Solución

Apartado a

Comencemos con la ecuación 3.23:

v(t)=r′(t)=2ti+2j+(6t−3)ka(t)=v′(t)=2i+6kaT=v⋅a∥v∥=(2ti+2j+(6t−3)k)⋅(2i+6k)∥2ti+2j+(6t−3)k∥=4t+6(6t−3)(2t)2+22+(6t−3)2=40t−1840t2−36t+13\begin{aligned} \bold{v}(t) &= \bold{r^{\prime}}(t) = 2t\bold{i} +2\bold{j} + (6t-3)\bold{k}\\ \bold{a}(t) &= \bold{v^{\prime}}(t)= 2\bold{i} + 6\bold{k}\\ a_{\bold{T}} &= \frac{\bold{v}\cdot\bold{a}}{\|\bold{v}\|}\\ &= \frac{(2t\bold{i}+2\bold{j}+(6t−3)\bold{k})\cdot(2\bold{i}+6\bold{k})}{\|2t\bold{i}+2\bold{j}+(6t−3)\bold{k}\|}\\ &= \frac{4t+6(6t-3)}{\sqrt{(2t)^2+2^2+(6t-3)^2}}\\ &= \frac{40t-18}{\sqrt{40t^2-36t+13}} \end{aligned}

Luego aplicamos la ecuación 3.14:

aN=∥a∥2−aT=∥2i+6k∥2−(40t−1840t2−36t+13)2=4+36−(40t−18)240t2−36t+13=40(40t2−36t+13)−(1600t2−1440t+324)40t2−36+13=19640t2−36t+13=1440t2−36+13\begin{aligned} a_{\bold{N}} &= \sqrt{\|\bold{a}\|^2-a_{\bold{T}}}\\ &= \sqrt{\|2\bold{i}+6\bold{k}\|^2-\Bigg(\frac{40t-18}{\sqrt{40t^2-36t+13}} \Bigg)^2}\\ &= \sqrt{4+36-\frac{(40t-18)^2}{40t^2-36t+13}}\\ &= \sqrt{\frac{40(40t^2−36t+13)−(1600t^2−1440t+324)}{40t^2-36+13}}\\ &= \sqrt{\frac{196}{40t^2-36t+13}}\\ &= \frac{14}{\sqrt{40t^2-36+13}} \end{aligned}

Apartado b

Debemos evaluar cada una de las respuestas de la parte a. en t = 2:

aT=40(2)−1840(2)2−36(2)+13=80−18160−72+13=62101aN=1440(2)2−36(2)+13=14160−72+13=14101\begin{aligned} a_{\bold{T}} &= \frac{40(2)-18}{\sqrt{40(2)^2-36(2)+13}}\\ &= \frac{80-18}{\sqrt{160-72+13}} = \frac{62}{\sqrt{101}}\\ a_{\bold{N}} &= \frac{14}{\sqrt{40(2)^2-36(2)+13}}\\ &= \frac{14}{\sqrt{160-72+13}} = \frac{14}{\sqrt{101}} \end{aligned}

Las unidades de aceleración son pies por segundo al cuadrado, al igual que las unidades de los componentes normal y tangencial de la aceleración.