# Vector position, velocity, coordinates and speed

1. Sep 15, 2013

1. The problem statement, all variables and given/known data
A particle initially located at the origin has an acceleration of vector a = 5.00j m/s2 and an initial velocity of vector v i = 8.00i m/s.

a)Find the vector position at any time t (where t is measured in seconds). (Use the following as necessary: t.)
Find the vector velocity at any time t.
b)Find the coordinates of the particle at t = 2.00 s.
x=....m
y=....m
Find the speed of the particle at this time.

2. Relevant equations

Not quite sure. Tried rf=ri+Vit+1/2at^2
Vf=Vi+at

3. The attempt at a solution
For a) I plugged what i thought was right into the first equation above with ri=0i+0j because it is at the origin(see question) but the answer (8.00it+2.50jt^2) was wrong

2. Sep 16, 2013

### vela

Staff Emeritus
Your answer to (a) looks fine to me. Why do you think it's wrong?

3. Sep 16, 2013

### lendav_rott

I would calculate the velocity of the object :
v(t) = [(8.00i)² + (5.00jt)²]^0.5 the i component remains constant where as the j component is accelerating from point of origin, so I am to assume that vj(0) = 0
and if you integrate v(t) from 0 to t you would get the position.

..or atleast this is how I understand the problem.

4. Sep 16, 2013

### yands

It's stated that you have to find the position vector as a function of time. you are correct writing R(0) = 0i + 0j , but this is for the displacement at time t = 0.
Think of this as a projectile motion in a gravitational field.
R here is a vector.

5. Sep 16, 2013

### vela

Staff Emeritus
Your expression for v(t) is for the speed of the object, not its velocity. If you integrate it, you'd get the distance the object moved, not its displacement.

If you look more closely at what nerdalert21 wrote, you'll see i's and j's in the answer, which are the Cartesian unit vectors. (It would have helped a lot if the OP had used the typesetting features of the forum.) The answer $\vec{r}(t) = (8.00t\,\hat{i} + 2.50t^2\,\hat{j})\text{ m}$ is correct, so it's unclear why the OP thinks it's wrong. I'm guessing this is for web-based homework, and the OP somehow typed it in incorrectly.

6. Sep 16, 2013

I got a, I had messed up on typing the answer in.
For finding the vector velocity at any time t (a part2) can I use the equation Vf=Vi+at?
And then plug in the Vi, a and 0 for t?

7. Sep 16, 2013

### Staff: Mentor

Yes. You have the right idea for part 2, but they are asking for the velocity vector at an arbitrary time t, rather than at time t = 0.

Chet

8. Sep 16, 2013

Thanks for all your help guys, I've almost got the question done
The last part says to find the speed of the particle at time t=2
Thats solving for V right?
But im confused on which equation i should use

9. Sep 16, 2013

### Staff: Mentor

You use your equation for the velocity vector evaluated at time t = 2, and then evaluate the magnitude of that vector to get the speed.

10. Sep 16, 2013

What do you mean by evaluate the magnitude?
Like, if i took the x and y components and put them like this:

R=√(x^2+y^2
??

11. Sep 16, 2013

### vela

Staff Emeritus
Yup, but with the components of the velocity.

12. Sep 16, 2013

ok so for the velocity vector at any time t i got 8.00i+5.00jt
and the coordinates at t=2 are
x=16.00m
y=10.00m

So i tried the r=√16^2 + 10^2 and got 18.9 but it says the answer is wrong
Did i do that right?

13. Sep 16, 2013

### vela

Staff Emeritus
Calculate the velocity vector at t=2 s. Then find the magnitude of that vector. $\vec{r}$ has nothing to do with the speed.

14. Sep 16, 2013

so i use Vf=Vi+at?
& I thought r is the magnitude?

15. Sep 16, 2013

### vela

Staff Emeritus
Every vector has a magnitude and direction. If you're talking about the displacement vector, $\vec{r} = x\,\hat{i}+y\,\hat{j}$, its magnitude $r=\|\vec{r}\| = \sqrt{x^2+y^2}$ would be the distance from the origin. If you're talking about the velocity vector, $\vec{v} = v_x\,\hat{i}+v_y\,\hat{j}$, its magnitude $v=\|\vec{v}\| = \sqrt{v_x^2+v_y^2}$ would be the speed.

16. Sep 16, 2013

ohhh ok so it would be V=√(8^2+5^2)?

17. Sep 16, 2013