Variable Acceleration

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a particle that starts from rest. Its acceleration is the biggest right at the start, and then it keeps decreasing as the particle moves ahead; later it even turns negative. Now tell me: where is the particle moving fastest? Right at the start, where the acceleration is the biggest? Hmm, think again. Today we settle it, with a little calculus. And look, JEE often asks questions where the acceleration keeps changing: with time, with velocity, even with position. So, let's begin.

0:38When the acceleration keeps changing

So, in our puzzle the acceleration changes as the particle moves. Can we still use our old friend from the last chapter, v equals u plus a t, where u is the initial velocity? No, not here! Remember the warning from the last chapter: v equals u plus a t, and all those equations of motion, work only when the acceleration is constant. Here the acceleration changes from moment to moment. So, we go back to calculus. Clear? Velocity is the rate of change of position, so v equals d x by d t. The upright d means a tiny change, so d x by d t is the derivative of x with respect to time, okay? Alright, and acceleration is the rate of change of velocity, so it equals d v by d t. And since v is itself d x by d t, this is also d square x by d t square, the second derivative of x. Got it? But in our puzzle, the acceleration depends on the position x, not on time. So d v by d t is not much help, right? We want a form with d x in it. Any idea how to get one? Yes, the chain rule! d v by d t equals d v by d x, into d x by d t; the dot here means multiply. And d x by d t is just v, isn't it? So the acceleration equals v d v by d x. Whenever the acceleration depends on position, this is your form. With me?

2:03First ask: a in terms of what?

So, first ask one thing: the acceleration is given in terms of what? Case one: acceleration as a function of time. We write it as a of t; the bracket just says it depends on t, okay? Then d v equals a d t: integrate it to get v. Then d x equals v d t: integrate again to get x. Two integrations, that's it. Case two is a little tricky. Here the acceleration depends on v itself, written a of v. Can we simply integrate a d t, like before? Not directly, because we do not yet know v as a function of t, right? So we separate the variables: bring everything with v to one side. d v over a of v equals d t: this links v with time. Or, v d v over a of v equals d x: this links v with position. Clear? And case three, our puzzle: the acceleration depends on x, written a of x. Use the form we just built: v d v equals a of x, d x. Integrate both sides, and you get v squared in terms of x. Very direct.

3:15Displacement or distance?

One last rule before the examples. To find the displacement, integrate v with respect to time. The tall S shape is the integral sign, okay? Now the distance. Same integral, right? Just integrate v over the interval? No, no, careful! This is a classic JEE trap. If the particle turns back, v becomes negative, and that part gets subtracted in the integral. Distance needs the integral of mod v, d t. The two straight bars around v are modulus bars: mod v is the speed, never negative. So, in practice: find the times when v is zero, and split the interval there. Integrate each part separately, and add them all as positive numbers. Got it?

4:01Example 1: a = 6t, find v and x at t = 2 s

Let's start with an easy one. The acceleration is six t metres per second squared; the m s to the power minus two on screen means metres per second squared, okay? At t equals zero, v is two metres per second, that is the m s to the power minus one, and x is zero. Find v and x at t equals two seconds; the s means seconds. Pause the video, and try it first. Done? Okay, let's check. At t equals two, the acceleration is twelve. So v equals u plus a t: two plus twelve into two, twenty six metres per second, right? No, no! That is the trap: the acceleration here is not constant. This is case one: d v equals six t d t. The double arrow means so, okay? Integrate: three t squared, plus a constant. At t equals zero, v is two, so the constant is two. So v equals two plus three t squared. Now integrate once more. d x equals two plus three t squared, d t. So x equals two t plus t cubed, with no constant, because x is zero at t equals zero. Clear? Now put t equals two. v equals two plus three into two squared: two plus twelve, fourteen metres per second. See, that is much less than the trap answer, isn't it? And x equals two into two, plus two cubed: four plus eight, twelve metres. That's it, done!

5:32Example 2: a = 4 - 2x, maximum speed and where it stops

Now a harder one, JEE level, and it is our puzzle with numbers. Here, a particle starts from rest at the origin, with acceleration four minus two x, in metres per second squared. Find its maximum speed, and where it stops again. Pause and try. Hint: the acceleration depends on x, the position. Done? Okay, let's check. Like the hint said, the acceleration depends on x. So which form of the acceleration do we use? Yes, v d v by d x! Exactly case three. So, v d v equals four minus two x, d x. Integrate both sides. At the origin it is at rest, so there is no constant. Half v squared equals four x minus x squared. Multiply by two: v squared equals eight x minus two x squared. Got it? Now the maximum speed, our puzzle! At x equals zero, the acceleration is the biggest it ever gets. So the particle is fastest there, right? No, no, careful! At x equals zero, it is at rest! See, while the acceleration is positive, the particle keeps speeding up. Once the acceleration turns negative, it slows down. So it is fastest exactly where the acceleration is zero: four minus two x equals zero, so x equals two metres. Put x equals two into v squared. Eight into two is sixteen, minus two into two squared, which is eight. So v squared is eight. Then v max equals root eight, that is two root two metres per second. The small max below v means maximum, and the tick-shaped sign is the square root, okay? Last part: where does it stop again? What do you put equal to zero? Yes, v equals zero! So eight x minus two x squared equals zero. x equals zero is just the start, so it stops again at x equals four metres. And our puzzle is solved: the particle is fastest not where the acceleration is biggest, but where it becomes zero. Do remember this one!

7:45Quick revision: a in terms of t, v or x, and distance

So, quick revision. First ask: is the acceleration given in terms of t, v or x? If it depends on x, use v d v equals a d x. And for distance, split the interval at the zeros of v. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!