# Engineering kinematics

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1. Oct 4, 2015

### romiomustdie

i have found out the second answer of the PHYSICS problem but unable to solve the first part.i tried all possible ways.:-(
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a test projectile is fired horizontally in to 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 travelled 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...)

Last edited by a moderator: Oct 4, 2015
2. Oct 4, 2015

### Bystander

What have you tried?

3. Oct 4, 2015

### 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.

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4. Oct 4, 2015

### Bystander

... and, for d?

5. Oct 4, 2015

### Staff: Mentor

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

6. Oct 6, 2015

### romiomustdie

sorry then its totally wrong i dont know what to do now...plz advice ,,, the first part to find distance feels impossible to find , getting no clues where to start

7. Oct 6, 2015

### 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 .......................... :-(

8. Oct 6, 2015

### Staff: Mentor

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

9. Oct 6, 2015

### romiomustdie

i have just started with calculas in class , and have not got in to depth yet so may be thats 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

10. Oct 6, 2015

### romiomustdie

I have to find the distance but how ? ?

11. Oct 6, 2015

### romiomustdie

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

what to do ? ? ? :-(

12. Oct 6, 2015

### 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 plz help ?????

13. Oct 6, 2015

### CWatters

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.

14. Oct 6, 2015

### Staff: Mentor

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

15. Oct 8, 2015

### romiomustdie

Could you find the distance?

16. Oct 8, 2015

### Staff: Mentor

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

17. Oct 9, 2015

### romiomustdie

18. Oct 9, 2015

### romiomustdie

Sorry I did mistake

19. Oct 9, 2015

### romiomustdie

It should be as follows

20. Oct 9, 2015

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