ 2.5t^3
Vy= V(y0) + a(y)t = 7 + 9t-1.4t^2
so V =(1+ 2.5t^3)*(i^) + (7 + 9t-1.4t^2) *(j^)?
or I should make one for the magnitude of velocity
since I am not sure I didn't do the one for the position until I know what the question meant
I only want to solve part a and b.

Last edited by a moderator: May 4, 2017
2. Jul 15, 2010

### fatra2

At first glance, it seems correct.

In this type of question, you have to deal with the x and the y component seperately. The common variable that joins both of them is the time.

Cheers

3. Jul 15, 2010

### tiny-tim

(try using the X2 and X2 icons just above the Reply box )
no, V(t) = V(0) + at only works if a is constant

you need to use V(t) = ∫a(t) dt (and V(0) will come in as a constant of integration)

4. Jul 15, 2010

( I tried working with it alot the integral sign won't show up right....)
so integral a x dt = integral (2.5 t^2)dt = 5/6 t^3+ vx0 = 5/6t^3 + 1

and for the y ay = 9-1.4t so integral (ay)dt = 9t -.7t^2 +vy0 = 9t -.7t^2 +7?
and then I do the same thing with the i and j?

5. Jul 15, 2010

### tiny-tim

yup!

(and have an integral: ∫ )

6. Jul 15, 2010

so Vr = vxi + vyj
= (5/6t^3 + 1)i + (9t -.7t^2 +7?) *(j^)?
I am pretty noob at integration to get the position should I integrate the vector equation or integral each dimension separately and since it start from the origin x0 and y0 = 0 right?

Last edited: Jul 15, 2010
7. Jul 15, 2010

### tiny-tim

it makes no difference (if you integrate the vector equation, you get a vector constant (x0,y0) instead of two equations with one constant each)
right!

8. Jul 15, 2010

r(t)=(5/24t^4 + t)i + (7t+9/2 t^2 -7/30 t^3)j
y(t)=7t+9/2 t^2 -7/30 t^3
vy(t)=9t -.7t^2 +7
I solved it by the quadratic formula and got t aprox=13.6
y(t)=340.6
I ignored the negative time because the rocket wasn't lunched so the equation it was always with y=0 before
EDIT
is it 340.6 or 341 I think its 341 but I am not good with significant figures

Last edited: Jul 15, 2010
9. Jul 15, 2010

I can sketch the path by using the critical points like graphing any function right? look for concavity and all that? I mean does sketching the graph for y vs t and x vs t satisfy sketching the path because I dont know how to sketch y vs x graph

10. Jul 15, 2010

### tiny-tim

the given values are 2.50 9.00 1.40 1.00 and 7.00, so I agree with 341 rather than 340.6

11. Jul 15, 2010

### tiny-tim

why bother, when you know the formulas for vx and vy anyway?

for example, y is a minimum or maximum when vy = 0