What is the magnitude of the acceleration of a skydiver at the instant she is falling at one-half her terminal speed?

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Hi, everyone. And to discuss this problem, we will have a box of food dropped from a humanitarian airplane. And we're being asked to determine the magnitude of the acceleration of that box when its speed actually reaches 30% of its terminal speed. The options given are a 0.88 m per second squared. B 8.93 m per second squared, C 9.81 m per second squared and D 10.7 m per second squared. So we will model the box as a point like particle and the box moves vertically downward because it is dropped from an airplane and it is going to be subjected to two different forces. The first one is going to be the gravitational force which is going to be vertically downward. And the second one is the drag force which is going to be vertically upward. So the drag force F D is going to be proportional to the square of the speed fee of the object. So F D which is going to be the drag force is equals to half multiplied by a row multiplied by C multiplied by a multiplied by V squared, the row is going to be the density of the air C is the drag coefficient A is the area of the box that is facing the air and V squared is the square of the speed fee of the object. So the way we want to tackle this problem is by applying Newton's second law on our system. So according to Newton's second law, sigma F net will equals to M multiplied by A. And in this case, we are looking at the vertical direction or the Y direction. So Sigma F Y will equals to M multiplied by A Y. So as I have described previously, the box is going to experience 22 different forces. The first one is the gravitational force or the weight pointing downward. And the second one is the direct force. So in this case, F D minus M G will equals to M multiplied by A Y. And we're using the convention at which the forces going upwards is positive. So at terminal speed, at terminal speed A Y is going to equals to zero m per second squared. And that is a common fact that we need to use to simplify our Newton Second Law equation. So then we have F D minus M G equals to zero, which mean that F D actually will equals to M multiplied by G. So I am going to employ our F D equation that we have recalled previously so that we have half multiplied by a row multiplied by C A V squared. Well equals to M G. What we are interested to look at is the velocity, which in this case is going to be our terminal velocity because we have employed the fact that A Y equals to zero m per second squared. And we wanna find the field uh the terminal velocity equation which in this case is going to be two M G divided by row C A. And we want to take the square root of that to get our actual terminal velocity equation to be the square root of two M G divided by a row C A Awesome. So we are asked to calculate the instantaneous acceleration when the velocity is actually 30% of its terminal speed. So in this case, we want to find the acceleration at V equals 0.3 of feet T. And in this case, we wanna reapply Newton's second law along the right direction to get the philosophy at 0.3 times terminal velocity. So are you applying Newton's second law in the right direction? Again, we will get pretty much the same thing which is F D minus M G equals M multiplied by A Y which in this case, our F D is going to still be our uh equation that we just recalled. But the velocity will be represented by the velocity or the terminal velocity multiplied by 0.3, which is the condition that we have listed here. So the F D will then be half multiplied by a row multiplied by C multiplied by A and the velocity is going to actually be 0.3 V P. So I'm gonna put open parentheses, 0.3 multiplied by P terminal, which is going to be square root of two M G divided by a row C A two M G divided by row C a close, close parentheses squared. And we wanna minus that with M G and equals that with M multiplied by A Y just like so awesome. So I am going to simplify this by simplifying the left side of our equation. So half row C A will stay that way. And 0.3 will then be uh 0.3 squared will then be 0.9 multiplied that by two M G divided by a row C A. And we can cross out the row C A out of the equation minus M G equals M multiplied by A Y. And then I'm gonna simplify this further by crossing out all the masses in our equation. So basically dividing the equation by mass so that our final equation will then just be the two here and half is going to cancel each other out as well. So our left side will then just be 0.9 multiplied by G minus G equals A Y. I'm gonna take the G out or pull it out of our equation so that A Y will then just be 0.9 minus one multiplied by G which will then equals to negative 0.91 multiplied that by G which is 9.81 m per second squared. And that will actually be negative 8.93 m per second squared. So that will be the final answer to our problem which will give us the instantaneous acceleration of the box when it reaches 30% of its terminal velocity to be demanded of 8.93 m per second squared without the negative sign. Because what is being asked is just the magnitude. So that will actually correspond to option B in our answer choices. So option B with the magnitude of the acceleration of the box when it reaches 30% of its terminal speed or terminal velocity being 8.93 m per second squared will be the answer to this practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be it for this video. Thank you.

What is the magnitude of the acceleration of a skydiver at the instant she is falling at one-half her terminal speed?

Verified Solution

Key Concepts

Terminal Velocity

Acceleration

Drag Force