kinematics problem

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ihatebluescrubs

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In this particular question:

An 80-kg parachutist strikes the ground at 4 m/s. If he decelerates uniformly over the entire 2-meter length of his body, what is the magnitude of the force of impact with the ground?

I understand that basically we need to plug and chug to:

1) Find the acceleration
2) plug into Force = mass * accel

I'm not sure why my method doesn't work when I attempt to solve for acceleration.

The answer key solved for acceleration using the VAX kinematics eq Vf^2 = Vi^2 +2a(change in distance)

However, I thought that I could solve using the definition of acceleration which is:
acceleration = (change in velocity)/(time). I found the time via using the initial velocity, 4m/s and using the given distance of 2m: so:

2m * 1/(4m/s) giving a total of 0.5 seconds. ---> plug this into the definition of acceleration:
(Vf-Vi which is 4m/s) / (time which was just solved above 0.5sec)


Why won't this method work? The question stem states that it is a uniform deceleration.
 
In this particular question:



I understand that basically we need to plug and chug to:

1) Find the acceleration
2) plug into Force = mass * accel

I'm not sure why my method doesn't work when I attempt to solve for acceleration.

The answer key solved for acceleration using the VAX kinematics eq Vf^2 = Vi^2 +2a(change in distance)

However, I thought that I could solve using the definition of acceleration which is:
acceleration = (change in velocity)/(time). I found the time via using the initial velocity, 4m/s and using the given distance of 2m: so:

2m * 1/(4m/s) giving a total of 0.5 seconds. ---> plug this into the definition of acceleration:
(Vf-Vi which is 4m/s) / (time which was just solved above 0.5sec)


Why won't this method work? The question stem states that it is a uniform deceleration.

Is the answer 320N?

If so,

use this equation: Vf^2 = Vi^2 + 2ax to find acceleration since it's the only thing you need to find the force (ma).

Vf is zero, so rearrange the equation to look like this: Vi^2 = 2ax (negative signs of a Vi cancels out with that of a, since a represents deceleration)

This will give you 4 for the acceleration. Plug and chug into F=ma and you'll get the answer.
 
Vavg= d/t
2m/s= 2m/t
t=1s

A=(vf-vi)/t
A=(0-4m/s)/1s
A= -4m/s/s. negative is decel

F=ma
F=80kg*4m/s/s
F=320 N

That's how I would do it if I forgot the relevant kinematic equation, otherwise just do what ibn alnafis did.
 
Vavg= d/t
2m/s= 2m/t
t=1s

A=(vf-vi)/t
A=(0-4m/s)/1s
A= -4m/s/s. negative is decel

F=ma
F=80kg*4m/s/s
F=320 N

That's how I would do it if I forgot the relevant kinematic equation, otherwise just do what ibn alnafis did.

I like your way better
 
I did not know you were the first to describe the pulmonary circulation of the blood. Kool stuff
 
I don't understand why we're using his body length as displacement? Wouldn't that imply that he travelled 2m downwards after impacting the ground?
 
In this particular question:



I understand that basically we need to plug and chug to:

1) Find the acceleration
2) plug into Force = mass * accel

I'm not sure why my method doesn't work when I attempt to solve for acceleration.

The answer key solved for acceleration using the VAX kinematics eq Vf^2 = Vi^2 +2a(change in distance)

However, I thought that I could solve using the definition of acceleration which is:
acceleration = (change in velocity)/(time). I found the time via using the initial velocity, 4m/s and using the given distance of 2m: so:

2m * 1/(4m/s) giving a total of 0.5 seconds.
---> plug this into the definition of acceleration:
(Vf-Vi which is 4m/s) / (time which was just solved above 0.5sec)


Why won't this method work? The question stem states that it is a uniform deceleration.

The bolded would be correct only for uniform (a=0) motion. You can see the rest of the posts for the proper solution.
 
d = ave V . t so t = 2/2 = 1 sec
a = V / t = 4 /1 = 4 m/s per second
F = ma = 80 x 4 = 320 N
 
Still don't understand their explanation of d. He's not travelling 2m after he hits the ground.

Yes, they have oversimplified that part of the problem. They assume that at the end of the impact his body is completely leveled with the ground. Using the center of mass is a better estimation but still estimation, since as the body "flattens", the center of mass moves up along the not-yet-flattened part of it.

Since they tell you what estimation to make as part of the question, I don't think that being somewhat off here is a big deal.
 
Yes, they have oversimplified that part of the problem. They assume that at the end of the impact his body is completely leveled with the ground. Using the center of mass is a better estimation but still estimation, since as the body "flattens", the center of mass moves up along the not-yet-flattened part of it.

Since they tell you what estimation to make as part of the question, I don't think that being somewhat off here is a big deal.

:shrug: I didn't get that from the question. I would've probably done it eventually out of frustration, having no other possible avenues, but the way the question was phrased didn't make me think "object travelled 2m" at ALL. I'm not worried about accuracy, I just think the question was confusing.
 
:shrug: I didn't get that from the question. I would've probably done it eventually out of frustration, having no other possible avenues, but the way the question was phrased didn't make me think "object travelled 2m" at ALL. I'm not worried about accuracy, I just think the question was confusing.

"If he decelerates uniformly over the entire 2-meter length of his body"

What else would you use? :shrug:
 
"If he decelerates uniformly over the entire 2-meter length of his body"

What else would you use? :shrug:

Nothing...that's why eventually I guess I'd just do it that way...but to me that phrasing just implies that his entire body decelerated uniformly, not that he moved 2 meters. Honestly, I took it as a way of saying that he didn't absorb the impact with his knees.

I can't even imagine any way for someone to collapse and move the same distance as their height. First of all, as you said, it'd be his center of mass. Second, how the hell do you collapse through yourself? I can put my head on the ground, sure, but not in a straight line. I dunno, I suppose that, when I'm considering the person as a solid mass (no bending of limbs, one homogenous mass, etc) I picture any movement as movement of the entire mass, not a contraction of it.

I just think it's a crappily-worded question. I'd eventually get the answer, but only by answering what I guessed they meant to ask, not what they really asked...and I hate that feeling.
 
Still don't understand their explanation of d. He's not travelling 2m after he hits the ground.

The equation is not d=v/t. The equation is d=(average v)/t. Of course when acceleration is zero the the second equation devolves into the first one, and students memorize the wrong equation, and then they get all confused.
 
The equation is not d=v/t. The equation is d=(average v)/t. Of course when acceleration is zero the the second equation devolves into the first one, and students memorize the wrong equation, and then they get all confused.

That's...not relevant at all to what I was saying. My point was simply that the phrase "decelerates uniformly over the entire 2-meter length of his body" would never, in reality, translate into "his mass travelled 2m". It's actually a pretty nonsensical phrase. His center of mass can't move 2m without burrowing into the ground, and even his head cannot continue downwards for 2m. When I first read that question and tried to picture the situation, I ran about 80 other possibilities through my head first before giving up and going with this, because the question is just really, really, terribly worded.
 
Youre overcomplicating a very simple question. There aren't 80 other possibilities, there's 2-3 valid possibilities if you keep your head on point and realize that they can only ask you certain types of questions. The center of mass argument is not going to significantly change the value uch a short deceleration distance, as such the impact force for the specified paramaters is relatively accurate.
 
I don't mean mathematical possibilities. I mean picturing it in my head. Jeez people, I understand the math and the question, I just think the wording is completely stupid. That is NOT an accurate description in ~3 different ways, and those ways are not the typical "ignore xyz complication" ways in which kinematics problems are often simplified, they're "pretend that when you picture a person decelerating after hitting the ground foot-first, you automatically assume that they end up with their head on the ground (lolno), then pretend the center of mass is on their scalp (not just a calculation problem, a conceptualization one too) and make weird assumptions about the way their body crumples which bring to mind ragdolls from online videogames rather than any living human you've ever seen...oh, and assume all of this from some pretty vague wording, because obviously it's the first thing you'd picture when presented with a scenario that's not clearly defined."

Also, I don't know what your standards are for "significantly different" but in my view, a factor of two is significant. If d is halved, a is doubled...if a is doubled, F is doubled. I'd call that significant :shrug: But that's an irrelevant tangent anyway because I was never once calculating it based on CoM, I just had a hard time visualizing it because I would never consider the movement of the head = to the movement of the person, which is what the 'correct' interpretation of the question implies.
 
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I guess the way I thought about it was that the parachuter lands on his feets, his knees bend and his body tucks as he decelerates. I guess i'm just having a hard time trying to explain all the things that are wrong conceptually with the question, because really, all the questions generally are much more complicated that our basic 100 level physics allows us to interpret, so that is why I don't even think about the complexities that are involved in a real-life scenario.

I guess my only advice in order to not get bogged down on these easy questions, is to realize it is just basic first year science being tested, and that it's going to be unlikely(not impossible) that the question will be some obscure complexity that basic physics doesn't usually test. Not to say things can't be compounded with other physics topics though.
 
I guess the way I thought about it was that the parachuter lands on his feets, his knees bend and his body tucks as he decelerates. I guess i'm just having a hard time trying to explain all the things that are wrong conceptually with the question, because really, all the questions generally are much more complicated that our basic 100 level physics allows us to interpret, so that is why I don't even think about the complexities that are involved in a real-life scenario.

I guess my only advice in order to not get bogged down on these easy questions, is to realize it is just basic first year science being tested, and that it's going to be unlikely(not impossible) that the question will be some obscure complexity that basic physics doesn't usually test. Not to say things can't be compounded with other physics topics though.

I wasn't getting bogged down; I was just complaining about bad wording.

Usually the questions compensate for our lack of complexity by simplifying their situations - i.e. "assume that he does not bend his knees upon landing" or "assume xyz ridiculous assumption that at least I can picture".

This one just made no sense; I wasn't trying to complicate it, I just wouldn't take their statement as an expression of distance travelled. It is my personal opinion that this question is one of the most crappily worded ones of its kind which I have encountered, which is saying something given that I've spent the last few weeks doing every question in the back of the kinematics chapters in my textbook.
 
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