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Position vector and velocity vector

Hi everyone,

Could someone please help me on this problem?

1. Homework Statement
A particle initially located at the origin has an acceleration of a=1.00jm/s^2 and an initial velocity of V=8.00im/s
a. Find the position vector at any time t (where t is measured in seconds)
b. Find the velocity vector at any time t

2. Homework Equations
r=xi+yj
v=vxi+vyj


3. The Attempt at a Solution

I know that the velocity is the first derivative of the position vector and the acceleration is the second derivative. However, I am not given the original equation or the velocity vector. Would I use these two equations?

Would it be something like this?

a. r=xi+yj
r=8.00i+1.00j

Could someone please show me what to do or give me a hint?

Thank you in advance
 

Answers and Replies

662
0
No, you've missed it a bit. The equation for r is its initial position, which is not where you said.

Also, you said you're not given the velocity vector, but you are - it's right there in the problem statement (V= ...) - that's a vector equation.

What you're missing are other equations that give the position and velocities as functions of time and acceleration. See if you can find them, because you must have seen them if you've been assigned this problem. Check back here once you've found them.
 
Thank you very much

I know that Vf=vi+at, so in this case the velocity vector is Vf=8i+1t, right? Is the "velocity vector at any time t" always the final velocity?

Would r just be 0? Or would it be the antiderivative of the final velocity or r=8it+1/2jt^2?

Thank you
 
Last edited:
662
0
Whoa ... I think you might be confusing your "i"s. The "i" in "Vf=vi+at" should be a subscript and stands for "initial". The "i" that appears in the equation giving the initial velocity is a unit vector in the x direction. Were you clear on that?

Yes, the initial position is at the origin, so that could be written as r = 0i + 0j.

I think it would help if you tried to write your vector equations more carefully, in other words, be very careful about the directions that the vectors point in. The acceleration is in the y direction, but the initial velocity is in the x direction. So, does the acceleration change the value of the velocity in the x direction? What about the y direction? What is the initial velocity in the y direction?

Maybe it would help you to write all your equations with both x and y components (i.e. include both the i and j unit vectors), even if one or both are zero. That would make it easier to see when you're mixing components incorrectly. Something like this:

[tex] r_i = r_x \hat{i} + r_x \hat{j} [/tex]
[tex] v_i = v_x \hat{i} + v_y \hat{j} [/tex]
[tex] a_i = a_x \hat{i} + a_y \hat{j} [/tex]

where some of those components are equal to zero.
 
Last edited:
Thank you very much

Can you please tell me if this is correct?

ri=8.00i+0j

vi=8.00i+0j

ai=0i+1.00j

vf=(8.00i+0j)+1.00jt

Is the "velocity vector at any time t" normally the final velocity?


Thank you
 
662
0
Nope, not quite.

You already had ri correct earlier - it's at the origin, remember. The equation in your last post puts it at 8 units out on the x-axis. (BTW - it's always a good idea to include the units of measurements in all your equations - you'll catch many mistakes if the units turn out wrong, like something that should be distance coming out as m/sec or something.)

The equations for vi and ai are correct!

Yes, "final" velocity just means the velocity at a later time that we're interested in. It might really be a final velocity, like when something reaches a given location of interest, but in any case, the expression you give tells you the velocity at time t. You just have to simplify the expression by adding the vectors, i.e. add the x-components and add y-components.

You're almost there!
 
Thank you very much

I understand why ri would be 0i+0j due to the fact that it begins at the origin, but the equation for ri is vxi+vyi. So, since vi is 8.00i, wouldn't ri be 8.00i+0j? I'm kind of confused on this.

Vf=vi+at
vf=8.00i+0j+(0i+1.00j)t
vf=8.00i+1jt

Does that look correct?

I also have to find the coordinates of the particle at t=6.00 s. I just need to plug in 6.00 s into the vf equation, right?

a. Find the position vector at any time t (where t is measured in seconds)

Since ri=0i+0j, is this just the answer for part a.?

Thank you
 
Last edited:
662
0
Thank you very much

I understand why ri would be 0i+0j due to the fact that it begins at the origin, but the equation for ri is vxi+vyi. So, since vi is 8.00i, wouldn't ri be 8.00i+0j? I'm kind of confused on this.

Vf=vi+at
vf=8.00i+0j+(0i+1.00j)t
vf=8.00i+1jt

Does that look correct?

I also have to find the coordinates of the particle at t=6.00 s. I just need to plug in 6.00 s into the vf equation, right?

a. Find the position vector at any time t (where t is measured in seconds)

Since ri=0i+0j, is this just the answer for part a.?

Thank you
Your result for vf is correct, but you're having trouble with r.

First, you say that "the equation for ri is vxi+vyi" - that can't possibly be correct, since you're equating a position to a velocity. It's like saying "my current location is 50 km/hr."

ri is simply what it is - it's given and you already identified it: ri = 0i + 0j. That' where this object begins its journey.

What you're missing is the equation that gives position as a function of time, initial position, initial velocity, and acceleration. This is the one whose integral, or antiderivative if you prefer, gives the velocity equation that you already used. Identify that equation and I think you'll be almost done.
 
Thank you very much

Can't I just find the antiderivative of Vf? So, wouldn't r just be 8it+1/2t^2j

In order to find the coordinates of the particle at t=6.00s, I would substitute t with 6 in part a, right?

x coordinate=48m
y coordinate=6m

speed of the particle=v=square root (vx^2+vy^2)
=square root (8^2+1^2)
=square root (65)

Thank you
 

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