Variable acceleration and terminal velocity

In summary, the conversation discusses the problem of calculating the terminal velocity and time of fall for a cube falling from a height of 200m with given information on its mass, cross-sectional area, and coefficient of drag. The conversation includes attempts at solving the problem using calculus and the realization that a function of time is needed. The correct formula for net acceleration is discussed and it is mentioned that velocity as a function of distance is necessary to calculate time of fall. The conversation concludes with a reminder to get enough sleep before taking a physics test.
  • #1
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4

Homework Statement



A cube is falling from a height of 200 m (initial velocity =0) , it reaches it's terminal velocity . Given that :its mass is 100 kg , cross sectional area 40 cm2 and coefficient of drag 2. Putting only weight and drag in consideration , calculate its terminal velocity and time of the fall (g=10)

Homework Equations



W=mg , Fd= Cd*A*1/2*ρ*v2 ( ρ
=1)

The Attempt at a Solution


it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , Fr=W-FD, after working it out , it was something like this , a=(100)/(1000-(0.08 v2) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time
 
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  • #2
abdo799 said:
it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , Fr=W-FD, after working it out , it was something like this , a=(100)/(1000-(0.08 v2) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time

$$a=\frac{dv}{dt}=\frac{dv}{dt}.\frac{dx}{dx}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dx}{dt}.\frac{dv}{dx}=v.\frac{dv}{dx}$$

Now, you have relation with which you don't need a function of time. And acceleration is a function of velocity.

Edit: check your acceleration again, it should be a little different.
 
  • #3
abdo799 said:

Homework Statement



A cube is falling from a height of 200 m (initial velocity =0) , it reaches it's terminal velocity . Given that :its mass is 100 kg , cross sectional area 40 cm2 and coefficient of drag 2. Putting only weight and drag in consideration , calculate its terminal velocity and time of the fall (g=10)

Homework Equations



W=mg , Fd= Cd*A*1/2*ρ*v2 ( ρ
=1)

The Attempt at a Solution


it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , Fr=W-FD, after working it out , it was something like this , a=(100)/(1000-(0.08 v2) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time

Recheck value of Fd.

Your expression of net acceleration also looks incorrect .

mdv/dt = Fd-mg .

Substituting value of Fd ,you will get mdv/(Fd-mg) =dt .

Integrate with proper limits and you will get velocity as a function of time.What do you get ?
 
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  • #4
Tanya Sharma said:
Integrate with proper limits and you will get velocity as a function of time.What do you get ?
He'll need to calculate velocity as a function of distance first, otherwise "velocity as a function of time" information would be useless.

But yes, he does need to calculate "velocity as a function of time".
 
  • #5
NihalSh said:
$$a=\frac{dv}{dt}=\frac{dv}{dt}.\frac{dx}{dx}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dx}{dt}.\frac{dv}{dx}=v.\frac{dv}{dx}$$

Now, you have relation with which you don't need a function of time. And acceleration is a function of velocity.

Edit: check your acceleration again, it should be a little different.

okay , this is how i got a
F= m/a
100/a = (100*10)-(1/2 * 1 * 2 * (0.42) v2)
100/a= 1000- (0.16 v2)
a=100/(1000-0.16v2)
if it's wrong please correct me
 
  • #6
NihalSh said:
He'll need to calculate velocity as a function of distance first, otherwise "velocity as a function of time" information would be useless.

But yes, he does need to calculate "velocity as a function of time".

I believe distance doesn't play a role in calculating terminal velocity ,unless you want to determine if the object achieves terminal velocity before reaching the ground.
 
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  • #7
abdo799 said:
okay , this is how i got a
F= m/a
100/a = (100*10)-(1/2 * 1 * 2 * (0.42) v2)
100/a= 1000- (0.16 v2)
a=100/(1000-0.16v2)
if it's wrong please correct me

Hey, ##F=m*a##.

Tanya already posted about this!...understanding the situation is more important. But here is the result:

taking +ve direction to be downwards, we have:

$$mg-F_{d}=ma$$
$$a=g-\frac{F_{d}}{m}$$
Substitute the correct values, you'll get acceleration as a function of velocity.
 
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  • #8
Tanya Sharma said:
I believe distance doesn't play a role in calculating terminal velocity ,unless you want to determine if the object achieves terminal velocity before reaching the ground.

terminal velocity is just the condition when the object is in equilibrium. ##F_{d}=mg## for the terminal velocity in this question. But the question asks the time of fall, not the time to reach terminal velocity. Time of fall is surely enough, dependent on the distance.

I hope you get the point!

Edit: Yes, distance doesn't play important role here to calculate terminal velocity. But I was talking about velocity in my previous posts. And its time of fall depends on velocity and not terminal velocity. But to calculate final velocity and hence calculate time, we need "velocity as a function of distance"......to reach 100% of terminal velocity you need infinite/very large time, because velocity reaches terminal velocity asymptotically with time.
 
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  • #9
NihalSh said:
Hey, ##F=m*a##.

Tanya already posted about this!...understanding the situation is more important. But here is the result:

taking +ve direction to be downwards, we have:

$$mg-F_{d}=ma$$
$$a=g-\frac{F_{d}}{m}$$
Substitute the correct values, you'll get acceleration as a function of velocity.

worst mistake I've ever made , sorry , i drank like 5 cups of coffee , didnt sleep , got a test in 2 days
 
  • #10
abdo799 said:
worst mistake I've ever made , sorry , i drank like 5 cups of coffee , didnt sleep , got a test in 2 days

get some sleep, sleepiness and physics don't go hand in hand!:tongue:
 
  • #11
NihalSh said:
get some sleep, sleepiness and physics don't go hand in hand!:tongue:

Will do, thanks
 
  • #12
abdo799 said:
Will do, thanks

:thumbs:
 

1. What is variable acceleration?

Variable acceleration is the change in an object's velocity over time. This means that the object's speed is not constant and is either increasing or decreasing as time passes.

2. How does variable acceleration affect an object's motion?

Variable acceleration can cause an object to speed up, slow down, or change direction. This is because the object's velocity is changing, so its motion is also changing.

3. What is terminal velocity?

Terminal velocity is the constant maximum velocity that an object reaches when falling through a fluid, such as air or water. This occurs when the force of air resistance equals the force of gravity on the object.

4. How does terminal velocity relate to variable acceleration?

When an object is falling through a fluid, it initially experiences variable acceleration as gravity pulls it downwards and air resistance slows it down. However, as the object's speed increases, the force of air resistance also increases until it equals the force of gravity and the object reaches its terminal velocity.

5. Can an object have variable acceleration and still reach terminal velocity?

Yes, an object can have variable acceleration and still reach terminal velocity. This can occur when an object is falling through a fluid with changing density or when the object's shape changes, causing the amount of air resistance to vary. However, once the object reaches terminal velocity, its acceleration will be zero, as its speed is no longer changing.

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