# Position, displacement, velocity, acceleration vectors

## Homework Statement

At time zero, a particle is at x = 4.0 m and y = 3.0 m and has velocity vector = (2.0 m/s)i + (-9.0 m/s)j. The acceleration of the particle is constant and is given by acceleration vector = (4.0 m/s^2)i + (3.0 m/s^2)j.
a) Find the velocity at t = 2.0 s
b) Express the position vector at t = 4.0 s in terms of i and j. In addition, give the magnitude and direction of the position vector at this time.

## Homework Equations

equations for velocity, position vector

## The Attempt at a Solution

So far I have done a, which I think is correct.

velocity vector = (10 m/s)i - (6 m/s)j

(velocity of x is 10 m/s and velocity of y is -6 m/s)

I am stuck at b.

So far I have the displacement of x is 32 m. I got to that by:

x = 4 + 2 + (1/2)(4)(4^2) = 38 m

38 (final position) - 4 (initial position) = 32

For the displacement of y:

y = 3 - 9 + (1/2) (3)(4^2) = 18 m

so 18 - 3 = 15 m

so my final answer here would be:

position vector = 32i + 15j

is this correct? It seems like the equation for the position of y should involve gravity, but I have acceleration given.. please help!

## Answers and Replies

well i didnt checked your questions but just use all vectors directly in equation s = so + ut + .5at^2

dont find mag and then use the eqn ... use vectors dorectly in eqn

well i didnt checked your questions but just use all vectors directly in equation s = so + ut + .5at^2

dont find mag and then use the eqn ... use vectors dorectly in eqn

that doesn't help me..

cepheid
Staff Emeritus
Science Advisor
Gold Member

## The Attempt at a Solution

So far I have done a, which I think is correct.

velocity vector = (10 m/s)i - (6 m/s)j

(velocity of x is 10 m/s and velocity of y is -6 m/s)

That doesn't look right to me. Double check your calculations.

So far I have the displacement of x is 32 m. I got to that by:

x = 4 + 2 + (1/2)(4)(4^2) = 38 m

38 (final position) - 4 (initial position) = 32

I'm pretty sure that 38 minus 4 equals 34.

is this correct? It seems like the equation for the position of y should involve gravity, but I have acceleration given.. please help!

Why do you think there should be gravity? Nobody said this was an object on Earth. You just have *some* particle travelling with *some* constant acceleration. So it's a bit of a more abstract problem. Just go with the information you are given.

Okay so the V vector = (10 m/s)i - (3 m/s)j (I had miscalculated there)

I don't know why I put 38 - 4 = 32, I have it written down correctly. I guess I didn't check my post for errors.

This is my first physics class and when we've dealt with the y direction we've always used gravity, which is why I assumed I would use it here. But it didn't say where this particle was so assuming isn't the way to go obviously. :-/

position vector = 34i + 15j

is that correct?

for magnitude I got: v = square root of 109

I did this by using v = square root of velocity of x squared + velocity of y squared

for direction: 163 degrees

by using theta = inverse of tan -3/10

does this look correct?

cepheid
Staff Emeritus
Science Advisor
Gold Member
I think part b wants the magnitude and direction of the position vector, not the velocity vector. That having been said, your methods for finding those things are correct.

the magnitude for the position vector is like the velocity vector, right? Just instead of using velocity of x and y, I am using their positions. But the equations are the same, yes?

Anyway, in that case I got magnitude = square root of 36^2 + 15^2 = 39 m

and direction = inverse tan (15/36) = 23 degrees