Position, displacement, velocity, acceleration vectors

In summary: Does this look correct?Yes, that looks good. Just be careful with your units - you should have meters for distance and meters per second for velocity. But otherwise, your calculations and reasoning are correct.
  • #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!
 
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  • #2
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
cupid.callin said:
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
jensson said:

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.

jensson said:
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.

jensson said:
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 traveling with *some* constant acceleration. So it's a bit of a more abstract problem. Just go with the information you are given.
 
  • #5
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
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
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
 

1. What is the difference between position and displacement?

Position refers to an object's location in space at a particular time, while displacement refers to the change in an object's position from its initial location to its final location.

2. How are velocity and acceleration related?

Velocity is the rate of change of an object's position with respect to time, while acceleration is the rate of change of an object's velocity with respect to time. In other words, acceleration is the change in an object's velocity over time, and velocity is the change in an object's position over time.

3. Can an object have a constant velocity and a changing acceleration?

Yes, an object can have a constant velocity and a changing acceleration. This occurs when the object is moving at a constant speed in a curved path, such as a car moving around a circular track. The direction of the velocity is changing, which means the acceleration is also changing even though the speed remains constant.

4. How do you calculate the displacement of an object?

To calculate displacement, you need to know the initial position and the final position of the object. You can then subtract the initial position from the final position to find the displacement. Displacement is a vector quantity, so it has both magnitude (the distance between the two positions) and direction.

5. What is the difference between speed and velocity?

Speed is a scalar quantity that refers to the rate of change of an object's position without regard to direction. Velocity, on the other hand, is a vector quantity that includes both speed and direction. This means that two objects can have the same speed but different velocities if they are moving in different directions.

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