What are the relevant equations for two dimensional motion in physics?

In summary, the conversation is about a first year college physics question on two dimensional motion. The student is unsure about the correct approach and equations for the problem. They discuss using integrals to find the velocity and sketching the path using critical points. In the end, they come to a solution for the problem.
  • #1
madah12
326
1
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

Homework Statement


http://img338.imageshack.us/img338/3289/rocketa.png [Broken]


Homework Equations


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


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.
 
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  • #2
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
hi madah12! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
madah12 said:
Vx= v(x0) +a(x)t =1+ 2.5t^3
Vy= V(y0) + a(y)t = 7 + 9t-1.4t^2

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) :wink:
 
  • #4
( I tried working with it a lot 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
yup! :biggrin:

(and have an integral: ∫ :wink:)
 
  • #6
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?
 
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  • #7
madah12 said:
so Vr = vxi + vyj
= (5/6t^3 + 1)i + (7 + 9t-1.4t^2) *(j^)?
I am pretty noob at integration to get the position should I integrate the vector equation or integral each dimension separately

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) :wink:
and since it start from the origin x0 and y0 = 0 right?

right! :smile:
 
  • #8
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
 
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  • #9
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 don't know how to sketch y vs x graph
 
  • #10
madah12 said:
is it 340.6 or 341 I think its 341 but I am not good with significant figures

the given values are 2.50 9.00 1.40 1.00 and 7.00, so I agree with 341 rather than 340.6 :wink:
 
  • #11
madah12 said:
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 don't know how to sketch y vs x graph

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

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

1. What is two dimensional motion?

Two dimensional motion refers to the movement of an object in a two-dimensional space, such as a plane or surface. It involves the object's displacement, velocity, and acceleration along two perpendicular axes or directions.

2. How is two dimensional motion different from one dimensional motion?

In one dimensional motion, an object only moves in one direction along a line or axis. In two dimensional motion, an object can move in two perpendicular directions, making it more complex and requiring the use of vectors to represent its motion.

3. What are the two types of two dimensional motion?

The two types of two dimensional motion are projectile motion and circular motion. In projectile motion, an object is launched into the air and moves along a curved path due to the influence of gravity. In circular motion, an object moves in a circular path around a fixed point.

4. How is velocity calculated in two dimensional motion?

In two dimensional motion, velocity is calculated using vectors. The magnitude of the velocity vector is the speed of the object, and its direction is the direction of motion. The velocity in the x and y directions can also be calculated separately using trigonometry.

5. What factors can affect two dimensional motion?

The factors that can affect two dimensional motion include the initial velocity, acceleration due to gravity, air resistance, and any external forces acting on the object. The shape and weight of the object can also affect its motion in certain cases.

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