How Do You Derive the Distance Traveled by a Projectile in a Viscous Liquid?

[romiomustdie]

i have found out the second answer of the PHYSICS problem but unable to solve the first part.i tried all possible ways.:-(
----------------------------------------------
a test projectile is fired horizontally into a viscous liquid with a velocity Vo . The retarding force is proportional to the square of the velocity, so that the accelaration becomes , a=-kV^2.

Derive an expression for the distance D traveled in the liquid and the corrosponding time required to reduce the velocity to Vo/2.Neglect any vertical motion.

Ans. 1. D=0.693/K
2. t=1/(kVo).(I solved for the second answer, but am unable to solve for the first one...)

 

[Bystander]

i have found out the second answer of the PHYSICS problem but unable to solve the first part.i tried all possible ways.:-(
----------------------------------------------
a test projectile is fired horizontally into a viscous liquid with a velocity Vo . The retarding force is proportional to the square of the velocity, so that the accelaration becomes , a=-kV^2.

Derive an expression for the distance D traveled in the liquid and the corrosponding time required to reduce the velocity to Vo/2.Neglect any vertical motion.

Ans. 1. D=0.693/K
2. t=1/(kVo).

What have you tried?

 

[romiomustdie]

a= - kVo^2

Since a=dv/dt

So,
Vo/t = - kVo^2
or, t= - 1/(kVo)
or, t = 1/(kVo) , as time can't be negative.

 

[Bystander]

... and, for d?

 

[Chestermiller]

a= - kVo^2

Since a=dv/dt

So,
Vo/t = - kVo^2
or, t= - 1/(kVo)
or, t = 1/(kVo) , as time can't be negative.

This solution to the differential equation is incorrect. The differential equation reads:

$$\frac{dV}{dt}=-kV^2$$

Note that there are V's on both sides of the equation. Do you know how to solve such an equation? If so, let's see.

Chet

 

[romiomustdie]

sorry then its totally wrong i don't know what to do now...please advice ,,, the first part to find distance feels impossible to find , getting no clues where to start

 

[romiomustdie]

and tomorrow is my exam , god knows if this comes in exam , its in the suggestion which means it can come in exam , i am striving with this math for four days ...... :-(

 

[Chestermiller]

and tomorrow is my exam , god knows if this comes in exam , its in the suggestion which means it can come in exam , i am striving with this math for four days ...... :-(

Are you taking, or have you taken a course in differential equations? If so, then you know you can solve the differential equation in post #5 using "separation of variables."

Chet

 

[romiomustdie]

i have just started with calculas in class , and have not got into depth yet so may be that's why not able to solve it , but how will i find distance using the differential equation
. i am not taking any course but i am in std X

 

[romiomustdie]

I have to find the distance but how ? ?

 

[romiomustdie]

normal v^2 = u^2 - 2aS , is not working .
where v= final velocity
u= initial velocity
a=accelaration
S= distance

what to do ? ? ? :-(

 

[romiomustdie]

dv/dt=- k v^2
or, 1/v^2 dv = -k dt
or, v^ -2 dv = - k dt
integrating both sides we get
v^(-2+1)/-2+1 = - kt
or,v^-1 = kt
or, t = 1/kv

if v = Vo
then , t = 1/kVo

this is the second part i think but what about the 1st part . i am so much in dispair ...what to do please help ?

 

[CWatters]

normal v^2 = u^2 - 2aS , is not working .
where v= final velocity
u= initial velocity
a=accelaration
S= distance

what to do ? ? ? :-(

That equation only applies when acceleration is constant which isn't the case in this problem.

See posts #5 and #8 and look up "separation of variables" as Chester suggests.

 

[Chestermiller]

dv/dt=- k v^2
or, 1/v^2 dv = -k dt
or, v^ -2 dv = - k dt
integrating both sides we get
v^(-2+1)/-2+1 = - kt
or,v^-1 = kt

This equation is incorrect. You forgot to include the constant of integration. You can determine the constant of integration by making use of the initial condition, v = Vo at t = 0.
So, using this, what equation do you get for the velocity?

Chet

 

[romiomustdie]

Could you find the distance?

 

[Chestermiller]

Could you find the distance?

You would have to next integrate the velocity to get the distance.

 

[romiomustdie]

 

[romiomustdie]

Sorry I did mistake

 

[romiomustdie]

Sorry I did mistake

It should be as follows

 

[romiomustdie]

 

[Chestermiller]

It should be as follows

Before you integrate the velocity to get the distance, you first need to get the equation for the velocity correct. So far it's not correct. See post #14.

Chet

 

[romiomustdie]

View attachment 89938

 

[romiomustdie]

Ya I had mistaken first bt how about the last upload

 

[Chestermiller]

Ya I had mistaken first bt how about the last upload

In that last upload, you starting equation is still incorrect. Did you upload the wrong file?

 

[romiomustdie]

oh i got it u mean to say the derived equation for velocity is incorrect as i ignored constant of integration . but taking v=Vo , will not the constant of integration be equal to 0

 

[romiomustdie]

or else how it should be ? please state

 

[Chestermiller]

or else how it should be ? please state

The problem statement says that the velocity is V₀ at time t = 0. You need to make use of this fact in your equation for the velocity. I don't see any V₀ in your equation for velocity.

Chet

 

[romiomustdie]

 

[romiomustdie]

View attachment 89940

Sorry the image turned

 

[Chestermiller]

Sorry the image turned

I can't read it. The equation for the velocity is:$$V=\frac{V_0}{(1+V_0kt)}$$