Motion in Space: Velocity and Acceleration

Physicsnoob90
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Homework Statement


For V(t) = (1+t)i + (t^2-2t)j find:

1. velocity
2.Acceleration
3. speed/length
4. Unit Tangent
5. tangential component
6. normal component at t=2

Homework Equations


The Attempt at a Solution



1. v'(t) = <1,2t-2> = <1, 2(t-1)>
2. a(t) = <0,2>
3. length = Square root of (1+4(t-1)^2)
4. Unit Tangent = <1,2(t-1)> / square root (1+4(t-1)^2)
from here I'm stuck trying to find the tangential component and normal component. [/B]
 
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Physicsnoob90 said:

Homework Statement


For V(t) = (1+t)i + (t^2-2t)j find:

1. velocity
2.Acceleration
3. speed/length
4. Unit Tangent
5. tangential component
6. normal component at t=2

Homework Equations


The Attempt at a Solution



1. v'(t) = <1,2t-2> = <1, 2(t-1)>
2. a(t) = <0,2>
3. length = Square root of (1+4(t-1)^2)
4. Unit Tangent = <1,2(t-1)> / square root (1+4(t-1)^2)
from here I'm stuck trying to find the tangential component and normal component. [/B]

First decide: normal and tangential component of WHAT? In general, if you have two vectors ##\vec{a}, \vec{b}## the component ##\vec{a}_{||}## of ##\vec{a}## parallel to ##\vec{b}## and the component ##\vec{a}_{\perp}## perpendicular to ##\vec{b}## are
\vec{a}_{||} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b} \\<br /> \vec{a}_{\perp} = \vec{a} - \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}
 
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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