Mixed Problems: Set 1 (Straight Line and Graphs)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here is a car. It covers the first half of its journey at forty kilometres per hour. You want its average speed for the whole journey to be forty eight kilometres per hour. So, how fast must it go in the second half? Fifty six, right? Forty and fifty six average to forty eight. Hmm, are you sure? Today we solve a mixed set on straight line motion and graphs. And remember, kinematics questions come up often in JEE Main, and each one should take you only a minute or two. So, let's begin.

0:39Tool 1: distance in the nth second

So, tool one: the distance in the nth second. s sub n equals u plus a by two, into two n minus one, okay? Here u is the starting velocity, a is the acceleration, and s sub n, with the small n below the s, is the displacement in the nth second alone, not in all n seconds together. If the body does not turn back during that second, this is also the distance. Clear? Try this one from our set. Starting from rest, a particle covers twenty seven metres in the fifth second. Find its acceleration. Right, from rest means u is zero, and here n is five. Two n minus one is nine. So, a by two, times nine, equals twenty seven. That gives a by two equal to three, and the acceleration is six metres per second squared, written m s to the power minus two; the double arrow means so. Simple, isn't it?

1:36Tool 2: stopping distance

Alright, tool two. Here, a car can stop in eight metres from forty kilometres per hour, written k m h to the power minus one. With the same braking, what is its stopping distance from eighty? Double the speed, so double the distance, sixteen metres, right? No, no, careful! This is the JEE trap. The stopping distance s equals u squared by two a, where u is the starting speed and a is the braking, the retardation. This curly sign means proportional to. So s is proportional to u squared, not just to u. Got it? So, speed times two means distance times four. Four times eight: thirty two metres. Not sixteen, thirty two!

2:21Tool 3: area under the v–t graph

Now, tool three: graphs. On a v t graph, velocity against time, the slope is the acceleration, and the area under the graph is the displacement. Remember this one; it saves a lot of time. Try this one. Here, a body speeds up uniformly from rest to thirty metres per second, written m s to the power minus one, in ten seconds; the arrow means goes to, and the s after each time means seconds. Then it slows down uniformly to rest in five seconds. We want the distance, and the average speed. First tell me, what shape is its v t graph? Yes, a triangle! Straight up for ten seconds, then straight down for five. Base fifteen seconds, height thirty. Half, times fifteen, times thirty: two hundred and twenty five metres. It never turns back, so this is also the distance. No equations of motion needed! And average speed is total distance over total time: two twenty five by fifteen, that is fifteen metres per second. Nice and simple.

3:27Topper's trick: speed up, then brake

Now, one topper's trick from the same kind of triangle. From rest, a car speeds up at alpha, then brakes at beta, taking a total time t to stop. Alpha and beta are just Greek letters for the two rates, okay? Then its top speed, v max, with max written small below the v, is alpha beta t, over alpha plus beta. One line, done. From our set: the car accelerates at two, and decelerates at four metres per second squared, taking fifteen seconds in all. What is its greatest speed? Okay, just put the numbers in. Two times four times fifteen, over two plus four: twenty metres per second. The dots here just mean times. And the distance? The v t graph is a triangle again, so half, times twenty, times fifteen: one hundred and fifty metres. With me?

4:18Example 1: second-half speed to average 48 km/h

Now, remember our car from the start? Let's finish it. It covers the first half of the distance at forty kilometres per hour. At what speed must it cover the second half, to average forty eight? Pause the video, and try it first. Done? Okay, let's check. Forty and fifty six average to forty eight, so the answer is fifty six, right? No, no, careful! Average speed is total distance over total time. It is not the plain mean of the speeds, unless the times are equal. Here the distances are equal. For equal distances, v bar equals two v one v two, over v one plus v two. The bar on top means average, and the small one and two below are the two speeds. This is the harmonic mean. So, put v one as forty, and v two as v, the speed we want. Two times forty times v, over forty plus v, equals forty eight. Cross multiply. Eighty v equals nineteen twenty, plus forty eight v. So thirty two v equals nineteen twenty. Divide, and v is sixty kilometres per hour. Not fifty six: sixty! Puzzle solved, and in under a minute.

5:33Example 2: distance vs displacement for x = 4t - t²

Now a harder one, JEE level. The position is x equals four t minus t squared, in S I units, so metres and seconds. Find the distance covered in the first three seconds. Pause and try. Hint: does it ever turn back? Done? Okay, let's check. At t equals three, x of three is twelve minus nine, that is three metres. So the distance is three metres, right? No, no, careful! Three metres is only the displacement, from start to finish. For the distance, first check: does it turn back? Velocity v is d x by d t, the rate of change of x, so v equals four minus two t. And v is zero at t equals two seconds. There it turns back. From zero to two seconds, x goes from zero to four: four metres forward. Then from two to three seconds, x comes back from four to three: one metre backward. So the distance is four plus one, five metres, while the displacement is just three. Remember, split the motion where v is zero, and add the pieces. That's the whole trick!

6:44Quick revision: harmonic mean, stopping distance, v-t area

So, quick revision. One: for equal distances, v bar is two v one v two over v one plus v two, not the plain mean. Two: stopping distance is proportional to u squared, and the area under a v t graph is the displacement. Three: for distance, split the motion where v is zero, and add the pieces. Now go and try the three practice problems on this page. Give each one a minute or two, like the real exam. All the best!