# A=V (dV/dx) LOST!

1. ### talaroue

303
1. The problem statement, all variables and given/known data
11.28 I have to solve acceleration given the equation for acceleration given the intial velocity=3.6 m/s, and the V equation V=.18V0/x

11.25 I have to find the distance it travels using the information in the picture below
2. Relevant equations

3. The attempt at a solution

2. ### ideasrule

2,321
For 11.28, you lost me after a=v*dv/dx. Since a=v*dv/dx and v=0.18v0/x, why not derive "v" to find dv/dx and multiply the result by 0.18v0/x? That'll give you "a".

For 11.25, what's the actual question? The distance travelled between what time and what time?

3. ### talaroue

303
For 11.28 I am not sure if I understand what you mean? I derived V using the relationship to a, which is a=V*dV/dx. making it V=a*dx/dV.

for 11.25 the actual question is determine the postion of the particle when V=6m/s

4. ### ideasrule

2,321
You know v already, don't you? It's 0.18v0/x. You don't know dv/dx, so you have to derive 0.18v0/x to figure it out.

Did I misunderstand the question?

Does it start at x=4 and v=0? If so, you set up the integral correctly, though I haven't checked your calculations. Note that an easier way to integrate v/(1-kv) is to rewrite it as 1/k * (1-kv+1)/(1-kv). You can split (1-kv+1)/(1-kv) into (1-kv)/(1-kv) + 1/(1-kv), both of which are trivial to integrate.

5. ### Larrytsai

228
hey for 11.25 the first one, look at this formula a ds = v dv, divide ds over so u have
a = v (dv/ds), u have v= 0.18vo/x , which is velocity as a function of position, or x. so if you derive that formula v u get dv/ds and u can sub it in for dv/ds and u will have all unknowns solved, let me knw if u understand it or not

6. ### Larrytsai

228
im sorry that was for 11.28 my mistake

7. ### Larrytsai

228
and to get the formula a ds = v dv , u have 2 equations a= dv/dt and v=ds/dt, solve for dt for both equations, and substitute, and u should be able to get a ds = vdv

8. ### ehild

12,385
Talaroue,

I do not understand your last line in problem 11.25. It is all right up to then, if you meant natural logarithm (ln) .

$$0.4x-1.6=1/k^2[1-kv-ln(1-kv)]_0^6$$

If I understand well, you substituted v=6 first, then v=0, and subtracted. But how did you get that 48.32?

1-6k-ln(1-6k)=0.1258+2.07317=2.1989, and divided by k^2: 103.6

ehild

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