Register to reply 
A=V (dV/dx) LOST! 
Share this thread: 
#1
Jan910, 12:19 AM

P: 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=.18V_{0}/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
Jan910, 12:48 AM

HW Helper
P: 2,322

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
Jan910, 02:19 PM

P: 303

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


#4
Jan910, 07:28 PM

HW Helper
P: 2,322

A=V (dV/dx) LOST!
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/(1kv) is to rewrite it as 1/k * (1kv+1)/(1kv). You can split (1kv+1)/(1kv) into (1kv)/(1kv) + 1/(1kv), both of which are trivial to integrate. 


#5
Jan910, 07:38 PM

P: 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
Jan910, 07:40 PM

P: 228

im sorry that was for 11.28 my mistake



#7
Jan910, 07:42 PM

P: 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
Jan910, 10:54 PM

HW Helper
Thanks
P: 10,632

Talaroue,
I sent this to your other thread, maybe you have not found. I do not understand your last line in problem 11.25. It is all right up to then, if you meant natural logarithm (ln) . [tex] 0.4x1.6=1/k^2[1kvln(1kv)]_0^6[/tex] If I understand well, you substituted v=6 first, then v=0, and subtracted. But how did you get that 48.32? 16kln(16k)=0.1258+2.07317=2.1989, and divided by k^2: 103.6 ehild 


Register to reply 
Related Discussions  
I need help with this i am lost  Introductory Physics Homework  1  
Lost without it  General Discussion  32  
Lost  General Discussion  7 