Question

Find the tangential and normal components of the acceleration vector. r(t) = (3 + t) i + (t2 − 2t) j

Answer 1

Acceleration can be broken into tangential and normal components, The tangential vector points in the direction which particle is moving and normal vector points in that direction in which curve of that object's path is turning.

Tangential component,$a_{T} = \frac{r'(t) . r''(t)}{|r'(t)|}$

And normal component,$a_{N} =\frac{|r'(t) X r''(t)|}{|r'(t)|}$

Given $r(t)=(3+t)i+(t^{2} -2t)j$

                      = < 3+t , $t^{2} -2t$ >

            r'(t) = <1 , 2t-2>

            r''(t) = <0,2>

Let's plugin these values in $a_{T}$ and $a_{N}$

$a_{T} = \frac{ . }{||}$

               = $\frac{(2t-2)2}{\sqrt{1+(2t-2)^{2} } }$

               =$\frac{4(t-1)}{\sqrt{4t^{2}-8t+5 } }$

$a_{N}=\frac{|X|}{|} = \frac{||}{\sqrt{4t^{2}-8t+5} }$

               = $\frac{2}{\sqrt{4t^{2}-8t+5 } }$