Crossing a River

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You are in a boat, and you have to cross a river. The water keeps pushing your boat downstream. You want to reach the other bank in the least possible time. So tell me: should you point the boat a little upstream, to fight the current, or straight across? Most students say, fight the current, obviously! Today we will find out who is right. And look, boat and river questions come up in JEE quite often. So, let's begin.

0:33Shortest time: head straight across

So, look at the figure. The grey lines are the banks. The blue arrows show the water flowing to the right, labelled v r, the small r for river. The orange arrow, v b, the small b for boat, is the boat's own velocity, relative to the water. And the green arrow, the resultant, is how the boat really moves: the boat's velocity plus the river's velocity, just like the plane and the wind in the last lesson. Clear? Now the letters we will use. The river has width d, and it flows at speed u: that is the size of the blue arrows, v r, okay? And the boat moves at speed v relative to the water, the orange arrow. Relative to the water simply means its speed in still water. So, u for the river, v for the boat. Don't mix them up! Now back to our puzzle. You want to cross in the least time. Where should the orange arrow point: a little upstream, or straight across? Straight across, yes! Like the left half of the figure. See, the current only pushes you along the river; it does not carry you across at all. So give all of v to crossing. The time is t equals d by v; the line means divided by. Pointing upstream would only waste a part of v on fighting the current. Got it? But there is a price. For that whole time t, the river carries the boat downstream at speed u. So it lands downstream of the point opposite, along the green arrow: this distance is the drift, written in grey at the top. Drift equals u times t, that is, u d by v. Okay?

2:09Shortest path: land exactly opposite

Alright, now a different wish. Forget the time. You want to land exactly opposite, with zero drift. Now where should the boat point? Upstream, yes! Against the current, like the orange arrow on the right. The angle there, the Greek letter phi, is measured from the normal: the line straight across the river, along the green arrow. The upstream part of the boat's velocity must exactly cancel the current, and that gives sine phi equals u by v. With me? And the time? The orange arrow v is the longest side of a right triangle: its upstream part is u, and its across part is the green arrow. So by Pythagoras, the across speed is root of v squared minus u squared; the tick-shaped sign is the square root. So t equals d by root of v squared minus u squared, and no drift at all. Now a classic JEE trap. This path is the shortest possible, just the width d. So it must also be the quickest crossing, right? No, no, careful! Root of v squared minus u squared is less than v; the small angle sign means less than. So this time is longer than d by v. Shortest path and shortest time need two different headings. Remember that, okay?

3:25Boat slower than the river

Now, one more case. What if the boat is slower than the river? Can it still land exactly opposite? No, it cannot. Sine phi equals u by v would be more than one, and no angle has a sine more than one. So zero drift is possible only if v is greater than u; this sign means greater than. Keep that in mind. So when v is less than u, the river wins, and you will drift. But you can keep it as small as possible. The least drift is d root of u squared minus v squared, by v. See, now u squared comes first, because u is bigger. And for that, head at cos inverse v by u to the upstream direction; the small minus one means inverse. Careful: this angle is measured from the upstream direction along the bank, not from the normal like phi, okay?

4:15Along the river: downstream and upstream

Last idea: the boat moves along the river, not across it. Going downstream, with the current, what is its speed relative to the bank? Yes, v plus u! The river helps you, so the speeds add. And upstream, against the current? The river fights you, so the speed is v minus u. Simple, right? But these two lines solve many JEE questions. Let's see one.

4:38Example 1: boat speed in still water and river speed

Okay, Example one. Our boat goes twelve kilometres downstream in one hour, and returns in three hours; km is kilometres, and h is hours. Find the boat's speed in still water, and the speed of the current. Pause the video, and try it first. Done? Okay, let's check. Twenty four kilometres in total, in four hours in total. So its speed in still water is twenty four by four, six kilometres per hour, right? The km h to the power minus one on screen just means kilometres per hour. So, six? No, no, careful! That six mixes up the two trips. Take each trip on its own. Downstream, the speeds add: v plus u equals twelve by one, twelve kilometres per hour. Upstream, they subtract: v minus u equals twelve by three, four kilometres per hour. Now add the two equations. Two v equals sixteen, so v is eight kilometres per hour. And subtract them: two u equals eight, so u is four kilometres per hour. That's it! Two lines, two unknowns, done.

5:50Example 2: least time, drift and straight-across time

Clear? Now a harder one, JEE level. The river is four hundred metres wide, and flows at three metres per second; m s to the power minus one means metres per second. The boat moves at five metres per second in still water. Find the least crossing time, and the drift then. And the time for crossing straight across, to the point exactly opposite. Pause and try. Hint: two different headings! Done? Okay, let's check. For the least time, point straight across, like in our puzzle. t equals four hundred by five, eighty seconds. Meanwhile, the river carries it downstream. Drift equals u times t: three times eighty, two hundred and forty metres; the cross means times. Now, to land exactly opposite, head upstream with sine phi equals three by five. So phi is about thirty seven degrees from the normal; the wavy equals sign means approximately, and the little circle means degrees. Now the time. The boat still moves at five metres per second, doesn't it? So the time is again four hundred by five, eighty seconds, right? No, no! Part of the five now goes into fighting the current. The across speed is root of five squared minus three squared, that is root sixteen, four metres per second. So t equals four hundred by four, one hundred seconds, longer than eighty seconds. And there you go, our puzzle, settled with real numbers! Fighting the current lands you exactly opposite, but straight across gets you to the other bank first.

7:29Quick revision: least time, zero drift, up and downstream

So, quick revision. For the least time, head straight across: t equals d by v, and the drift is u d by v. For zero drift, only if v is greater than u, head upstream with sine phi equals u by v, and t equals d by root of v squared minus u squared. And along the river, v plus u downstream, v minus u upstream. Now go and try the practice problems on this page. You will find them easy. All the best!