Uniformly Accelerated Motion

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here is a particle that starts at ten metres per second, with an acceleration of minus two metres per second squared. How far does it go in eight seconds? One student uses the formula and gets sixteen metres. Another student says, no, thirty four metres! Same particle, same eight seconds. Who is right? Hmm, can both be right? Sounds strange, doesn't it? By the end of this class, you will see exactly what is going on. And look, JEE uses these equations again and again. So, let's begin.

0:41Equations of motion (constant a)

So, what does uniformly accelerated mean? Simply that the acceleration stays the same, all the time, okay? We call the initial velocity u, the final velocity v, and this constant acceleration a, okay? Clear? And in a time t, the particle has a displacement s, okay? Now think. Constant acceleration a means the velocity grows by a, every second, right? So, starting from u, what is the velocity after time t? Yes, u plus a t! So the first equation is v equals u plus a t, okay? Really simple. Second, the displacement. s equals u t, plus half a t squared, okay? That small one by two on screen is just half, and the little two up top means squared. Write this down. Third: v squared equals u squared, plus two a s, okay? Toppers love this one; you will soon see why. And fourth: s equals u plus v by two, into t, okay? Look, u plus v by two is just the average of u and v, right? Four equations, four tools. Got it?

1:55Topper's trick: which equation to use?

Now, the topper's trick. Count the quantities: u, v, a, s and t, okay? Five of them. But each equation uses only four. Look, v equals u plus a t has no s in it. So when the displacement is neither given nor asked, this is your equation. Next, s equals u t plus half a t squared: no v, okay? Now you tell me. In v squared equals u squared plus two a s, which quantity is missing? Time, yes! There is no t in it. So when a question says nothing about time, go straight to v squared equals u squared plus two a s, every time, okay? And the last one, s equals u plus v by two into t, has no a in it. So, read the question, find the one quantity that is neither given nor asked, and pick the equation without it. With me?

2:54Distance in the nth second

Alright, now a formula that comes up often in JEE problems. s n, with the small n below, is the distance covered in the n-th second: only that one single second, not all n seconds, okay? s n equals u, plus a into the bracket, n minus half. Now think. The body starts from rest, so u is zero. Put n equals one, two and three. What pattern do you get? Let's see. With u zero, s one is a by two, s two is three a by two, and s three is five a by two. One, three, five on top. Yes, odd numbers! So, in the first, second and third seconds, the distances are in the ratio one is to three is to five. The colons mean, is to, and the dots mean, and so on, okay? But careful, don't mix this up. In the first one, two and three seconds, all together, s is half a t squared, so the distances go as t squared: one is to four is to nine. In the n-th second, odd numbers. In the first n seconds, squares. Clear?

4:06Common trap: displacement, not distance

Now the most important warning of today. All these equations hold only when the acceleration is constant. If a keeps changing, don't use them, okay? And one more question. The s we get from them: is it the distance the particle travels? No! It is the displacement: the change in position, with its sign. So if the particle turns back on the way, the s from the formula is not the distance. So always check: does the velocity change sign during the interval? If yes, the particle has turned back. First find the turning point, where v becomes zero. Then find the distance before the turning point and after it, and add the two parts. Remember our puzzle from the start? This is exactly the key to it. Now, let's see it in the examples.

4:56Example 1: final speed and distance of an accelerating car

So, our first example. Here, a car moving at ten metres per second accelerates at two metres per second squared, for five seconds. Find its final speed and the distance covered. Pause the video, and try it first. Done? Okay, let's check. First the data: u is ten, the acceleration a is two, and t is five. On screen, m s to the power minus one means metres per second, and m s to the power minus two means metres per second squared. And the upright s after the five is seconds, the unit, not our displacement s, okay? Now, which equation gives the final speed straight away? Yes, v equals u plus a t, the one with no s in it. Ten plus two into five: twenty metres per second. Now the distance. s equals u t, plus half a t squared. Ten into five is fifty, and half into two into five squared is twenty five. Fifty plus twenty five: seventy five metres, and this m means metres. The velocity goes from ten to twenty and never changes sign, so here the displacement is the distance, okay? And a quick check with the fourth equation. Ten plus twenty, by two, into five: seventy five again. Exactly the same!

6:20Example 2: displacement vs distance when it turns back

Now a harder one, and it is our puzzle from the start. Here, a particle has u equal to ten metres per second, and a equal to minus two metres per second squared. Find its displacement and the distance covered in eight seconds. Pause and try. Hint: check whether it turns back. Done? Okay, let's check. Displacement first. s equals u t plus half a t squared. Ten into eight is eighty, and half into minus two into eight squared is minus sixty four. Eighty minus sixty four: sixteen metres. Easy, isn't it? So the distance covered is also sixteen metres, right? No, no, careful! This is the classic trap. The acceleration is negative, so the velocity keeps falling, and at some moment it becomes zero. v equals ten minus two t, equals zero. This arrow means, so: t equals five seconds. At five seconds the particle stops and turns back, but we were asked about eight seconds, remember? So, the first five seconds, going forward. Ten into five is fifty, and half into minus two into five squared is minus twenty five. So, twenty five metres forward. In the last three seconds, it starts from rest at the turning point and comes back with the same two metres per second squared. Half into two into three squared: nine metres back. So the distance is twenty five plus nine: thirty four metres. And the displacement is twenty five minus nine: sixteen metres, same as before! See, puzzle solved: both students were right. Sixteen is the displacement, thirty four is the distance. The lesson is simple: always look for a turning point first.

8:11Quick revision: three equations and the turning point

So, quick revision. For constant acceleration: v equals u plus a t, s equals u t plus half a t squared, and v squared equals u squared plus two a s, okay? Pick the equation without the quantity you neither know nor need. And these give displacement: if the velocity changes sign, split at the turning point. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!