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Terminal velocity explanation?

How does terminal velocity work, like the process?

This usually comes up as 6 markers
Feel free to use examples:u:
Reply 1
Original post by tamanna___
How does terminal velocity work, like the process?

This usually comes up as 6 markers
Feel free to use examples:u:


Normally, you are given the example of a sky diver:

1) As soon as the parachutist jumps, only weight acts, so no drag and they are accelerating downwards at the acceleration of freefall (9.81ms2). Resultant force is not zero, and it is downwards.

2) Later in the jump, they are still accelerating downwards but at an acceleration less than 9.81ms2 as drag is acting now. Weight > Drag so they still move downwards with a resultant force which is also downwards.

3) When they first open the parachute, the large surface area increases the drag force enourmously, so they accelerate upwards, with drag > weight, and resultant force now upwards.

4) A while after opening the parachute, the drag = weight which means they are at terminal velocity as the net force is 0, because the 2 forces are equal and they are travelling at a constant speed in the downward direction.

So, terminal velocity occurs when the drag force is equal to the weight, when the net force is 0 (no acceleration, constant velocity).

Any questions, feel free to ask:smile:
Reply 2
Original post by derpz
Normally, you are given the example of a sky diver:

1) As soon as the parachutist jumps, only weight acts, so no drag and they are accelerating downwards at the acceleration of freefall (9.81ms2). Resultant force is not zero, and it is downwards.

2) Later in the jump, they are still accelerating downwards but at an acceleration less than 9.81ms2 as drag is acting now. Weight > Drag so they still move downwards with a resultant force which is also downwards.

3) When they first open the parachute, the large surface area increases the drag force enourmously, so they accelerate upwards, with drag > weight, and resultant force now upwards.

4) A while after opening the parachute, the drag = weight which means they are at terminal velocity as the net force is 0, because the 2 forces are equal and they are travelling at a constant speed in the downward direction.

So, terminal velocity occurs when the drag force is equal to the weight, when the net force is 0 (no acceleration, constant velocity).

Any questions, feel free to ask:smile:


Thankyou! How would you explain it for a cyclist cycling? That question has come up once.. How would you start it off?☺️
Reply 3
Original post by tamanna___
Thankyou! How would you explain it for a cyclist cycling? That question has come up once.. How would you start it off?☺️


Is it a cyclist travelling horizontally, or downhill?
Reply 4
Original post by derpz
Is it a cyclist travelling horizontally, or downhill?


This is the question
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Reply 5
Original post by tamanna___
This is the question
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Im just unsure about what the drag force would be? Increase in wind or something?
Original post by derpz
Im just unsure about what the drag force would be? Increase in wind or something?


Air resistance is typically proportional to velocity (or at least modelled that way), so the faster you go the stronger the drag. Eventually, the drag = force from pedalling and you've hit terminal velocity.
(edited 7 years ago)
Reply 7
Solve the DE mx¨=mgkvm\ddot{x} = mg - kv for x˙\dot{x} and take the limit as tt \to \infty.
Reply 8
Original post by Implication
Air resistance is typically proportional to velocity (or at least modelled that way), so the faster you go the stronger the drag. Eventually, the drag = acceleration from pedalling and you've hit terminal velocity.


Yeah, I just never came across this example but this makes sense, so gradually, the acceleration would decrease until the air resistance is equal to the force provided by the pedals by the rider...correct?
Original post by derpz
Yeah, I just never came across this example but this makes sense, so gradually, the acceleration would decrease until the air resistance is equal to the force provided by the pedals by the rider...correct?


Exactly :smile: It would work the same way in the case of a skydiver before their parachute deploys - they still hit terminal velocity, it's just a much higher terminal velocity than without the parachute!
Reply 10
Original post by derpz
Yeah, I just never came across this example but this makes sense, so gradually, the acceleration would decrease until the air resistance is equal to the force provided by the pedals by the rider...correct?


You've worded it a bit weirdly but, essentially - yeah, the air resistance equals the driving force which then means no resultant and hence zero acceleration.
Reply 11
Original post by Implication
Exactly :smile: It would work the same way in the case of a skydiver before their parachute deploys - they still hit terminal velocity, it's just a much higher terminal velocity than without the parachute!


Original post by Zacken
You've worded it a bit weirdly but, essentially - yeah, the air resistance equals the driving force which then means no resultant and hence zero acceleration.


Got the hang of it, I was just unsure what the actual drag force would be. Thanks! :smile:
Original post by Implication
Air resistance is typically proportional to velocity (or at least modelled that way), so the faster you go the stronger the drag. Eventually, the drag = force from pedalling and you've hit terminal velocity.


Original post by derpz
Yeah, I just never came across this example but this makes sense, so gradually, the acceleration would decrease until the air resistance is equal to the force provided by the pedals by the rider...correct?


Original post by Zacken
You've worded it a bit weirdly but, essentially - yeah, the air resistance equals the driving force which then means no resultant and hence zero acceleration.


Thankyou all for helping!! I think i understand it better now

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