# Will Two Ships Collide Based on Different Methods?

• brotherbobby
In summary: Method 1 : Conclusion - Ships will not collide .Imagine the ships collide after a time ##t##.Since they both start at the same time, the distances traveled by them would be ##3t## (A) and ##t## (B), respectively. (I have shown the two distances in the diagram, in terms of ##t##).In the right angled ##\triangle ABP##, using the Pythagorean theorem, we have ##(3t)^2 = t^2 + 4^2 \Rightarrow 9t^2 = t^2 + 16 \Rightarrow 8t^2 = 16 \Rightarrow t^2 = 2 \Rightarrow t =
brotherbobby
Homework Statement
Two ships A and B begin at the same time from their respective points 4 kms apart along the ##y## axis with speeds 3 km/h and 1 km/h (uniformly) respectively, as shown in the figure below. The motion of ship B is parallel to the ##x## axis while that of A is at an angle of ##60^{\circ}## to the ##x## axis. ##\textbf{Will the two ships collide at point P}?##
Relevant Equations
For unaccelerated motion, distance travelled ##d = v_0 t##, where ##v_0## is the uniform speed. Also, in a right angled triangle, ##\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}##.

There are two ways to solve. My problem is, when I apply the two, I end up with a contradiction.

Method 1 : Conclusion - Ships will not collide .

Imagine the ships collide after a time ##t##.

Since they both start at the same time, the distances traveled by them would be ##3t## (A) and ##t## (B), respectively. (I have shown the two distances in the diagram, in terms of ##t##).

Using parallel-line and angles, we see that ##\angle BPA = 60^{\circ}##. Thus if we take the cosine of the angle using trigonemetrical ratios of a right angled triangle, we have ##\cos 60^{\circ} = \frac{t}{3t} = \frac{1}{3}##!

We know that ##\cos 60^{\circ} = \frac{1}{2}##, hence it's clear that two two ships will not collide .Method 2 : Conclusion - Ships will collide.

Imagine the ships collide after a time ##t##.

Since they both start at the same time, the distances traveled by them would be ##3t## (A) and ##t## (B), respectively. (I have shown the two distances in the diagram, in terms of ##t##).

In the right angled ##\triangle ABP##, using the Pythagorean theorem, we have ##(3t)^2 = t^2 + 4^2 \Rightarrow 9t^2 = t^2 + 16 \Rightarrow 8t^2 = 16 \Rightarrow t^2 = 2 \Rightarrow t = \sqrt{2}\; \text{hr} \approx 85 \text{minutes}## .

Hence the two ships will collide after a time of 85 minutes from start.

One possible way out is to note that nowhere in my second solution did I consider the fact that Ship A was moving at an angle of 60 degrees to the horizontal. What I did show was that with the speeds as given, the two ships can collide after the given time (85 minutes). If that was to happen, then the angle at which Ship A should move would be ##\cos^{-1} \frac{t}{3t} = \cos^{-1} \frac{1}{3} = 70.5^{\circ}##. It becomes a different problem really : Given the speeds of 1 and 3 km/h, calculate the angle at which ship A should move in order that they collide.

Am I right? When I drew the relative velocity of B with respect to A (##\vec v_{BA}##), I found that it was pointing away from A, clearly showing that ship B would not be moving in A's direction and that no collision would take place.

brotherbobby said:
Am I right?
Yes
brotherbobby said:
I found that it was pointing away from A
You mean: it was not pointing directly at A:
if it were pointing away from A, the ships would not come closer than the original 4 km, but they do!

BvU said:
You mean: it was not pointing directly at A:
if it were pointing away from A, the ships would not come closer than the original 4 km, but they do!

Yes, I paste a small diagram of how the relative velocity of B with respect to A (##\vec {v}_{BA}##) looks like :

As you can see above, ##\vec {v}_{BA}## is given by the red arrow. Yes, of course I meant the relative velocity of B w.r.t. A is not pointing directly at A. I suppose that is enough to imply that the ships won't collide.

AP is the distance of shortest approach. I suppose it will be instructive of me to calculate how much AP is?

brotherbobby said:
I suppose it will be instructive of me to calculate how much AP is?
Yes: in practice (boats, planes) a safety margin is to be observed. Do you know how to find this distance ? I get 0.76 km

brotherbobby said:
Method 1 : Conclusion - Ships will not collide .

Imagine the ships collide after a time ##t##.Method 2 : Conclusion - Ships will collide.

Imagine the ships collide after a time ##t##.

I think it would be much simpler to
1. determine the coordinates (x,y) of point P (this is simple trig). This allows you to figure the distance AP and the distance BP.
2. determine how long Ship A takes to reach P
3. determine how long Ship B takes to reach P

If the times are the same, then they arrive together (ie, collide). If the times are different, one ship gets there first (ie, they do not collide).

Just a minor point that might prevent confusion and frustration in the future. In the drawing in post #3 all arrows representing vectors should be labeled with their magnitudes which, in this case, are speeds. There should be no -3 km/h label. The convention is that a vector drawn graphically as an arrow needs a label to convey the length of its shaft (magnitude) and an arrowhead to convey its direction. Negative signs come in when you transit from the graphic to the algebraic representation. In this case you would write,
##\vec v_1=3~\mathrm{km/h}~\cos(60^o)~\hat i+3~\mathrm{km/h}~\sin(60^o)~\hat j##
and
##\vec v_2=3~\mathrm{km/h}~\cos(240^o)~\hat i+3~\mathrm{km/h}~\sin(240^o)~\hat j##

It is clear that, when you do it formally and measure angles counterclockwise from the positive ##x##-axis, the negative sign is associated with the component(s) and not the magnitude of a vector. The arrowheads are there to help you figure out the correct angle. Had you used vector symbols instead of scalar numbers it would have been OK to label one of the arrows with ##\vec v_1## and the other with ##-\vec v_1##. Then it's clear that you are labeling the whole arrow and not just its shaft.

Last edited:
BvU
BvU said:
Yes: in practice (boats, planes) a safety margin is to be observed. Do you know how to find this distance ? I get 0.76 km

Yes. I have found two ways to find the distance of closest approach and the answer matches yours. Nonetheless, please see my working below to let me know if they are ok.

[Find the distance of closest approach for the two ships in the problem above and the time at which they are at their closest]

First Method : (Mathematical)

(Please note that all time given above are in hours and all distances in kilometers. Speeds are in km/h).

After a time period of ##t##, the coordinates of ship B are ##(t,4)## at the point B' and those of ship A are ##(\frac{3}{2}t, \frac{3\sqrt{3}}{2}t)## where the distance traveled by A is ##AA' = 3t## and elementary trigonometry is used to find its coordinates at A'.

The distance between the ships ##A'B' = s(t)## where ##s^2 = \frac{t^2}{4}+\left(\frac{3\sqrt{3}}{2}t - 4\right)^2##.

Differentiating s with respect to t and putting ##\frac{ds}{dt} = 0## for conditions of extrema, we are left with,
##0 = \frac{t}{2}+\frac{27}{2}t-12\sqrt{3} \Rightarrow 14t = 12\sqrt{3} \Rightarrow t = \frac{6}{7}\sqrt{3}\Rightarrow \boxed{t = 1.48\; \text{hr}}##.

The closest distance ##s_{\text{min}} = \left(\frac{1.48^2}{4} + \left(\frac{3\sqrt{3}}{2}\times 1.48-4\right)^2 \right)^{\frac{1}{2}}\Rightarrow \boxed{s_{\text{min}}= 0.76\; \text{km}}##.

(The necessity to prove that the distance so found is a minima is perhaps superfluous. Clearly the distance cannot be a maxima because the two ships can continue moving further and the maximum distance of their separation is ##\infty##. However, if the reader does want to really show that the distance is a minima, the above expression for ##s## needs to be differentiated again to find ##\frac{d^2s}{dt^2} = \frac{7}{s_{\text{min}}} >0## (minima)

Second Method : (Physical)

First let's find the velocity of B relative to A ##\vec v_{BA}##. In the right angled triangle PQR in figure (a) above, we know from elementary trigonometry that ##v_{BA} = (v_A^2 + v_B^2 - 2 v_A v_B \cos \theta)^{\frac{1}{2}} = (9+1-2\times 3\times 1\times \frac{1}{2})^{\frac{1}{2}} = 7^{\frac{1}{2}} = 2.645\; \text{km/h}##.
Using the "sine rule" to triangle PQR and noting the angle ##\phi##, we have ##\frac{3}{\sin(90+\phi)} = \frac{2.465}{\sin 60}\Rightarrow \phi = 10.81^{\circ}##.

Moving over to diagram on side(b), the distance of closest approach must be the perpendicular from A = AP. Using trigonometry, ##AP = 4 \sin\phi = 4\times \sin 10.81 \Rightarrow \boxed{AP = 0.75\; \text{km}}##.

Using trigonometry again, the distance traveled by B in A's frame (Very important = This is not the same as the distance traveled by B in the Earth's frame at closest approach. Please note: the distance between non-simultaneous events is not an invariant between frames in uniform motion) : ##BP = 4 \cos 10.81 = 3.93\; \text{km}##.

Hence the time for closest approach : ##t = \frac{BP}{v_{BA}} = \frac{3.93}{2.645} \Rightarrow \boxed{t = 1.48\; \text{hr}}##.

## 1. Will the two ships collide?

Without proper navigation and communication between the ships, there is a possibility that they may collide. However, with advanced technology and skilled navigation, the chances of collision can be significantly reduced.

## 2. What factors determine whether the ships will collide?

The speed, direction, and location of each ship are crucial factors in determining whether they will collide. Other factors such as weather conditions, visibility, and the presence of other objects in the water also play a role.

## 3. How can the collision of two ships be prevented?

To prevent a collision, both ships must follow the rules and regulations set by the International Maritime Organization (IMO). This includes maintaining a safe speed and distance, proper communication and signaling, and using advanced navigation equipment.

## 4. What are the potential consequences of a ship collision?

The consequences of a ship collision can be severe, ranging from damage to the ships and cargo to environmental pollution and loss of life. It can also lead to legal and financial repercussions for the involved parties.

## 5. How often do ship collisions occur?

The frequency of ship collisions varies depending on various factors such as the number of ships in a particular area, weather conditions, and human error. However, with strict regulations and advanced technology, the number of ship collisions has significantly reduced in recent years.

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