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: (c) Find the flea's acceleration at 0.5 ms, 1.0 ms, and 1.5 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: (c) Find the flea's acceleration at 0.5 ms, 1.0 ms, and 1.5 ms.

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

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

Accepted Answer

Hey everyone in this problem, we're told that the video is a collection of motion pictures measured in frames per second. Okay. Ordinary cameras can attain 60 frames per second. Um cameras used to capture and analyze movements occurring very fast to the order of milliseconds. Okay, so we have a nearly vertical launch of a 450 g and centimeter long projectile. Okay. And what we're asked to do is to determine the acceleration of the projectile at 10 milliseconds, 25 milliseconds and 40 milliseconds. Okay, we're told to assume that the curve is a straight line. Alright, so let's get started. First thing we're looking for acceleration. What is our graph giving us? Okay, well this graph is a V. T. Graph, velocity versus time. Now recall the acceleration is related to velocity by the following. The acceleration is equal to D V D. T. Or the change in velocity over change in time. And we can just say that this is equal to the slope of a beachy graph. Okay, so we're given a VT graph. We're looking for acceleration. So we want to be looking for the slope at the point skip. Okay, so let's start with the acceleration at 10 m/s. Okay, and again we're looking for the change in the velocity over. The change in time velocity is on the Y axis. So this if you want to think of it is like your rise over run. Alright, so we're told to approximate this with a straight line. Okay? We see that it kind of gets flat at the bottom here. So we want to make sure that we're not Using that. Okay, we're gonna be going we want to look for a tangent at 10. So I think if we draw the straight line kind of between these two points, we're gonna get a pretty good estimate of the tangent there. Okay. All right. Now, when we do this, our rise, if we consider the rise of this straight line, we go from zero milliseconds up to or sorry, zero m per second up to seven m per second. Okay. And so we have seven m per second minus zero m per second. That's our change in the velocity or change in the y divided by how far we're going in the access the time. Okay, We're at 32.5 milliseconds. Okay, minus here is about 7.5. No, it's it means All right, so if we work this out now pay attention, we're in meters per second for velocity, but we're in milliseconds for time. So we're gonna want to convert those times two m per second or two seconds, Sorry? Okay, so we get seven m per second divided by 32.5 minus 7.5 is going to give us and instead of milliseconds we're gonna have times 10 to the negative three seconds here. We have meters per second divided by second. That's going to give us meters per second squared, which is a unit for acceleration. Okay, so the units work out there And this gives us 280 m/s squared. And so the acceleration at 10 milliseconds Is 280 m/s squared. Now we want to look at 25 milliseconds at 25 milliseconds. Well, let's take a look at 25 words. We can approximate again by the same line that we've already drawn. We're assuming that we have a straight line here and so are tangent at that point can be approximated by that same line. And so our acceleration at 25 milliseconds is just gonna equal the acceleration at milliseconds, which we found to be 280 m per second squared. Okay, so that's two of our accelerations done one more to go. The last one we're asked for is 40 milliseconds. And so a the acceleration at 40 milliseconds is going to be equal to what? Well, at 40 milliseconds, we're here on our graph and we're on this portion. That's this straight line, This horizontal flat line. Okay. Now remember when we talked at the beginning that the acceleration is just the slope of a VT graph. Okay, well if we're at this horizontal line then our slope is zero. And so our acceleration at 40 milliseconds is just zero m per second squared. Okay. Alright, So now let's look at our answer choices and find the one that matches, scroll down so we can see them all and we found that a 10 is m per second squared. A 25 is 280 m per second squared, and a 40 is zero m per second squared. And so we're going to have answer D. That's it for this one. I hope this video helped see you in the next video.

Verified Solution