# Objects moving in 2D

1. Dec 5, 2006

### ch00se

1. The problem statement, all variables and given/known data

the coordinates of an object moving in the 2D space are given by x=2t and y=4t^2. what are the dimensions of the numerical coefficients 2 & 4, respectively? What are the objects velocity, speed and acceleration and the magnitude of the acceleration at t=0. what is the angle between the velocity and the acceleration at t=0

2. Relevant equations

have been through my text book god knows how many times and cannot decide which to use

3. The attempt at a solution

i have honestly tried for hours and havent managed to get anywhere

however, for the dimensions i substituted 2 and 4 in respectively, but i doubt this is correct

Last edited: Dec 5, 2006
2. Dec 5, 2006

### neutrino

I think you meant y = 4t^2 in the second equation.

For the fist equation, what are the dimensions of x and t. Having found that, what should be the dimension of 2, for the equation to be valid?

3. Dec 5, 2006

### ch00se

edited, thanks

i got :

x=2 = 4
x=4 = 8

y=2 = 64
y=4 = 256

however, like i said, i doubt this is correct....or is it?

4. Dec 5, 2006

### m2287

with those equations, should you put them together to get a y=mx+c styled equation?

i get y = x^2 and then go from there maybe?

5. Dec 5, 2006

You can write the following: $$\vec{r}(t) = x(t)\vec{i}+y(t)\vec{j}$$. Plug the given functions in, and use the following relations: $$\frac{d\vec{r}(t)}{dt}=\vec{v}(t)$$, $$\frac{d\vec{v}(t)}{dt}=\vec{a}(t)$$.

6. Dec 5, 2006

### neutrino

Nope, it's not.

The DIMENSION of x is Length(L) and the dimension of t is Time(T). Now, what is the coefficient's dimension to make a physically meaningful equation?

Whichever textbook you may be using, the first or second chapter probably covers the topic of units and dimensions. Give it a read.

7. Dec 5, 2006

### ch00se

i think for x(t)

you put 2 into x=2t, so you get 4?

then for y( t)

you put 4 into y=4t^2, so you get 74 (4x16)

so r( t) = 4i + 74j ?

Last edited: Dec 5, 2006
8. Dec 5, 2006

### HallsofIvy

Staff Emeritus
t is the TIME, measured, say, in seconds. Why should the time be 2 seconds for x and 4 seconds for y? Those equations give the (x,y) coordinates of the point at time t seconds.

The first question, which I think you are trying to answer, asks "the coordinates of an object moving in the 2D space are given by x=2t and y=4t^2. what are the dimensions of the numerical coefficients 2 & 4, respectively?
What kind of UNITS should they have, not numerical values. In x= 2t, t will have TIME units, x will have DISTANCE units. Think of it as multiplying fractions: (Distance/Time)(Time)= Distance. That "2" factor must have units of Distance/Time (so it's really a "speed" factor), for example meters/sec if t is measured in seconds and x in meters, or miles/hour if x is measured in miles and t in hours.

Now look at y= 4t^2. Again, t is a TIME, y is a DISTANCE. t^2 will be Distance^2. To get Distance= ( )(Time^2), that ( ) must look like Distance/Time^2 so that the "Time^2" terms cancel and leave "Distance". The "4" must have units of Distance/Time^2 (so it's really an 'acceleration' factor), for example meters per second per second or miles per hour per hour.

9. Dec 5, 2006

### ch00se

thanks, that was a very useful post

for the velocity, how would you get a direction on the force?

also can you reach an answer or should the answer be displayed in equation format?

thanks!

10. Dec 5, 2006

### Kurdt

Staff Emeritus
If you go back to post number 5 by radou he explains exactly how to complete the question. The speed is the magnitude of the velocity and is given by using pythagoras' theorem.

$$|\mathbf{v}|= \sqrt{x(t)^2+y(t)^2}$$

Of course its made rather simple by the fact that its all at time t=0.

Last edited: Dec 5, 2006