Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Somebody brakes a car. In the first two seconds, it covers thirty metres. In the next two seconds, only ten metres. How much further does it go? Now, most of you start guessing a pattern, right? Hold on to that thought; we will crack it at the end. Today is the JEE Main part of the kinematics problem bank. I will give you three quick ideas from it, and then two full problems. So, let's begin.

0:31Idea 1: greatest velocity where a=0

Idea number one. Here, a particle starts from rest, and its acceleration a equals six minus two t, where t is the time in seconds. The m s to the power minus two means metres per second squared. Find its greatest velocity, okay? First, the velocity. Acceleration is the rate of change of velocity, so we integrate. Starting from rest, v equals six t minus t squared. Clear? Now think. At which instant is v the greatest? Yes, exactly when a becomes zero! Till then, a is positive, so v keeps growing. After that, a turns negative, and v starts falling. So, six minus two t equals zero: t equals three seconds. Put t equals three. Six into three, minus three squared: eighteen minus nine, so nine. The little max at the foot of v means its maximum value, and m s to the power minus one means metres per second. Just three short lines, and done!

1:35Idea 2: does v=0 mean turning back?

Alright, idea number two, and this one hides a trap. The position is x equals t cubed, minus three t squared, plus three t. The S I in brackets means S I units: metres and seconds, okay? Does the particle ever turn back? And what is its displacement in the first two seconds? First, the velocity. v equals d x by d t, the rate of change of x: three t squared, minus six t, plus three. Take three common, and look, a perfect square: three into t minus one, whole squared. So at t equals one second, v becomes zero. Velocity zero means the particle turns back, right? No, no, careful! This is the JEE trap. To turn back, v must change sign. But three into t minus one, whole squared, is a square, and the sign on screen means, greater than or equal to zero. It is never negative. So the particle only stops for an instant at t equals one second, and it never turns back. Clear? Now the displacement. x of zero, the position at t equals zero, is zero. So the displacement is just x of two: eight minus twelve plus six, two metres. Remember: v equals zero alone is not enough. Check whether v changes sign. Got it?

2:58Idea 3: two bodies in free fall

Now idea number three, a topper's trick. One ball is dropped from twenty metres. At the same instant, another is thrown up at twenty metres per second from the ground below it. When and where do they meet? Most students write two long equations. Toppers first ask one question: what is the acceleration of one ball, as seen from the other? Zero, absolutely! Both fall freely with the same g, the acceleration due to gravity. So relative to each other, there is no acceleration at all. Each ball sees the other coming at a constant velocity. So the gap is twenty metres, and it closes at a steady twenty metres per second: the dropped ball starts from rest, and the other comes up at twenty. Time equals twenty by twenty: one second. Where? In that one second, the dropped ball falls half g t squared, with g equal to ten: half into ten into one squared, five metres. So they meet fifteen metres above the ground. One idea, and no long equations. Neat trick, isn't it?

4:04Example 1: river crossing in least time, and the drift

Now, example one, an easy one. Here, a man who swims at five kilometres per hour crosses a river one kilometre wide, flowing at three kilometres per hour. The k m h to the power minus one means kilometres per hour. He wants the least time. Find the time, and the drift. Pause the video, and try it first. Done? Okay, let's check. Look at the figure. The blue arrows are the river's velocity, v r; the little r is for river. In orange is v b, the swimmer's own velocity in still water; the little b is for boat, here our swimmer. The green arrow is the resultant, his actual path. Now, the river pushes him downstream. So, to cross fastest, he should aim a little upstream, against the current, right? No, no! That is the trap. Heading upstream, at the angle phi, the Greek letter in the right-hand picture, gives the shortest path, not the least time. For the least time, head straight across, like the left-hand picture. Then his whole swimming speed goes into crossing, okay? So t equals d by v: the width d, over his swimming speed v, the v b of the figure. One kilometre at five kilometres per hour is one fifth of an hour, and h means hours. That is twelve minutes; min is short for minutes. But while he crosses, the river carries him along. That is the drift, marked at the top of the figure. Drift equals the river's speed into the time: the cross means times, so three times one fifth: zero point six kilometres. And we are done!

5:43Example 2: braking car, how much further does it go

Now, our puzzle from the start, and it is a little harder. The braking car covers thirty metres in the first two seconds, and ten metres in the next two seconds. How much further does it go? Pause the video, and try it. Hint: take the first four seconds together. Done? Okay, let's check. Let u be the starting speed and a the acceleration. The distance s equals u t, plus half a t squared. Put t equals two: two u plus two a equals thirty. Now the trick. Don't treat the next two seconds separately. Take the first four seconds together: thirty plus ten, forty metres. Put t equals four: four u plus eight a equals forty. With me? Solve the two equations together: u equals twenty metres per second, and a equals minus five metres per second squared. Minus, because the car is braking. Now the real question. It covered ten metres in the last two seconds, so it must roll on a little more, right? Wait, wait! First check the velocity. v equals u plus a t, at t equals four seconds: twenty minus twenty, zero! So it goes no further at all: it has just stopped. Remember our puzzle? Guessing a pattern would have given you a wrong answer. Always find the velocity at the end first. Got it?

7:07Quick revision: max velocity, turning back, free fall

So, quick revision. Greatest velocity: look where a equals zero. And v equals zero alone does not mean turning back: check whether v changes sign. Two bodies in free fall together: relative acceleration zero. And for the least time across a river, head straight across. In braking problems, find v first: the car may have already stopped. Now try the practice problems on this page, and then attempt the whole JEE Main part as a test, with a clock on the table. You can do this, I promise. All the best!