High-speed motion pictures (3500 frames/second) of a jumping, 210–μg flea yielded the data used to plot the graph in Fig. E2.54. (See 'The Flying Leap of the Flea' by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (b) Find the maximum height the flea reached in the first 2.5 ms.

Question

High-speed motion pictures (3500 frames/second) of a jumping, 210–μg flea yielded the data used to plot the graph in Fig. E2.54. (See 'The Flying Leap of the Flea' by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (b) Find the maximum height the flea reached in the first 2.5 ms.

Velocity-time graph showing a flea's jump, with velocity in m/s and time in ms.

Speed-time graph of a flea's jump, with speed in cm/s and time in ms.

Accepted Answer

Hey everyone in this problem, we're told that leaps are very common among animals. So scientists use high motion pictures at 3000 frames per second to analyze the motion of a frog. They can leap 20-50 times their length. Okay. And we're told the size and length of a frog. Okay. They have a near vertical takeoff. Okay. And we're given this graph of their motion. And what we want to do is we want to determine the maximum height reached by the frog in the 1st 75 milliseconds and 225 milliseconds. All right, so, if we're talking about the height, Okay, we're talking vertical distance or displacement. Let's take a look at our graph we have here. V. T. Okay, so this is a velocity versus time graph. V. T graph. When we're relating displacement to the velocity, we know that the displacement or the distance is going to be the area under the velocity curve. Okay, Now, because we're told that this is a near vertical takeoff, that means the displacement or the distance is going to be all in the Y direction. Ok, so when we take that area under the curve, that's going to be all in the Y direction which is going to correspond to the height. Alright, so let's get started. Let's do 75 milliseconds first. Okay, so it's 75 milliseconds. Let's take a look at our graph. Okay, so we want to find the area under the curve that's going to give us the displacement from a velocity time graph from 75 milliseconds. Okay, so 75 milliseconds is here. So the area we want to find is this red area here. Okay, so this area under the curve up to 75 milliseconds. And let's go ahead and approximate this by a triangle so that we can find that area. The area is going to be again, we're approximating by a triangle. So we have one half the base times the height. One half the base. Okay, well that's 75 milliseconds. Let's convert this into seconds because our velocity is in meters per seconds. We want consistent units. So we're gonna have 75 times 10 to the negative three seconds. Okay, times the height. Well this goes all the way up to one m per second. So we get one theater per second. Okay, and multiplying this out, we are going to get 37. times 10 to the negative three meters. Okay. And you'll notice that the possible answer choices are in centimeters. So let's go ahead and convert it to centimeters. So we get 3.75 centimeters. Okay, so that's the area under the curve which means that the max height at 75 millis milliseconds is 3.75 centimeters. Okay, because the area under the curve, is that displacement in this case? It's all in the y direction. And so that is going to be our height. Alright, great. So we're done now with 75 milliseconds let's move on to 225 milliseconds in this case, let's get rid of the red here. In this case. 225 milliseconds is all the way down here. So we need the area under the whole curve. Okay, well we can break that into two parts. Okay, we can break that into this big rectangle here. Okay, and into the little triangle over here and now the area under our curve is going to be the area of the blue part, which is going to be the length times the width because it's a rectangle. Okay, plus the area of the green park which we're gonna approximate by a triangle. So this is gonna be one half base right? Okay. Alright, I'm gonna switch to black so that we don't have to switch back and forth, but you can see where each of these parts of the area come from. Alright, so the length times the width. Well the width is, we start at 100 we go to 225. So that's going to be 100 and 25 milliseconds. And again let's convert this Into second, so we get 10 to the negative three seconds. Okay, times the width or sorry, times the length which in this case is the height, we're going from 0 to 1.751 point m per second. Okay, let's give ourselves some more room to work here. Alright and we're just going to add the second part down here. We get one half the base of this triangle. Well that's 100. Okay, and again, converting into seconds times 10 to the negative three seconds Times the height of the triangle. 1.75 meters for a second. Alright, so now we have the formula for our area. We just need to go ahead and do the calculation. We're going to get that. The area is equal to 0.21875 m. Okay, we have meters per second times seconds we're gonna be left with meters plus 0.875 m. So we get the contribution from the rectangle and from the triangle. And this is going to give us a total area of 0.30625 m. And again, the answer choices are in cm. So let's convert this into cm. We're going to get 30.6 centimeters. That is the area under the curve. At 225 milliseconds. Which is going to tell us the maximum height that we reached at that point. Okay. And so the answer is going to be D at t equals 75 milliseconds. Our height is 3.75 centimeters and at 225 milliseconds, our height is 30.6 centimeters. Thanks everyone for watching. See you in the next video

Verified Solution