Trying to understand why I got this problem wrong

[Bruh]

Homework Statement

So the problem states: "A skydiver jumps from a high-flying plane. As her velocity of fall increases, her acceleration:
A.) Increase B.) Decreases C.) Remains unchanged regardless of air resistance.

Relevant Equations

The only equations provided for this situation are:
A = Change in V/Time interval & V = A * Time(only in free fall motion)

I selected A.) but the answer is B.)

My logic was this: V1 = 10m/s, T = 5s, so A = 2
V2 = 20m/s, T = 5s, so A = 4

Am I applying the wrong equation to the situation, or simply what am I getting wrong?

 

[hmmm27]

You chose the simplistic parameters of basic physics texts.

[edit: true, but what I missed was that the OP answered the question wrongly, stating that acceleration would increase]

The question appears to be specifically requiring something more realistic.

As the fall of a skydiver progresses, does downwards acceleration :
a) increase,
b) decrease,
c) stay the same.

May as well include "why ?"

 

[Bruh]

You chose the simplistic parameters of basic physics texts. The question appears to be specifically requiring something more realistic.

As the fall of a skydiver progresses, does downwards acceleration :
a) increase,
b) decrease,
c) stay the same.

May as well include "why ?"

I'm unsure what you mean. It was a question on one of my previous exams, and I need to understand why I got it wrong because I am sure a similar question will be on the final. :/

 

[PAllen]

Your computation is wrong, even in a vacuum. However, the wording of choice C suggests you are supposed to think about air resistance. What does that suggest?

 

[Bruh]

Your computation is wrong, even in a vacuum. However, the wording of choice C suggests you are supposed to think about air resistance. What does that suggest?

I guess it means that they would be reaching terminal velocity and that would make sense, but I considered each of the situations absent of air resistance. Can you explain where I went wrong computationally?

 

[hmmm27]

I guess it means that they would be reaching terminal velocity and that would make sense, but I considered each of the situations absent of air resistance. Can you explain where I went wrong computationally?

Firstly, my bad for not noticing that your answer was wrong on another level.

You chose option 'A'. Why ?

 

[haruspex]

My logic was this: V1 = 10m/s, T = 5s, so A = 2
V2 = 20m/s, T = 5s, so A = 4

Not sure how you are getting those numbers. If the acceleration is $2m/s^2$ then after 5s the velocity will be 10m/s. After a further 5s the velocity will be 20m/s. The acceleration in the second 5s is the increase in velocity divided by the duration, (20-10)/5=2.
You seem to have divided the velocity gain over 10s by a duration of 5s.

 

[Bruh]

Firstly, my bad for not noticing that your answer was wrong on another level.

You chose option 'A'. Why ?

I realize now I was thinking about the question wrong in 2 ways. I didn't take into account air resistance because I thought that C made it a nonfactor for the other answers(just didn't take my time to think it through), and I also was considering 2 different people falling at 2 different velocities (V1/V2) but it was really just one scenario. I chose A because I applied their 2 initial velocities and didn't take the difference of one persons initial velocity and ending velocity. Right? Could I just assume that once they reach terminal velocity their acceleration would inevitably be 0, so the answer is B without any math?

 

[Bruh]

Not sure how you are getting those numbers. If the acceleration is $2m/s^2$ then after 5s the velocity will be 10m/s. After a further 5s the velocity will be 20m/s. The acceleration in the second 5s is the increase in velocity divided by the duration, (20-10)/5=2.
You seem to have divided the velocity gain over 10s by a duration of 5s.

I realize now I was thinking about the question wrong in 2 ways. I didn't take into account air resistance because I thought that C made it a nonfactor for the other answers(just didn't take my time to think it through), and I also was considering 2 different people falling at 2 different velocities (V1/V2) but it was really just one scenario. I chose A because I applied their 2 initial velocities and didn't take the difference of one persons initial velocity and ending velocity. Right? Could I just assume that once they reach terminal velocity their acceleration would inevitably be 0, so the answer is B without any math?

 

[haruspex]

Could I just assume that once they reach terminal velocity their acceleration would inevitably be 0, so the answer is B without any math?

That works, but it does mean you have to use the knowledge that there is a terminal velocity, rather than deducing the answer from the physics.

 

[Bruh]

That works, but it does mean you have to use the knowledge that there is a terminal velocity, rather than deducing the answer from the physics.

That is true. I have a lot of material to review, so as long as I can have some way of understanding it, then its better than nothing :) Thanks for the help brother!

 

[PAllen]

Have you ever learned that g, in equations like $s=-(1/2)gt^2+v_0t+s_0$ is taken to be a constant (a simplification valid for distances less than many kilometers near earth)? This answers the question without air resistance. Have you ever seen an equation like $F_{drag} \propto -v$ , for air resistance? (This is an approximation ignoring turbulence, useful for many common situations). Without any detailed math, this is sufficient to answer the question.

 

[Bruh]

Have you ever learned that g, in equations like $s=-(1/2)gt^2+v_0t+s_0$ is taken to be a constant (a simplification valid for distances less than many kilometers near earth)? This answers the question without air resistance. Have you ever seen an equation like $F_{drag}=-v$ , for air resistance? (This is an approximation ignoring turbulence, useful for many common situations). Without any detailed math, this is sufficient to answer the question.

Oh lord, nah I haven't learned those concepts yet. It is a conceptual physics course, but only an introductory one. Thank you for pointing out the problem I had with the air resistance. That really helped a lot!

 

[jbriggs444]

Oh lord, nah I haven't learned those concepts yet. It is a conceptual physics course, but only an introductory one. Thank you for pointing out the problem I had with the air resistance. That really helped a lot!

At the conceptual level, you do not need anything as fancy as a proportionality. All you need is "the faster you go, the greater air resistance becomes".

Then ask yourself whether air resistance increases your downward acceleration or decreases it.