Two Bodies and Two Stages

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Car A is at rest. Just as it starts speeding up, car B passes it at a steady speed. At first, car B pulls ahead, and the gap grows. But car A keeps getting faster, so later the gap shrinks, and car A overtakes. Now tell me: at the exact moment when the gap is the biggest, which car is faster? Hmm, think about it. We will answer it today with one simple line. And you know, chase questions like these are common in kinematics practice: a police jeep and a speeding car, two trains on one track. So, let's begin.

0:41Topper's trick: one clock for both bodies

So, here is the topper's trick for every two-body question. Two bodies on one straight line. Body A is at position x A, and body B is at position x B, okay? The small A and B written below the x are just labels, okay? Now, each body has its own motion. So should we use two different stopwatches, one for each body? No, no! Just one clock. Write both positions with the same clock time t. Why? Because meeting means the same place at the same instant, right? So both bodies must read the same clock. Clear? Alright, but what if body B starts a little later, t naught seconds after the clock starts? That is t with a small zero, the delay. When the clock shows t, for how long has body B actually been moving? Yes, t minus t naught! So a body that starts later has t minus t naught in its equations, wherever its time appears. First it waits, then it moves: two stages, one clock. Simple as that.

1:50Meeting, or the biggest gap?

Now, with one clock, two questions come again and again. First: when do the two bodies meet? What condition do you write? Yes, exactly! Same place at the same time. x A equals x B: the positions are equal. Put the two positions equal, and solve for t. That gives you the meeting time. Second question, and this is our puzzle. When is the gap between the two bodies the largest? When their velocities are equal! v A equals v B; v stands for velocity, okay? Look, in our puzzle, while car A is slower, car B pulls away and the gap grows. Once car A is faster, the gap shrinks. So the gap is largest at the moment the velocities are equal. And when a gap first shrinks and then grows, the same moment gives the smallest gap. So the puzzle is solved: at the biggest gap, neither car is faster. Now a quick one. At that moment of equal speeds, car A has also caught car B, right? No, no, careful! This is the classic trap. Equal velocities do not mean they meet. At that moment the gap is the biggest, not zero! Meeting needs equal positions; the biggest or smallest gap needs equal velocities. Two different conditions.

3:08Example 1: when car A overtakes car B, and the largest gap

Got it? Now, remember our puzzle? Here it is with numbers. Car A starts from rest with two metres per second squared, just as car B passes it at a steady twenty metres per second. The m s to the power minus two means metres per second squared, and m s to the power minus one means metres per second, okay? When does car A overtake, and what is the largest gap before that? Pause the video, and try it first. Done? Okay, let's check. Start the clock as car B passes, so both start from the same point. Car A starts from rest, so its position is half into two into t squared, which is just t squared. Car B moves steadily, so x B equals twenty t. With me? Now, when does car A overtake? Which condition do you use? Yes, positions equal! t squared equals twenty t. The double arrow on screen means so, okay? Then t equals zero, but that is just the start, when car B passes. The other root is t equals twenty seconds; the s means seconds. And by then, each car is four hundred metres from the start. Now the largest gap before the overtaking. Which condition do we use this time? Equal velocities, exactly! The gap is x B minus x A: twenty t minus t squared. And the velocities: car A gains two metres per second every second, so v A is two t, and car B stays at twenty. Put them equal: two t equals twenty, so t equals ten seconds. See, our puzzle again: at the biggest gap, both cars move at twenty metres per second! Now put t equals ten into the gap: twenty into ten, minus ten squared. Two hundred minus one hundred: one hundred metres. So car A overtakes after twenty seconds, and the largest gap before that is one hundred metres. Got both? Very good!

5:16Example 2: police jeep catching a speeding car

Now a slightly harder one. Here, a car speeding at thirty metres per second passes a parked police jeep. The jeep starts two seconds later, with an acceleration of five metres per second squared. How long after starting does the jeep catch the car? Pause and try. Hint: first decide where your clock starts. Alright, notice the two stages here: first the jeep waits two seconds, then it chases. The question asks, how long after starting, so let's start our clock when the jeep starts. Now, when the clock shows t, for how long has the car been moving? Two seconds longer, yes! t plus two, in brackets. Look, this is our trick from the other side. Earlier, the later body got t minus t naught. Here the clock belongs to the later body, so the earlier one gets t plus two. Same idea. Now the positions, both from where the jeep was parked. x J is the jeep, and x C is the car. The jeep starts from rest: half into five into t squared, that is five by two t squared. The car moves steadily: thirty into t plus two. Okay? Catch means same place, same time, so put them equal. Five by two t squared equals thirty into t plus two. Multiply both sides by two by five, and bring everything to one side: t squared minus twelve t minus twenty four equals zero. Now the quadratic formula. The plus or minus sign gives two roots, and the tick-shaped sign is the square root, okay? Twelve, plus or minus root of one hundred forty four plus ninety six, all over two. That is six, plus or minus root sixty. Two roots. Which one do we keep? Six minus root sixty is negative: a time before the jeep even started, so we throw it out. So t equals six plus root sixty, which is six plus two root fifteen, about thirteen point seven seconds. The wavy equals sign means approximately. And the question asked the time after the jeep starts, which is exactly what our clock measured. And we are done!

7:35Quick revision: meet at equal x, max gap at equal v

So, quick revision. One clock t for both bodies, and a body that starts later uses t minus t naught. They meet when the positions are equal. The gap is largest or smallest when the velocities are equal. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!