Ship A leaves port sailing north at a speed of 30 mph. A half hour later, Ship B leaves the same port sailing east at a speed of 35 mph. Let t (in hours)

Distance A has travelled north at time t is  y = 30t + (1/2)*30 miles or y = 30t + 15 miles

Distance B has travelled east at time t is x = 35t

Use Pythagoras to get distance between them:  

$D=\sqrt{x^2+y^2}=\sqrt{(35t)^2+(30t+15)^2}=\sqrt{2125t^2+900t+225} \text{ miles}$

When t=3,  $D=\sqrt{2125*9+900*3+225}=148.49 \text{ miles}$

Oops! This is the distance between them 3 hours after B leaves port.  See heureka's reply for the distance 3 hours after A leaves port.

Ship A leaves port sailing north at a speed of 30 mph. A half hour later, Ship B leaves the same port sailing east at a speed of 35 mph. Let t (in hours) denote the time ship B has been at sea.

(a) Find an expression in terms of t giving the distance D between the two ships. 

(b) Use the expression obtained in part (a) to find the distance between the two ships 3 hr after Ship A has left port. (Round your answer two decimal places.)

(a)

$\small{ \begin{array}{lrcl} & \vec{a} &=& \begin{pmatrix} 0 \\ 30\cdot (t+\frac12) \end{pmatrix} \\ & \vec{b} &=& \begin{pmatrix} 35\cdot t \\ 0 \end{pmatrix} \\ \\ \hline \\ \text{distance: } & \vec{d} &=& | \vec{a}-\vec{b} | \\ & \vec{d} &=& | \begin{pmatrix} 0 \\ 30\cdot (t+\frac12) \end{pmatrix} -\begin{pmatrix} 35\cdot t \\ 0 \end{pmatrix} | \\ & \vec{d} &=& | \begin{pmatrix} 0 -35\cdot t \\ 30\cdot (t+\frac12) -0 \end{pmatrix} | \\ & \vec{d} &=& | \begin{pmatrix} -35\cdot t \\ 30\cdot (t+\frac12) \end{pmatrix} | \\ & d &=& \sqrt{ 35^2\cdot t^2 + (30\cdot t + 15)^2 } \\ & d &=& \sqrt{ 35^2 \cdot t^2 + 30^2\cdot t^2 + 2\cdot 30 \cdot 15 \cdot t + 15^2 } \\ & d &=& \sqrt{ 15^2 + 30^2\cdot t + (30^2 + 35^2) \cdot t^2 } \\ \boxed{~ d = \sqrt{ 15^2 + 30^2\cdot t + (30^2 + 35^2) \cdot t^2 } ~} & \qquad t = 0 \rightarrow d = 15 ~\text{miles}\\ \end{array} }$

(b)

$\small{ \begin{array}{rcl} t+\frac12 &=& 3~\text{hours} \\ t &=& 3-\frac12\\ t &=& 2.5 ~\text{hours} \\ \\ \hline \\ d(2.5~\text{hours} ) &=& \sqrt{ 15^2 + 30^2\cdot 2.5 + (30^2 + 35^2) \cdot 2.5^2 } \\ d(2.5~\text{hours} ) &=& \sqrt{ 15^2 + 30^2\cdot 2.5 + 2125 \cdot 2.5^2 } \\ d(2.5~\text{hours} ) &=& \sqrt{ 15^2 + 30^2\cdot 2.5 + 2125 \cdot 6.25 } \\ d(2.5~\text{hours} ) &=& \sqrt{ 225 + 2250 + 13281.25 } \\ d(2.5~\text{hours} ) &=& \sqrt{ 15756.25 } \\ d(2.5~\text{hours} ) &=& 125.523902106 \\ \end{array} }$

The distance between the two ships 3 hr after Ship A has left port is 125.52 miles