# Acceleration as a function of velocity - HW problem help

by lonsalot
Tags: acceleration, function, velocity
 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?
 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.
P: 365
 Quote by lonsalot ...I don't think this is right because when I integrated the right side, it ends up being undefined?
No it isn't!

$$\alpha=-0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=-0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=-\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big|_{v_0}^v=-0.05\,s\Rightarrow v=v_o\,e^{-0.05\,s}$$

P: 1
Acceleration as a function of velocity - HW problem help

 Quote by Rainbow Child No it isn't! $$\alpha=-0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=-0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=-\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big|_{v_0}^v=-0.05\,s\Rightarrow v=v_o\,e^{-0.05\,s}$$
How can you just cancel "v" on both sides in second step , where it can take a value zero...
P: 1
Just to set the record straight on Rainbow child's answer:
 No it isn't! LaTeX Code: \\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}
This slightly wrong, it should be a == d(0.5v2)/ds not d(v2)/ds.
 P: 7 Hi.. just to go one step further, would the distance (with respect to time) equation be :- $$\int{{{v}{}_{0}\over{e^{0.05\,t}}}}\,dt = -{{20\,v{}_{0}}\over{e^{{{t}\over{20}}}\,\log e}}$$
P: 671
 Quote by jayvastani How can you just cancel "v" on both sides in second step , where it can take a value zero...
Easy, you just say that the step was valid for all non-zero values of v.
$$a=v\frac{dv}{dx}=-kv^2$$ (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).

$$v=\frac{v_0}{1+kv_0 t}$$

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