# Given acceleration relation, find variables

1. Jan 7, 2010

### talaroue

1. The problem statement, all variables and given/known data
The acceleration of a particle is defined by the relation a=0.4(1-kV), where k is a constant. Knowing t=0 the particle starts from rest at x=4m and that when t=15s, V=4 m/s, determine....
a)the constant k
b)the postion when V=6m/s
c) the maximum velocity of the particle

2. Relevant equations

3. The attempt at a solution

a=0.4(1-kV)
0.4(1-kV)=dV/dt.........im lost after this, can someone just simply point me in the right direction.

2. Jan 7, 2010

### ehild

This is a differential equation for v, you can solve it by separation of variables:

$$\int{\frac{dv}{1-kv}=\int{0.4 dt}$$

ehild

3. Jan 7, 2010

### talaroue

let me see if I can take it from here if I can't ill post what work I have done.

4. Jan 7, 2010

### talaroue

I am stuck again, I solved the right side of the equation you created and got .6s. But on the left side i solved by substitution. Where u=1-kV du=k dV so the left side becomes....

(1/k)ln(1-kV) over the intgral of V0=0 V1=4 m/s .....after everything cancels i have (1/k)(ln4k)=.6 is this right? Sorry if it is hard to follow.

5. Jan 7, 2010

### talaroue

which is wrong because those sides do not equal each other when I plug in a number for k

6. Jan 7, 2010

### talaroue

i tried solving with out using u, and i got ln(0) at one point which is impossible to solve so i am once again stuck.

7. Jan 8, 2010

### ehild

You need the function v(t) to get x(t). Integrate with respect to v between 0 and v(t), and between 0 and t on the right side.

$$\int_{0}^{v(t)}{\frac{dv}{1-kv}=-\frac{1}{k}(\ln(1-kv(t))-\ln(1-0))=0.4t$$

$$-1/k\ln(1-kv(t))=0.4t\rightarrow 1-kv(t)=\exp(-0.4kt)\rightarrow v(t)=1/k(1-\exp(-0.4kt))$$

At t=15 s, v=4 m/s, so 4=1/k(1- exp(-6k)), k=0.25(1-exp(-6k)) It is a bit tricky, to get k from here. Apply some numerical method.

ehild

8. Jan 8, 2010

### Winzer

There are two things you can do from here:
1) Solve for k numerically via computer box
or
2) You can get an approximation by expanding the exponential with a Taylor Series.
$$e^{-0.4 k *15} \approx 1 - 0.4 k*15 + \frac{(-0.4 k*15 )^2}{2}$$
Throw this in your eq. and you should get a decent approximation for k.

9. Jan 8, 2010

### ehild

it is: k=0.1458...

ehild

10. Jan 9, 2010

### talaroue

I was right, i just couldnt figure out how to solve. That mkes me feel better! Thanks for everything you guys!

11. Jan 9, 2010

### talaroue

for the second part I am stuck.....here is my work.... this is problem 11.25

12. Jan 9, 2010

### ehild

I do not understand your last line. It is all right up to then.

After integration, you got:

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

Substitute first v=6, then v=0, and subtract. 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