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enter image description here

The first part i) I can solve correctly, but I need some advice and intuition on how to solve the second part ii).

Here is the mark-scheme for the question: enter image description here

But for part ii) I do not understand their logic as shown in red i need to know why this must be the case for closest approach.

Could someone please talk me through this step by step in simple English as I have no idea the mark-scheme means?

Thank you,

with kind regards.

BLAZE
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2 Answers2

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At any time, you can decompose your velocity vector into two components, one toward/away from $S$, and one perpendicular to the direction toward $S$. If the former component is nonzero, then you are not at closest approach: If the component is positive, your distance from the object is decreasing; if it is negative, your distance from the object is increasing. Either way, you're not at closest approach, which therefore requires that your distance toward the object is instantaneously zero.

Travis Willse
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  • Hi there, thank you for your answer. You said that "At any time, you can decompose your velocity vector". When you say 'your velocity vector' do you mean the velocity vector of the patrol boat $B$? You also said "into two components, one toward/away from $S$, and one perpendicular to the direction toward $S$". I'm still a little confused I need to know why the velocity of patrol boat $B$ relative to ship $S$ MUST be at a right angle to the velocity of patrol boat $B$. I didn't or failed to see and explanation of this in your answer. Do you know why the text boxed in red must be true? – BLAZE Aug 28 '14 at 18:21
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From a physics perspective -- (here we go again, answering physics questions on a math site! ;) ) -- it may make more sense if you go into the inertial reference frame where the ship being approached is stationary.

This involves subtracting the (vector) velocity of the ship being approached from the velocities of both ships. Or put another way: pretend you're in a blimp that's directly above that ship at all times. You look down, you see that ship, and it's always in the same position.

Now that one ship is stationary, its velocity is zero, so it stays at the same place all the time. It's at a point. (The red point.)

The other ship moves along some straight line path (the blue path) past that point. (If the ship was moving at constant velocity in one inertial reference frame, it's moving at constant velocity in any other inertial reference frame.)

The minimum distance from the point to the line is a perpendicular. (The dashed line is my poor attempt at dropping the perpendicular from the point to the line.)

enter image description here

John
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  • I am unable to distinguish between maths and physics, unless you care to define them both accurately. For this reason I always ask my questions here as the answers given are generally better:). In any case "This involves subtracting the (vector) velocity of the ship being approached from the velocities of both ships." I do not understand this part, could you please clarify? When i first asked this question i thought it would be quite straightforward to get a simple answer. Thanks for your reply. – BLAZE Aug 28 '14 at 21:03
  • (A little secret: I have a degree in physics.) I edited my question to try to explain it differently. – John Aug 29 '14 at 05:23
  • Yes forgive me John since this question is actually from a maths exam and since i'm studying theoretical physics i sometimes find it hard to distinguish between physics and maths. If the blue arrow represents the velocity vector of the ship $S$, I assume the dotted line represents the velocity of the patrol boat $B$ relative to the ship $S$ as these are perpendicular. Is this correct? – BLAZE Aug 29 '14 at 15:53
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    My apologies for not labeling the diagram per your question. In the reference frame we're in, the velocity of the ship $S$ is zero. So, for now, label the red dot with an $S$. The velocity of the patrol boat $B$ is constant, so its path is a straight line. That's the solid blue arrow that should be labeled with $B$. I moved into the reference frame where $S$ is stationary because, in my mind, it's easier to explain that the shortest distance from a point to a line is the perpendicular from the point to the line. – John Aug 29 '14 at 16:12
  • Okay John. But I still have to know what the dashed blue line represents? I know you said this is the distance of closest approach and I can see from your explanation why this must be the case. But the question i'm asking John is marked in red above. The two lines in your diagram are perpendicular, the solid blue line represents the velocity vector of the patrol boat $B$ as you rightly said. But this must mean that the dashed blue line is a relative velocity, (as well as being the shortest distance). Can you explain if the dashed line represents the velocity of $B$ relative to $S$? – BLAZE Aug 29 '14 at 19:02