Problem $3[40$ pts$]$

Given a rectangular box $A-B-C-D-E-F-G-H$ with the orthogonal triad hat$(x),$hat$(y),$hat$(z)$ defined

as shown in the figure below. Point $Q$ is the center point of the top face $C-D-H-G.$

The lengths of sides $AB,BC,BF$ are $40,20,10,$ respectively.

A vector $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)$ lies along the direction $\frac{?}{\text{b}\text{a}\text{r}}(AQ)$ as shown in the figure. The tail of

the vector $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)$ is fixed at A but the head of the vector is free to move along the

line $AQ.$ The length of the vector $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)$ grows over time according to $|)/(b|=t+1$

for time $t0.$ Once the length of vector $J(t)$ reaches the length of $AQ,$ it stops

growing any further and remains a constant vector. Let $\frac{?}{\text{b}\text{a}\text{r}}(V)$ be a vector of magnitude

$1$ lying in the hat$(x)-$hat$(z)$ plane as shown in the figure. The angle between $\frac{?}{\text{b}\text{a}\text{r}}(V)$ and hat$(z)$ is $30.$

$($i$)$ Find the time, $t^{**},$ at which the vector $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)$ reaches the top face.

$($ii$)$ Find $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)$ for $t{t}^{**}.$

$($iii$)$ Compute $\frac{?}{\text{b}\text{a}\text{r}}(U)(t)\frac{}{b}ar(V)$ for $t{t}^{**}$