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Acceleration as a function of velocity  HW problem help 
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#1
Jan2408, 01:31 PM

P: 1

1. The problem statement, all variables and given/known data
The acceleration of a particle is defined by the relation a = 0.05v^2, where a is expresssed in m/s^2 and v in m/s. The particle starts at s=0 m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will travel before it comes to rest. 2. Relevant equations a = v (dv/ds) 3. The attempt at a solution I don't think this is right because when I integrated the right side, it ends up being undefined? 


#2
Jan2408, 03:47 PM

P: 39

not sure if what i'm thinking is correct but its an idea:
a= dv/dt so we know dv/dt = .05v^2. We can integrate that to get the velocity function. once we have the velocity function we can integrate again since v(t) = ds/dt. we have initial conditions. its just an idea. 


#3
Jan2408, 09:46 PM

P: 365

[tex]\alpha=0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big_{v_0}^v=0.05\,s\Rightarrow v=v_o\,e^{0.05\,s}[/tex] 


#4
Aug2008, 09:57 AM

P: 1

Acceleration as a function of velocity  HW problem help



#5
Nov208, 06:47 AM

P: 1

Just to set the record straight on Rainbow child's answer:



#6
Sep910, 03:04 PM

P: 7

Hi.. just to go one step further, would the distance (with respect to time) equation be :
[tex]\int{{{v}{}_{0}\over{e^{0.05\,t}}}}\,dt = {{20\,v{}_{0}}\over{e^{{{t}\over{20}}}\,\log e}}[/tex] 


#7
Sep910, 04:21 PM

P: 671

[tex]a=v\frac{dv}{dx}=kv^2[/tex] (In this case, k=0.05) From this step to the next, you have two options. Either v is always 0, in which case you have a solution for v, and you are not allowed to divide by v. Or v is different from 0, and you are allowed to divide by it. suchara, the velocity as a function of time is not what you posted. Note that you used the expression for the velocity as a function of distance and replaced the x with t (A mistake). [tex]v=\frac{v_0}{1+kv_0 t}[/tex] 


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