Motion Under Constant Power

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An engine pushes a body from rest, delivering the same power every second. After two seconds, the body has some speed. When will its speed be double that? After four seconds, right? Hold that answer; a surprise is coming. Constant power comes up often in JEE, and the usual equations of motion will not help here. So, let's begin.

0:27Speed under constant power

So, here is our setup. An engine gives a body a constant power, capital P on screen, okay? The body starts from rest, and there is no resistance at all. Our main results depend on these three conditions. Now think. Power is work per second, so the engine does P joules of work every second. In t seconds, how much work in total? P times t! Yes, exactly! The power is constant, so the work done in time t is simply W equals P t, with me? Where does this work go? No resistance, so all of it becomes kinetic energy. By the work energy theorem, half m v squared equals P t, where m is the mass and v the speed. Okay? So v is the square root of two P t over m; that is the root sign on screen. The time sits under the root, so v is proportional to t to the power one half. This curly sign, like a stretched alpha, means proportional to. Clear? Now, back to our puzzle. The speed grows as the square root of the time. So to double the speed, how much time do we need? Four times the time! Correct! Double the speed means four times the kinetic energy, and the engine supplies energy at a steady rate. So the speed at two seconds doubles only at eight seconds, not four! The double arrow means it follows that. Surprised? And if the body already has a speed u when we start the clock? The engine's work still equals the gain in kinetic energy, so v squared equals u squared plus two P t over m. Put u equals zero, and you are back to our first result. Simple, isn't it?

2:14The trap: the force keeps falling

Now, a trap that catches many students. The power is constant, so the force and the acceleration are constant too, right? No, no, careful! Power is force times velocity: P equals F v. So the force is P over v. And v grows as root t, so F is proportional to t to the power minus one half. Got it? So as the speed rises, the force falls, and so does the acceleration. Same energy every second, but less speed added each second. Right? So the constant acceleration equations, like v equals u plus a t, do not apply here at all. Use energy, or calculus.

2:57Distance: integrate the speed

Okay? Now, how far does the body go? Speed is the rate of change of position: v equals d x over d t, where x is the distance from the start. And v is the root of two P over m, times t to the power one half. Right? Integrate with respect to time, starting from x equals zero. t to the power one half integrates to two thirds of t to the power three over two. So x equals two thirds, times the root of two P over m, times t to the power three over two. Or, taking the two thirds inside, the root of eight P over nine m. With me? So the distance is proportional to t to the power three over two. Remember this power; we use it soon. Clear?

3:43Speed against distance

Sometimes a question gives the distance, not the time. Then write the acceleration as v d v over d x. So the force is m v d v over d x, and the power, F times v, is m v squared d v over d x. Okay? Separate the variables: P d x equals m v squared d v. Integrate from rest, and P x equals m v cubed over three. Right? So v cubed goes up in step with x. Then v itself goes as which power of x? One third! Exactly! v cubed equals three P x over m, so v is proportional to x to the power one third, the cube root of x. Clear?

4:27Topper's trick: just read the powers

Now, a trick that toppers use. Put the four results in one line: v goes as t to the half, x as t to the three halves, v as x to the one third, and F as t to the minus half. Ratio questions need nothing more. Okay? Try one. Same body, same power, from rest. If the time doubles, the distance becomes how many times? Two root two times! Very good! x goes as t to the three halves, so the factor is two to the power three over two, which is two root two. The little arrow means becomes. No calculus. That's it!

5:06Example 1: speed of a 2 kg body after 9 s at 4 W

Now, example one. So, a two kilogram body starts from rest and receives a constant power of four watts. Find its speed after nine seconds. On screen, k g is kilograms, capital W is watts, and s is seconds. Pause the video and try it. Done? Okay, let's check. Constant power, from rest, no resistance mentioned. Energy given in nine seconds: P t, four times nine, thirty six joules; capital J is the joule. Okay? All of it becomes kinetic energy. So half times two times v squared equals thirty six. Right? Half times two is one, so v squared is thirty six. Same as the root of two P t over m: two times four times nine, over two. So v is six metres per second; m s to the power minus one means metres per second. Easy!

6:06Example 2: time and distance for a 20 kW car

Now a harder one. So, a one thousand kilogram car has an engine giving a constant twenty kilowatts, written k W on screen. It speeds up from ten to twenty metres per second on a level road; ignore resistance. Find the time taken and the distance covered. Pause the video and try it. Done? Okay, let's check. Let v sub one be ten and v sub two be twenty; the small one and two below are subscripts. The engine's work, P t, equals the gain in kinetic energy: half m, times v sub two squared minus v sub one squared. That is half times one thousand, times four hundred minus one hundred. One point five times ten to the power five joules. Okay? Twenty kilowatts is twenty thousand watts. So t equals one point five times ten to the power five, over twenty thousand: seven point five seconds. Clear? Now the distance. The speed goes from ten to twenty, so the average speed is fifteen. Fifteen times seven point five: one hundred and twelve point five metres, right? No, no, wait! Averaging the speeds works only for constant acceleration, and here the force keeps falling. So integrate P equals m v squared d v over d x, from v sub one to v sub two. x equals m over three P, times v sub two cubed minus v sub one cubed. Got it? Twenty cubed is eight thousand and ten cubed is one thousand, so the bracket is seven thousand. And three P is sixty thousand. One thousand times seven thousand, over sixty thousand: about one hundred and seventeen metres. The wavy equals sign means approximately. Not one hundred and twelve point five! That was the trap.

8:00Quick revision: v, x and force under constant power

So, quick revision. One: constant power, from rest, no resistance: half m v squared equals P t. Two: v goes as t to the half, x as t to the three halves, v as x to the one third, and the force, P over v, as t to the minus half. Three: the force keeps falling, so never use the constant acceleration equations. Now try the practice problems on this page. All the best!