Motion in a Plane with Constant Acceleration

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Imagine a particle with a constant acceleration: the same size, and the same direction, all the time. Constant acceleration, so its path must be a straight line, right? Hmm, are you sure? Today we check it, with vectors. And remember, in JEE, projectile motion, which comes next in this chapter, is built entirely on this one idea. So, let's begin.

0:30The same equations, now with arrows

So, you know the equations of motion for a straight line, from the last chapter. Now the particle moves in a plane: left and right, and up and down, at once. What changes in our equations? Yes, they become vectors! The little arrow on top means a vector. So, v vector equals u vector, plus a vector into t. Same equation, just with arrows. Clear? For position, we use r vector, the position vector, from the origin to the particle. r vector equals r nought vector, plus u vector t, plus half a vector t squared. The small zero below r means the starting position; we say r nought, okay? But note, there is one condition. These equations are valid only when the acceleration vector is constant, in size and in direction. And our puzzle particle has exactly that.

1:24Split it: the x motion and the y motion

Alright, but how do we calculate with a vector equation? In a plane, every vector has an x part and a y part, right? So this one vector equation hides how many ordinary ones? Two, one for each direction! For x: x equals x nought, plus u sub x t, plus half a sub x t squared. The small x below u and a means the x component, okay? The straight-line equation, with only the x parts. And the same for y. y equals y nought, plus u sub y t, plus half a sub y t squared. Only the y parts here. Got it? Velocity splits the same way. v sub x equals u sub x plus a sub x t, and v sub y equals u sub y plus a sub y t. Each component obeys the one-dimensional equations, separately. Now suppose the acceleration points only along y, so a sub x is zero. Then v sub y keeps changing. But the particle is accelerating, so v sub x must change too, right? No, no, careful! Look at the velocity line. v sub x equals u sub x plus a sub x t: there is no a sub y anywhere in it. So with a sub x zero, v sub x stays equal to u sub x, all the time. The two motions share only the time t. This is the heart of today's lesson. With me? Very good.

2:55Topper's trick: two 1D problems, one clock

Now a trick that toppers use. Step one: write u vector and a vector in i cap, j cap form. The little cap means a unit vector: i cap points along x, and j cap along y, okay? The number with i cap is the x component; the number with j cap, the y component. Step two: solve x and y separately, like two straight-line problems. They share only the clock: the same t in both. Clear? Step three: combine. r vector equals x i cap plus y j cap, and v vector equals v sub x i cap, plus v sub y j cap. That's it! Two easy one-dimensional problems, instead of one scary two-dimensional one.

3:43Example 1: u = 3i, a = 2j, position and velocity at 2 s

Let's start with an easy one. Here, a particle starts at the origin with u vector three i cap, in m s to the power minus one, that is, metres per second. Its acceleration is two j cap, in m s to the power minus two, metres per second squared. Find its position and velocity at t equals two seconds; the s means seconds. Pause the video, and try it first. Done? Okay, let's check. Step one of our trick: the components. u vector is three i cap only, so u sub x is three, and u sub y is zero. a vector is two j cap only, so a sub x is zero, and a sub y is two. Right? Now the x motion, alone. It starts at zero, and a sub x is zero, so x is just u sub x into t. Three into two: six metres. And the y motion. u sub y is zero, so y equals half, into two, into two squared: four metres. So the position is the point six comma four, in metres; the double arrow means so, okay? Now the velocity. v sub x stays three, as a sub x is zero. v sub y is two into two, four. So v vector equals three i cap plus four j cap. So, the speed is three plus four, seven metres per second, right? No, no! Classic JEE trap. The components are at right angles, so we never just add them. Speed is the size of the velocity vector: the root of, three squared plus four squared; the tick-shaped sign is the square root. Nine plus sixteen is twenty five, and root twenty five is five metres per second. Now look back at our puzzle. At the start, the velocity was three i cap, along x. At two seconds, it is three i cap plus four j cap, tilted upwards. The direction of motion changed, though the acceleration never did! So constant acceleration does not always mean a straight path. Got it?

5:54Example 2: boat from rest, straight path and distance in 4 s

Now a harder one, JEE level. Here, a boat starts from rest at the origin, with acceleration two i cap plus j cap, metres per second squared. Show that it moves in a straight line, and find how far it goes in four seconds. Pause and try. Hint: it starts from rest. Done? Okay, let's check. In Example one, the path bent. And here, the acceleration has an x part and a y part, both. So this boat must also curve, right? Hold on, not so fast! It starts from rest, so u vector is the zero vector; that is the arrow over the zero, okay? It starts at the origin, so r nought vector is zero too. So our big equation shrinks to r vector equals half a vector t squared. Now look closely. After the start, half t squared is just a positive number. So r vector is always a positive number times the acceleration, a vector, which never changes, right? The position always points in the direction of the acceleration, so the boat moves in a straight line. So, from rest the path is straight; in Example one, it bent. So the answer to our puzzle: it depends on how the particle starts. Now, how far in four seconds? On this straight line, we use half a t squared, as in one dimension. But which number is a here? Two? One? Or two plus one, three? Careful, none of those! We need the size of the acceleration, written mod a vector; the two straight bars mean the magnitude. It is the root of, two squared plus one squared: root five metres per second squared. So the distance equals half, into mod a vector, into t squared. That is half, into root five, into four squared. Clear? Half of sixteen is eight: eight root five metres. The wavy equals sign means approximately, so about seventeen point nine metres. That's it, done!

8:00Quick revision: solve x and y separately, same time t

So, quick revision. Constant a vector: v vector equals u vector plus a vector t, and r vector equals r nought vector, plus u vector t, plus half a vector t squared. Solve x and y separately; they share only the time t. And the speed is the root of, v sub x squared plus v sub y squared, never v sub x plus v sub y. Now go and try the practice problems on this page. All the best!