Answer
$a=\dfrac{4}{3} \ T + \dfrac{2 \sqrt 5}{3} \ N$
Work Step by Step
$v(t)=\dfrac{dr}{dt}=i+2j+2tk $
or, $|v(t)|=\sqrt {1^2+(2)^2+(2t)^2}=\sqrt {4t^2+5}$
Next, $a(t)=\dfrac{d \ v(t)}{dt}= \dfrac{4t}{\sqrt {4t^2+5}} $
or, $|a(1)|=\dfrac{4(1)}{\sqrt {4 (1)^2+5}}=\dfrac{4}{3}$
$a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {2^2 -(\dfrac{4}{3})^2}=\sqrt {4 -\dfrac{16}{9}}=\dfrac{2 \sqrt 5}{3}$
So, $a=a_T T+a_{N}=\dfrac{4}{3} \ T + \dfrac{2 \sqrt 5}{3} \ N$
