Vectors in the Description of Motion

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a particle moving along the x axis at three metres per second. Two seconds later, it is moving along the y axis at four metres per second. Its speed went up by just one metre per second, in two seconds. So the average acceleration is half a metre per second squared, right? Hmm, no! The correct answer is two point five. Five times bigger! Where does the extra come from? Today we find out, by describing motion with vectors. And this idea comes up often in JEE; even projectile motion is built on it. So, let's begin.

0:44Position as a vector: r(t)

So, where is the particle at any moment? We describe it with one arrow from the origin, called the position vector. On screen, r with a small arrow on top: we read it as r vector. The t in brackets means it changes with time, so we say r of t, okay? And i cap, j cap and k cap, the letters with little caps, are the unit vectors along x, y and z. Right? Their coefficients, x of t, y of t and z of t, are simply the coordinates of the particle at time t. As the particle moves, these three numbers keep changing. Clear?

1:20Velocity and acceleration: differentiate

Now, velocity. Velocity tells us how fast the position is changing. So tell me, if you know r of t, how do you get the velocity? Yes, exactly! Differentiate it with respect to time. v vector equals d r vector by d t, the rate of change of the position vector, okay? And in components, it is really simple. See the small dot over x? That dot means the derivative with respect to time, and we read it as x dot. So, v vector equals x dot i cap, plus y dot j cap, plus z dot k cap. Just differentiate each component separately. With me? Then, one more step. Acceleration is the rate of change of velocity. So, a vector equals d v vector by d t. Position, differentiate, velocity. Differentiate again, acceleration. That's it!

2:15Average velocity and average acceleration

Alright, now averages. This little triangle is the Greek letter delta, and it means change in, okay? So, delta r vector is the change in position: final position minus initial position. Average velocity equals delta r vector divided by delta t, the time taken. In the same way, average acceleration equals delta v vector divided by delta t: the change in velocity, over the time taken. Got it? Now, remember our puzzle? Let's finish it. The velocity goes from three i cap to four j cap, in two seconds. The m s to the power minus one is metres per second, and the s alone is seconds, okay? So the change in velocity is four minus three, just one, right? No, no, careful! Velocity is a vector, so we subtract the vectors, not just their sizes. Delta v vector equals four j cap minus three i cap: final minus initial. With me? Now its size. The two straight bars mean magnitude, so we read this as mod delta v vector. Under the square root sign, sixteen plus nine, that is twenty five, and its root is five metres per second. One, or five? Very different, right? Divide by two seconds: five by two is two point five metres per second squared. That is the m s to the power minus two. Mystery solved! The direction changed, and that change counts too. This trap comes up often in JEE questions, so subtract the vectors first, then take the magnitude.

3:51Topper's trick: one direction at a time

Okay? Now, a trick that toppers use. Look back at how we differentiated: x gave x dot, y gave y dot, z gave z dot, each one on its own. So the x, y and z motions can be treated separately. Three easy straight line problems, instead of one hard one! And this is exactly the idea behind projectile motion, which you will meet in chapter four. Keep it in mind. Clear?

4:18Example 1: velocity, acceleration and speed from r(t)

Now, example one. r vector equals three t i cap, plus four t squared j cap, in metres; that is the m at the end, okay? Find the velocity, the acceleration, and the speed at t equals one second. Pause the video, and try it first. Done? Okay, let's check. Differentiate each component. Three t gives three, and four t squared gives eight t. So, v vector, which is d r vector by d t, equals three i cap plus eight t j cap. Differentiate again. Three is a constant, so it gives zero, and eight t gives eight. So, a vector, d v vector by d t, equals eight j cap metres per second squared: constant, and along y only. Right? Now put t equals one. The velocity becomes three i cap plus eight j cap. So the speed is three plus eight, eleven metres per second, right? Oh no, careful! Speed is the magnitude of the velocity, mod v vector. The two components are at right angles, so square them, add, and take the root. Nine plus sixty four is seventy three, so the speed is root seventy three metres per second. Never just add the components, okay?

5:37Example 2: when is velocity perpendicular to acceleration

Now a slightly harder one. r vector equals two t cubed i cap, minus three t squared j cap, in metres. When is the velocity perpendicular to the acceleration? Pause the video, and try it. Done? Okay, let's check. First, the velocity. Two t cubed gives six t squared, and minus three t squared gives minus six t. So, v vector equals six t squared i cap, minus six t j cap. Differentiate again. Six t squared gives twelve t, and minus six t gives minus six. So, a vector equals twelve t i cap, minus six j cap. With me? Now, the key question. When two vectors are perpendicular, which product of the two must be zero? Yes, exactly! The dot product, from our earlier lesson. This double arrow means implies. So, perpendicular implies v vector dot a vector equals zero, with the dot in the middle. Clear? Multiply the matching components and add. Six t squared times twelve t is seventy two t cubed. Then minus six t times minus six. Careful with the signs: minus times minus is plus, so plus thirty six t. Take thirty six t common: thirty six t, times two t squared plus one. So, when is this zero? Thirty six t is zero at t equals zero. But what about two t squared plus one? Never! t squared can never be negative, so two t squared plus one is always greater than zero; that is the greater than sign on screen. So the velocity is perpendicular to the acceleration only at t equals zero. See? Differentiate twice, take the dot product, factorise. Very neat.

7:27Quick revision: differentiate r for v, then v for a

So, quick revision. Differentiate r vector to get v vector, and differentiate v vector to get a vector. For averages, use delta r by delta t and delta v by delta t, and always subtract the vectors. Speed is mod v vector: square the components, add, take the root. And treat the x, y and z motions separately. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!