# Dynamic problem

1. May 25, 2007

### hastings

1. The problem statement, all variables and given/known data
an object with m=100kg is moving on a straight line. for t=0 its speed is v0=20 m/s. There's a force which is opposing to the movement F=C*v^2 ( C=1kg/m and v is the speed (v(t)=...) ). Find the time t1 in which the speed v=v0/2.

2. Relevant equations
F=m*a (?)
v(t)=a*t+v0

3. The attempt at a solution
v1=a*t1+v0 --> v0/2=a*t1+v0 --> t1=(-v0/2)*1/a

assuming a (accelation) constant
m*a = F = C*v1^2= C*(v0/2)^2
a= [C*(v0/2)^2]/m
Putting a in
t1= (-v0/2)/a should be getting the answer

Unfortunately the answer I get is not correct (should be 5second)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 25, 2007

### cepheid

Staff Emeritus
The acceleration is not constant because the force is not constant. Think about it. The drag force is proportional to the square of the velocity.This means that as the object slows down (due to the drag force), the drag force itself decreases. So not only does it decelerate, but it decelerates at an ever decreasing rate. You need a new way to approach this problem. Using the dynamical equations derived for the special case of motion under a constant force doesn't cut it here.

3. May 25, 2007

### hastings

then how to solve it?

4. May 25, 2007

### arunbg

Big hint: Think Newton's Second Law and think calculus....

5. May 25, 2007

### hastings

F=m*a=C*v^2
so a=(C*v^2)/m

a=dv/dt ==> v=integral ( (C*v^2)/m )= C*v^2/m * t

so then

t= (v*m)/(C*v^2)=m/(C*v)

I then say
t0=0
t1 -t0=t1
t0=m/(C*v0)
t1= 2m/(C*v0)
t1= t1-t0= m/(C*v0)=5 seconds

6. May 25, 2007

### arunbg

Umm..not quite.
v is actually dependent on t, and so you can't treat it as a constant.
You should write
$$\frac{dv}{dt}=Cv^2/m$$

$$\frac{dv}{v^2}=Cdt/m$$
And integrate on both sides.

And don't forget the constant of integration whose value can be determined with the given data.

7. May 25, 2007

### hastings

$$\int{\frac{1}{v^2}}dv$$? How to do it? I dunno
Maybe $$-\frac{1}{v}$$?

Then what about $$\int \frac{C}{m}dt= \frac{C}{m} \cdot t+ c_0$$?

Last edited: May 25, 2007
8. May 25, 2007

### arunbg

You are perfectly correct, why did the doubts arise?
Anyway can you proceed now?

9. May 25, 2007

### hastings

So is it ....

$$-\frac{1}{v}+v_0=\frac{C}{m} \cdot t$$

$$t=\frac{m}{C} \cdot ( -\frac{1}{v}+v_0)$$ ?

Then I substitute $$v=\frac{v_0}{2}$$ in the last expression and I should get the answer, right?
result should be 5 seconds.
Let's see...
Well...it's not right

Last edited: May 25, 2007
10. May 25, 2007

### arunbg

I notice that in the question, it is mentioned that the mass is decelarated. This means you should assign a negative value to acceleration.

Not quite, v_0 refers to velocity at time t=0.

Integrating, you should get
-1/v + K(constant of integration) = -Ct/m

Can you find K?

11. May 25, 2007

### hastings

Thought it was K=v0; wouldn't know How can I get it ?

12. May 25, 2007

### arunbg

At t=0, v=20. For LHS=RHS, K= ....

13. May 25, 2007

### hastings

What's LHS=....?
Never heard
Is it K=20? but that's what i said previously: K=v0=20

14. May 25, 2007

### arunbg

LHS - Left hand side RHS - Right hand side (for the equation -1/v + K(constant of integration) = -Ct/m) )
What I'm saying is put t=0 and v=20 in this equation and find K.

15. May 25, 2007

### cepheid

Staff Emeritus
LHS - left-hand side (of the equation), that's all

16. May 26, 2007

### hastings

so k=1/20
$$-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t$$

Is it right?
Now I substitute t=t1 and v=v0/2 in the expression?

$$-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t1$$

$$t1=(-\frac{m}{C}) \cdot (-\frac{1}{v} +\frac{1}{20}) =$$

$$=(-\frac{m}{C}) \cdot (-\frac{2}{v_0}+\frac{1}{20})=$$

$$=(-\frac{m}{C}) \cdot (\frac{-2 \cdot 20 + v_0}{20 \cdot v_0})=$$

$$=(-\frac{m}{C}) \cdot (\frac{-40+v_0}{20 \cdot v_0})=$$

$$=(-100) \cdot (\frac{-40+20}{20 \cdot 20})=$$

$$=(-100) \cdot (\frac{-20}{400})= 5 \mbox{ seconds }$$

OK! GREAT! Thank you! :rofl:

Last edited: May 26, 2007