# Differentiation Problem

1. May 15, 2004

### mcintyre_ie

I’m having some difficulty with this question:

(b) A particle starts with a speed of 20 m/s and moves in a straight line. The particle is subjected to a resistance which produces a retardation which is initially 8 m/s^2 and which increases uniformly with the distance moved, having a value of 9 m/s^2 when the particle has moved a distance of 5 m.
If v m/s is the speed of the particle when it has moved a distance x m:

(i) prove that, while the particle is in motion,

v dv/dx = - ( 8 + x/5 )

(ii) Calculate the distance moved by the particle in coming to rest

I can’t seem to get out part (i), which is probably key to solving part (ii). I’m having problems with the force equation, more specifically with the resistance increasing with time/distance.

Thanks in advance for any help.

2. May 15, 2004

### jcsd

express v in terms of x.

edited: sorry that's a rather silly way to go about it. What does d2v/dx2 equal? so what does dv/dx equal?

Last edited: May 15, 2004
3. May 15, 2004

### mcintyre_ie

The way that I usually do it is to get a force equation for the question first... then work out dv/dx etc. Any help with actually getting a force equation?

4. May 15, 2004

### jcsd

I'll give you a start: d2v/dx2 is going to be linear (the question tells you this), therefore d2v/dx2 = k (k = a constant), what is k?

5. May 16, 2004

### mcintyre_ie

I’m still not getting it out… am I supposed to be using equations of motion to get a, which is equal to v(dv/dx) and (d^2x)/(dx^2)? I’ve been trying to do it that way, but I’m still getting nowhere. I still can't figure out where the 8m/s^2 and 9m/s^2 retardations come in. I’d be grateful for a little more help.

6. May 16, 2004

### arildno

Hint:
The acceleration of an object is given by $$\frac{dv}{dt}$$.
Assuming that the velocity may be written as a function of the position x(t), we have, by the chain rule:
$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$, since by definition of velocity, we have $$v=\frac{dx}{dt}$$

7. May 16, 2004

### jcsd

Okay, a = dv/dx,

As I siad before we know d2v/dx2 = a constant as the question tells us this:, therefore from the question we can work out that: d2v/dx2 = [(-9) - (-8)]/5 = -1/5

So now integrate to find dv/dx

8. May 16, 2004

### arildno

$$a\neq\frac{dv}{dx}$$, since the dimensions are unequal.
Rather, we have:
$$v(x(t))=\frac{dx}{dt}\rightarrow{a}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

9. May 16, 2004

### jcsd

D'oh, yes your're right of course acceltrtion doesn't equal dv/dx

10. May 16, 2004

### mcintyre_ie

Ok, so now I'm just as lost. a = v(dv/dx), right? But I still dont know what I should be doing next...

11. May 16, 2004

### arildno

Well, we're halfway, so you'r just halfway as lost.
By Newton's 2. law, you have:
$$a=\frac{F}{m}$$
Now, you have been given that the resistance(force) induces a retardation of absolute value (8+x/5) (right?), so by assuming v is in the positive direction, the induced acceleration (F/m) must be -(8+x/5)

12. May 16, 2004

### mcintyre_ie

Yes, I know how to apply the force equation (F=Ma, etc), but what I dont understand is where the (8 + x/5) part comes from - I don't know how to firgure out that. That's my original question, and I still don't know how to do it.

13. May 16, 2004

### arildno

Initially (that is, when you have moved zero distance!), the resistance force induces an acceleration of -8 m/s^(2) (that is, a retardation by assuming v positive)
This is given in the exercise.
You are then informed that the resistance increases in a uniform manner with the distance travelled so far.
Now uniform manner means that the resistance must increase in a linear relation with x (the distance travelled).
Hence, the induced acceleration must be on the form: -(8+sx), where s is a constant.

You are then informed that when you have traveled 5 meters (x=5), the induced acceleration is -9m/s^(2).

Hence, you may conclude that s=1/5

14. May 17, 2004

### mcintyre_ie

Thanks, that's really all I wanted to know.

15. May 17, 2004

### arildno

Well, I had to take my time, you know..

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