Chapter 6, Problem 6.1P

Use the relative velocity equation and solve graphically or analytically.
a. A ship is steaming due north at 20 knots (nautical miles per hour). A submarine is laying in wait 1/2 mile due west of the ship. The sub tires a torpedo on a course of 85 degrees. The torpedo travels at a constant speed of 30 knots. Will ii strike the ship! If not, by how many nautical miles will it miss?
b. A plane is flying due south at 500 mph at 35.000 ft altitude, straight and level. A second plane is initially 40 miles due east of the first plane, also at 35.000 feet altitude, flying straight and level and traveling at 550 mph. Determine the compass angle at which the second plane would be on a collision course with the first. How long will it take for the second plane to catch the first?

To determine

Whether the torpedo will strike the ship or not and by what distance will it miss the ship.

Answer to Problem 6.1P

The torpedo will miss the ship by 1.91 miles.

Explanation of Solution

Given:

Speed of ship$\left(v_{\text{ship}}\right)=20\text{ knots}$

Speed of torpedo $\left(v_{\text{tp}}\right)=30\text{ knots}$

Course of firing $\left(\theta \right)={85}^{\circ }$

Separation between ship and submarine(R) = 0.5 mile

Concept Used:

$\begin{array}{l}\text{distance(d) = speed(v)×time(t)}\\ \text{tanθ =}\frac{\text{perpendicular}}{\text{base}}\\ \text{cosθ =}\frac{\text{base}}{\text{hypotenuse}}\end{array}$

Calculation:

Drawing diagram,

From the above figure,

  $\begin{array}{l}\cos \theta =\frac{0.5\text{ mile}}{\text{h}}\\ \text{h}=\frac{0.5\text{ mile}}{\cos \theta }\end{array}$

Now put $\theta ={85}^{\circ }$ simplify further

$\begin{array}{c}\text{h}=\frac{0.5\text{ mile}}{\cos {85}^{\circ }}\\ =5.74\text{ miles}\end{array}$

It is also calculated that,

$\begin{array}{c}\tan \theta =\frac{\text{p}}{0.5\text{ mile}}\\ \text{p =}\left(0.5\text{ mile}\right)\left(\tan {85}^{\circ }\right)\\ \text{p}=5.72\text{ miles}\end{array}$

Let, time to cover 5.74 miles by torpedo is, $t_1$ , then,

$\begin{array}{l}{t}_1=\frac{5.74\text{ miles}}{v_{\text{tp}}}\\ \text{substituting values,}\\ t_1=\frac{5.74\text{ miles}}{30\text{ knots}}=0.19\text{ h}\end{array}$

In this duration, the distance travelled by ship is given by,

$\begin{array}{l}\text{distance(d) = }{v}_{\text{ship}}\times t_1\\ \text{substituting values}\\ \text{distance(d) = }\left(20\text{ knots}\right)\left(0.19\text{ h}\right)=3.8\text{ miles}\end{array}$

As the torpedo missed the target. Let it miss the target by a distance s, then,

$\begin{array}{l}s=\text{p - d}\\ \text{substituting values,}\\ s=\left(5.72\text{ miles}\right)\text{ - }\left(3.8\text{ miles}\right)\\ s=1.92\text{ miles}\end{array}$

Therefore, torpedo misses the target by 1.92 miles.

To determine

The compass angle at which the second plane would be on a collision course with the first and the time taken by the second plane to catch the first plane.

Answer to Problem 6.1P

The compass angle at which the second plane would be on a collision course with the first is ${65.5}^{\circ }$ andthe time taken by the second plane to catch the first plane is 0.176 h.

Explanation of Solution

Given:

Speed of first plane, $v_{\text{p1}}=500\text{ mph}$

Speed of second plane, $v_{\text{p2}}=550\text{ mph}$

Distance of separation between the planes, $R=40\text{ miles }\left(\text{due east}\right)$

Concept Used:

$$

$\begin{array}{l}\text{distance(d) = speed(v)×time(t)}\\ \text{tanθ =}\frac{\text{perpendicular}}{\text{base}}\\ \text{cosθ =}\frac{\text{base}}{\text{hypotenuse}}\end{array}$

Calculation:

Drawing diagram,

From the diagram,

$\begin{array}{l}\cos \theta =\frac{40\text{ miles}}{\text{h}}\\ \text{Or, h =}\frac{40\text{ miles}}{\cos \theta }\\ \text{also,}\\ \text{tan}\theta \text{=}\frac{\text{P}}{40\text{ miles}}\\ \text{P}=\left(40\text{ miles}\right)\text{tan}\theta \end{array}$

To collide head on, the two plane must cover the distances P and h in same time interval. Let it takes time t to cover these distances, then

$\begin{array}{l}t=\frac{\text{P}}{v_{\text{p1}}}=\frac{\text{h}}{v_{\text{p2}}}\\ \text{substituting values,}\\ t=\frac{\left(40\text{ miles}\right)\tan \theta }{500\text{ knots}}=\frac{\left(40\text{ miles}\right)}{\cos \theta \left(550\text{ knots}\right)}\\ \tan \theta \cos \theta =\frac{500\text{ knots}}{550\text{ knots}}\end{array}$

Further simplify,

$\begin{array}{l}\sin \theta =0.91\\ \theta ={\sin }^{-1}\left(0.91\right)\\ \theta ={65.5}^{\circ }\end{array}$

Therefore, the compass angle at which the second plane would be on a collision course with the first is ${65.5}^{\circ }$ .

Now, time taken, t is given as,

$\begin{array}{l}t=\frac{\left(40\text{ miles}\right)\tan \theta }{500\text{ knots}}\\ \text{substituting }\theta \text{ = 65}{\text{.5}}^{\circ }\\ t=\frac{\left(40\text{ miles}\right)\tan {65.5}^{\circ }}{500\text{ knots}}\\ t=0.176\text{ h}\end{array}$

Therefore, the time taken by the second plane to catch the first plane is 0.176 h.