Motion Graphs

Transcript

0:00Introduction

Hello everyone! Here is one small puzzle before we start. Two students read the same velocity time graph, for the same body, over the same six seconds. One says the body moved eighteen metres. The other says thirty metres. And the teacher gives both of them full marks! How can one graph give two answers? Strange, isn't it? By the end of this video, you will see it in one look. And remember, JEE often asks graph questions in kinematics. So, let's begin.

0:34The x–t graph: slope is velocity

So, here are three graphs of one motion: a body starting from rest, with uniform acceleration. Left, position x against time t: a parabola. Middle, velocity v against t: a straight line. Right, acceleration a against t: a flat line, constant. The little dash between two letters just means against, okay? The left graph gets steeper and steeper. What does that tell you about the body? Yes, it is moving faster and faster! Steepness is slope: change in x over change in t, and that is velocity. So, the slope of the x t graph equals the velocity v. Clear? So if an x t graph is a straight line, the slope never changes, right? Same velocity all the time: uniform velocity. Our parabola bends upwards: its slope, the velocity, keeps growing. So bending upwards means positive acceleration, and the flat line on the right, sitting above zero, agrees. Makes perfect sense. Alright, now suppose an x t graph rises, reaches a peak, and comes down. At the very top of that peak, what is the velocity? Zero, exactly! At the peak the graph is flat for one instant, so the slope, v, is zero. That is a turning point: the body stops and turns back. Same at a valley, okay?

1:57The v–t graph: slope and area

So, now the most important one: the v t graph, velocity against time, the middle graph. Its slope is change in velocity over time: acceleration. The figure writes it on top: slope equals a, for acceleration. Simple, right? Now the shaded area under it. Area is velocity into time, and that gives displacement. That is why the figure says area equals s, where s is the displacement. To be exact, the signed area, with a plus or minus sign, okay? But what if a v t graph goes below the time axis? There the velocity is negative. So which way is the body moving? Backwards, yes! So the area below the time axis counts negative for the displacement. Got it? So the distance is also just the signed area, right? No, no, careful! This is the classic JEE trap. Distance can never be negative. So for the distance, count every area as positive, the total area. Clear?

3:02One graph, two answers

Now, remember our two students? Here is their graph. The velocity starts at twelve, falls in a straight line, crosses the time axis at four seconds, and reaches minus six at six seconds. The m s to the power minus one on the v axis means metres per second, and the s in brackets on the t axis means seconds, okay? Ignore the notes on the right for a moment; we will build them ourselves. Above the axis, the green triangle: half into four into twelve, plus twenty four metres. The dots just mean multiply. Below the axis, the pink triangle: half into two into six, with a minus sign. Minus six metres. Signed area: twenty four minus six, eighteen metres. That is the displacement: the body ends up eighteen metres from its start. Same as the first note on the right. And counting both as positive: twenty four plus six, thirty metres, the distance. See, mystery solved! One student found the displacement, the other the distance. Both right. The notes on the right also give the slope: an acceleration of minus three metres per second squared, the m s to the power minus two. Twelve down to zero in four seconds, spot on.

4:17Two more graphs: a–t and v–x

Two more graphs, quickly. First, the acceleration time graph. Acceleration into time is change in velocity, right? So the area under it equals the change in velocity. Not the velocity itself, only the change, okay? Next, the v x graph: velocity against position x. Now tell me, is its slope the acceleration? No! Its slope is d v by d x: how fast v changes with x, not with time. The upright d's are the d of calculus. The acceleration is v into d v by d x: velocity times the slope, not the slope itself. Many students fall here. Got it?

4:55Graphs that cannot happen

Last idea: graphs that cannot happen. An x t graph can never have two values of x at one time t: a body cannot be at two places at once, can it? And speed and distance can never be negative. Now, can a v t graph jump, from one value to another, at a single instant? No! That jump means the velocity changes in zero time: infinite acceleration. An x t graph cannot jump either; that would need infinite velocity. And the total distance can never go down. But an acceleration time graph can jump: switch on a force suddenly, and the acceleration changes at once. That is perfectly allowed. With me?

5:36Example 1: distance and average speed from a v-t graph

Now, Example one. Here, a car's velocity rises uniformly from zero to twenty metres per second in four seconds, stays at twenty for six seconds, and then falls uniformly to zero in two seconds. Find the distance and the average speed. Pause the video and try it. Hint: sketch the v t graph first. Done? Okay, let's check. The v t graph goes up, stays flat, and comes down: a trapezium. It never dips below the axis, so the distance is just the area. Three pieces: a triangle, half into four into twenty; a rectangle, six into twenty; and a triangle, half into two into twenty. Forty, plus one hundred and twenty, plus twenty: one hundred and eighty metres. Nice and simple! Average speed is total distance over total time. Total time: four plus six plus two, twelve seconds. One eighty by twelve: fifteen metres per second. Not twenty, the top speed; the slow parts pull it down. Clear?

6:42Example 2: from an a-t graph to displacement and distance

Now a harder one, JEE level. Starting from rest, a body has an acceleration of four metres per second squared for two seconds, then minus two metres per second squared for six seconds. Find its displacement, and the distance covered in the eight seconds. Pause the video and try it. Hint: velocity first, then the v t graph. Done? Okay, let's check. The area under the acceleration time graph is the change in velocity: four into two, eight. It started from rest, so v of two, the velocity at two seconds, is eight metres per second. Alright, at two seconds the acceleration becomes negative. So the body starts coming back at two seconds, right? No, no, careful! That is the trap. At two seconds the velocity is still plus eight; negative acceleration only slows it down. Now v equals eight minus two into the bracket, t minus two. It hits zero at six seconds: the turning point. And v of eight is minus four metres per second: coming back. Now the v t graph: three triangles. Half into two into eight: eight. Half into four into eight: sixteen. And below the axis, half into two into four, with a minus: minus four. Displacement: eight plus sixteen minus four, twenty metres. Distance, all positive: eight plus sixteen plus four, twenty eight metres. See, it is the same story as our two students: the graph dips below the axis, so displacement and distance differ. Got it?

8:24Quick revision: slopes and areas of x-t, v-t, a-t graphs

So, quick revision. Slope of x t is velocity; slope of v t is acceleration. Under v t, the signed area is displacement, and the total area is distance. Under the acceleration time graph, the area is the change in velocity, and on a v x graph, the acceleration is v into d v by d x. Now go and try the practice problems on this page. All the best!