How Do You Determine the Initial Velocity Vector Given Speed and Direction?

Bipolarity
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Homework Statement



At time t = 0, a particle is located at the point (1, 2, 3). It travels
in a straight line to the point (4, 1,4), has speed 2 at (1, 2, 3) and
constant acceleration 3i - j + k. Find an equation for the position
vector r(t) of the particle at time t.

Homework Equations


The Attempt at a Solution


If we integrate the acceleration vector twice, we get the position vector (along with some constants we must determine). Plugging in the initial position vector gives us

r(t) = <1.5t^{2},-.5t^{2},.5t^{2}> + v_{0}t + <1,2,3>

The problem is I don't know how to find the initial velocity. That's all I need to complete the problem. I know the speed at time t=0, but not the velocity. I also know that the particle travels in a straight line, meaning its unit tangent vector is constant and that its motion is nonperiodic, but how does that help?

BiP
 
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Bipolarity said:

Homework Statement



At time t = 0, a particle is located at the point (1, 2, 3). It travels
in a straight line to the point (4, 1,4), has speed 2 at (1, 2, 3) and
constant acceleration 3i - j + k. Find an equation for the position
vector r(t) of the particle at time t.

Homework Equations





The Attempt at a Solution


If we integrate the acceleration vector twice, we get the position vector (along with some constants we must determine). Plugging in the initial position vector gives us

r(t) = <1.5t^{2},-.5t^{2},.5t^{2}> + v_{0}t + <1,2,3>

The problem is I don't know how to find the initial velocity. That's all I need to complete the problem. I know the speed at time t=0, but not the velocity. I also know that the particle travels in a straight line, meaning its unit tangent vector is constant and that its motion is nonperiodic, but how does that help?

BiP

If a particle travels in a straight line, how are the velocity and acceleration vectors related?

RGV
 
Bipolarity said:

Homework Statement



At time t = 0, a particle is located at the point (1, 2, 3). It travels
in a straight line to the point (4, 1,4), has speed 2 at (1, 2, 3) and
constant acceleration 3i - j + k. Find an equation for the position
vector r(t) of the particle at time t.

Homework Equations





The Attempt at a Solution


If we integrate the acceleration vector twice, we get the position vector (along with some constants we must determine). Plugging in the initial position vector gives us

r(t) = <1.5t^{2},-.5t^{2},.5t^{2}> + v_{0}t + <1,2,3>

The problem is I don't know how to find the initial velocity. That's all I need to complete the problem. I know the speed at time t=0, but not the velocity. I also know that the particle travels in a straight line, meaning its unit tangent vector is constant and that its motion is nonperiodic, but how does that help?

BiP

That equation doesn't make any sense because you can't add vectors and scalars. And remember you can always express a velocity vector as the speed times a unit vector in the correct direction.
 
LCKurtz said:
That equation doesn't make any sense because you can't add vectors and scalars. And remember you can always express a velocity vector as the speed times a unit vector in the correct direction.

Where in my equation did I add vectors with scalars? Every term in my equation is a position vector...

BiP
 
Bipolarity said:
the particle travels in a straight line, meaning its unit tangent vector is constant and that its motion is nonperiodic
If it travels in a straight line then its acceleration must always be collinear with the velocity.
 
Bipolarity said:
r(t) = <1.5t^{2},-.5t^{2},.5t^{2}> + v_{0}t + <1,2,3>


BiP

LCKurtz said:
That equation doesn't make any sense because you can't add vectors and scalars. And remember you can always express a velocity vector as the speed times a unit vector in the correct direction.

Bipolarity said:
Where in my equation did I add vectors with scalars? Every term in my equation is a position vector...

BiP

Well, you mentioned that the initial speed was 2. Since there was no definition given for ##v_0## and it looks like a scalar, I assumed it was ##2##. And there is no notation to distinguish vectors from scalars, I assumed ##v_0 t## was a scalar.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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