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Position, displacement, velocity, acceleration vectors

  • Thread starter jensson
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  • #1
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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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,137
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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
 
  • #3
25
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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..
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36

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.
 
  • #5
25
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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?
 
  • #6
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
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.
 
  • #7
25
0
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
 

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